Conic Sections: Class 11 Maths Chapter 10 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
3 simulations
44 practice questions
23 MCQs

Is chapter mein aap kya seekhenge

Samjhana ki ek double cone ko kaatne wala plane apne angle ke hisaab se circle, ellipse, parabola, hyperbola ya degenerate conic kaise deta hai.
Centre aur radius se circle ki equation likhna, aur completing squares se equation se centre aur radius nikaalna.
Parabola ki focus–directrix definition aur ellipse aur hyperbola ki two-focus definitions batana, aur unki standard equations derive karna.
Standard equation se focus, vertices, axes, eccentricity aur latus rectum padhna, aur aise data se equation banana.
Ellipse ke liye a² = b² + c², hyperbola ke liye c² = a² + b² aur e = c/a use karke ek parameter se doosre tak jaana.
Reflectors, arches, cables aur sliding rods ko conics se model karna aur unse bane problems solve karna.

NCERT Mathematics Class 11, Chapter 10, pp. 176–207 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

10.1 Introduction

Torch ko seedha floor ki taraf neeche point karo, to light ka patch ek gol disc hota hai. Torch ko thoda tilt karo, to disc khinch kar oval ban jaata hai. Aur tilt karo, to patch ka door wala sira floor par bhaagta chala jaata hai aur kabhi band nahi hota. Beam light ka ek cone hai aur floor ek flat plane hai, isliye tumne jo bhi outline dekhi woh ek plane dwara cone se kaata gaya curve hai.

2/3Tilt karne par: patch khinch kar oval ban jaata hai.

Fig. 10.1Torch tilt karne par uska light ka circle pehle ellipse aur phir khula parabola ban jaata hai.

Upar: level floor ke upar ek torch ka side view, uski beam dashed edges wale halke cone ki tarah bani hai aur floor ka roshan hissa green mein mark hai. Neeche: wahi roshan patch upar se dikhta hua. Step 1: torch seedha neeche point karti hai aur patch ek gol circle hai. Step 2: torch tilt hai aur patch khinch kar ellipse ban jaata hai. Step 3: torch itni tilt hai ki beam ka upar wala edge floor ke parallel chalta hai, aur patch kabhi band nahi hota: uski outline ek parabola hai.

Chapter 9 mein tumne straight lines ko equations se describe kiya tha. Yeh chapter chaar curves ke liye wahi karta hai: circle, ellipse, parabola aur hyperbola. Kyunki yeh chaaron cone ko kaatne se aate hain, inhe conic sections (शंकु परिच्छेद) ya simply conics kehte hain.

Dhyaan se dekho to yeh curves har jagah hain. Planets aur comets ellipses par chalte hain. Maidaan mein phenki gayi ball parabola follow karti hai (agar hum air resistance ko ignore karein), aur parabolic shapes DTH dish mein signals ikattha karti hain aur car headlamp ki beam ko focus karti hain. Kai power-station cooling towers ki kamar jaisi sides hyperbolas hain, aur satellite navigation se pehle hyperbolas ne ships ko radio se apni position pata karne mein bhi madad ki.

Quick check

2 mein se 1

Ek torch seedha level floor ki taraf neeche point karti hai. Uske light ke patch ka edge hai

ek ellipseek circleek parabolaek hyperbola

Answer dekho

Answer (B) ek circle

Jab beam floor ke square hai, to har edge ray barabar distance tay karti hai, isliye patch gol hota hai. Torch tilt karne se woh khinch kar ellipse banta hai aur phir khul kar parabola ban jaata hai.

10.2 Cone ke Sections

Double-napped right circular cone do khokhle cones hain jo nok se nok jude hain, aur dono bina ruke aage badhte jaate hain (Fig. 10.2). Ise banane ke liye do lines kaafi hain. Ek vertical line $l$ ko fixed rakho. Uske ek point $V$ se ek line $m$ banao jo $l$ se acute angle $α$ par ho. $m$ ko $l$ ke around ghumao, jaise ek spoke axle ke around ghoomti hai, aur $α$ ko fixed rakho: $m$ ki saari positions milkar surface banati hain.

3/3Bani hui surface: do nappes jo vertex $V$ par milte hain.

Fig. 10.2Line

m

ko axis

l

ke around ghumane se ek double cone banta hai.NCERT Figs 10.1, 10.2 · p. 176

Ek vertical dashed axis l aur ek tirchhi line m jo use point V par kaatti hai; dono ke beech ka angle alpha mark hai. Step 1 mein sirf do lines hain. Step 2 mein m ki kai positions par halki copies aur upar-neeche dashed rings judte hain, saath mein ek arrow jo dikhata hai ki m, l ke around ghoom rahi hai aur alpha fixed hai. Step 3 bani hui surface ko shade karta hai: V par milte do cones, jin par upper nappe aur lower nappe likha hai, V ko vertex, l ko axis aur m ko generator mark kiya gaya hai.

Kuch naam: milne wala point $V$ vertex hai, fixed line $l$ axis hai, aur apni kisi bhi position mein $m$ ek generator hai. $V$ ke upar aur neeche surface ke do halves hain, upper aur lower nappes.

Ab ek flat plane lao. Yeh cone ko jahan bhi kaate, kaate gaye hisse ka edge ek conic section hota hai, aur uska shape sirf do cheezein decide karti hain: plane ka axis se angle $β$ (jise cone ke angle $α$ se compare karte hain), aur yeh ki plane $V$ se guzarta hai ya nahi.

10.2.1 Circle, ellipse, parabola aur hyperbola

Pehle plane ko vertex se door rakho. Plane ko axis ke square rakh kar shuru karo aur use thoda-thoda tilt karo:

$β = 90^{\circ}$ : plane axis se right angle par hai aur cut ka edge ek circle hai.
$α < β < 90^{\circ}$ : plane tilt hai lekin ab bhi ek nappe ko poora paar karke kaatta hai, isliye cut ek band oval hai, yaani ellipse.
$β = α$ : plane ek generator ke parallel hai, isliye us side par woh cone se dobara kabhi nahi milta aur curve khula rehta hai; yeh parabola hai.
$0 \leq β < α$ : plane cone ki side se zyada steep hai aur dono nappes ko kaatta hai, jisse hyperbola ki do alag branches milti hain.

4/7Cone se zyada steep: dono nappes kat jaate hain, ek hyperbola.

Fig. 10.3Kaatne wale plane ka angle decide karta hai ki kaun sa conic milega.NCERT Figs 10.4–10.10 · p. 177

Left: side se dikhta double cone, jismein uska axis aur vertex V par angle alpha mark hai. Ek dashed line plane ko edge-on dikhati hai aur ek moti green line dikhati hai ki plane cone se kahan milta hai. Right: ek panel cut ka asli shape, uska naam aur beta ki condition dikhata hai. Step 1: horizontal plane se circle milta hai. Step 2: ek nappe ke aar-paar tilt plane se ellipse milta hai. Step 3: generator ke parallel plane se khula parabola milta hai. Step 4: vertical plane dono nappes ko kaatta hai aur do branches wala hyperbola deta hai. Steps 5 se 7 mein plane V se guzarta hai: horizontal plane cone se sirf V par milta hai (ek point), generator ke saath wala plane ek line deta hai, aur axis se guzarta vertical plane do kaatti lines deta hai.

Pattern dekho: plane jitna steep kaatta hai, curve utna zyada khulta hai. Circle aur ellipse band hain aur ek nappe par hain; parabola ek nappe par hai lekin khula hai; hyperbola khula hai aur use dono nappes chahiye.

10.2.2 Degenerated conic sections

Ab plane ko vertex $V$ se guzarne do. $β$ ki wahi teen ranges conics ke pichke hue versions deti hain, jinhe degenerate conics kehte hain (Fig. 10.3 ke steps 5 se 7):

$α < β \leq 90^{\circ}$ : plane cone se sirf $V$ par milta hai, isliye section ek single point hai.
$β = α$ : plane ek generator ke saath-saath padta hai, aur section wahi straight line hai. Yeh degenerate parabola hai.
$0 \leq β < α$ : plane do generators ke saath dono nappes se guzarta hai, jisse pair of intersecting lines milta hai. Yeh degenerate hyperbola hai.

Yahan se hum cone ko ek taraf rakhte hain. Har curve ko plane mein distances ke ek simple rule se describe kiya ja sakta hai, aur woh rules seedhe equations tak le jaate hain.

Quick check

3 mein se 1

Plane vertex se nahi guzarta. Angle

β

ko curve se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$β = 90^{\circ}$ : circle
$α < β < 90^{\circ}$ : ellipse
$β = α$ : parabola
$0 \leq β < α$ : hyperbola

Plane ko square position se tilt karne par curve khulta jaata hai: circle, phir ellipse, phir parabola jab plane generator ke parallel ho, phir hyperbola jab woh itna steep ho ki dono nappes tak pahunch jaaye.

10.3 Circle

Definition

Circle

Plane mein ek point $C$ aur ek positive number $r$ fix karo. Centre $C$ aur radius $r$ wala circle us plane ke un sabhi points se banta hai jo $C$ se distance $r$ par hain.

Is rule ko equation mein badalne ke liye centre ko $C (h, k)$ par rakho, radius ko $r$ kaho, aur $P (x, y)$ ko plane ka koi bhi point maano. $P$ circle par tabhi hai jab $C P = r$ . Distance formula ek right triangle ke do legs se $C P$ deta hai, aur square karne se square root hat jaata hai (Fig. 10.4).

2/3Legs hain $x - h$ aar-paar aur $y - k$ upar.

Fig. 10.4Radius ke neeche bane triangle mein Pythagoras se circle ki equation milti hai.NCERT Fig. 10.12 · p. 180

Coordinate axes ke saath ek circle jiska centre C(h, k) first quadrant mein hai. Step 1 mein C ko circle ke ek point P(x, y) se radius r dwara joda gaya hai aur CP = r likha hai. Step 2 radius ke neeche ek right triangle banata hai: x minus h length ka horizontal leg aur y minus k length ka vertical leg, aur panel batata hai ki (x minus h) squared plus (y minus k) squared equals r squared. Step 3 circle par teen aur points jodta hai, har ek C se same length r ke radius se juda hai, yeh dikhane ke liye ki circle ka har point same equation follow karta hai.

Circle: centre (h, k), radius r

\begin{aligned} (x - h)^{2} + (y - k)^{2} \\ = r^{2} \end{aligned}

(10.1)

Circle ka har point (10.1) ko satisfy karta hai, aur jo bhi point ise satisfy karta hai woh $C$ se distance $r$ par hai, isliye yeh equation circle ko bilkul sahi describe karti hai. Centre origin par ho to yeh $x^{2} + y^{2} = r^{2} .$ ban jaati hai.

(10.1) ko expand karne par $x^{2} + y^{2} - 2 h x - 2 k y + (h^{2} + k^{2} - r^{2}) = 0$ milta hai. Isliye circle ki equation mein $x^{2}$ aur $y^{2}$ ke coefficients barabar hote hain aur koi $x y$ term nahi hota. Ulti taraf se, neeche wale form ki equation ko completing squares se padha jaata hai:

General form

\begin{aligned} x^{2} + y^{2} + 2 g x \\ + 2 f y + c = 0 \end{aligned}

(10.2)

jahan

$(- g, - f)$: centre
$\sqrt{g^{2} + f^{2} - c}$: radius, jab $g^{2} + f^{2} - c > 0$

Agar $g^{2} + f^{2} - c = 0$ ho to “circle” sikud kar single point $(- g, - f)$ ban jaata hai, aur agar yeh negative ho to koi bhi real point equation ko satisfy nahi karta. Agar $x^{2}$ aur $y^{2}$ ka coefficient 1 nahi hai, to pehle usse divide karna padega.

SimulationC se distance 5 par points

Dono axes par minus 8 se 8 tak ek coordinate grid. Ek fixed point C (1, minus 2) par hai aur ek draggable point P (4, 2) se shuru hota hai; segment CP apni length ke saath bana hai, aur move karne par P grid points par snap hota hai aur apne coordinates dikhata hai.

Fig. 10.5

P

ko drag karo aur distance

C P

dekho. Circle woh saare points hain jahan yeh 5 dikhata hai.

Try karo

$P$ $(4, 2)$ par shuru hota hai aur $C P = 5$ hai. Woh saare baarah grid points dhoondho jinke liye $C P = 5$ ho.
Har mile hue point ke liye check karo ki $(x - 1)^{2} + (y + 2)^{2} = 25$ hai.
$P$ ko $(3, 0)$ par rakho. Kya $C P$ 5 se zyada hai ya kam? $P$ circle ke andar hai ya bahar?

Example 10.1NCERT Example 1 · p. 180

Centre origin par

Origin par centre wala ek circle $(5, - 12)$ se guzarta hai. Uski equation likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.2NCERT Example 2 · p. 180

Centre aur radius diye hain

Us circle ki equation likho jiska centre $(2, - 5)$ aur radius 3 hai, phir use expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.3NCERT Example 3 · p. 180

Completing squares se circle padhna

Circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ ka centre aur radius nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.4NCERT Example 4 · p. 181

Do points aur centre ke liye ek line

Ek circle $A (1, 3)$ aur $B (5, 1)$ se guzarta hai, aur uska centre line $x - 2 y = 2$ par hai. Circle nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Circle

x^{2} + y^{2} - 2 x + 6 y - 15 = 0

ka radius kya hai?

Answer dekho

Answer 5

Squares complete karo:

(x - 1)^{2} + (y + 3)^{2} = 15 + 1 + 9 = 25

, isliye

r = 5

. Answer 15 dene ka matlab hai squares complete karte waqt jode gaye 1 aur 9 ko bhool jaana.

Practice · 9 questions

Practice 10.1: answers ke saath

1
Short answerEasyNCERT Exercise 10.1 Q1–Q4
Har circle ki equation uske centre aur radius se expanded form mein likho.
1. Centre $(0, - 3)$ aur radius 3
  Solution dikhao
  Answer $x^{2} + y^{2} + 6 y = 0$
  $x^{2} + (y + 3)^{2} = 9$
  $x^{2} + y^{2} + 6 y = 0$
2. Centre $(4, - 1)$ aur radius 5
  Solution dikhao
  Answer $x^{2} + y^{2} - 8 x + 2 y - 8 = 0$
  $(x - 4)^{2} + (y + 1)^{2} = 25$
  $x^{2} + y^{2} - 8 x + 2 y - 8 = 0$
3. Centre $(\frac{1}{3}, - \frac{1}{2})$ aur radius $\frac{1}{6}$
  Solution dikhao
  Answer $3 x^{2} + 3 y^{2} - 2 x + 3 y + 1 = 0$
  $(x - \frac{1}{3})^{2} + (y + \frac{1}{2})^{2} = \frac{1}{36}$
  Expand karo: $x^{2} + y^{2} - \frac{2}{3} x + y + \frac{1}{9} + \frac{1}{4} - \frac{1}{36} = 0$ , aur $\frac{1}{9} + \frac{1}{4} - \frac{1}{36} = \frac{1}{3} .$ 3 se multiply karo:
  $3 x^{2} + 3 y^{2} - 2 x + 3 y + 1 = 0$
4. Centre $(- 2, - 2)$ aur radius $\sqrt{2}$
  Solution dikhao
  Answer $x^{2} + y^{2} + 4 x + 4 y + 6 = 0$
  $(x + 2)^{2} + (y + 2)^{2} = 2$
  $x^{2} + y^{2} + 4 x + 4 y + 6 = 0$
Short answerMediumNCERT Exercise 10.1 Q5
Centre $(p, - q)$ aur radius $\sqrt{p^{2} + q^{2}}$ wale circle ki equation nikaalo. Kaun sa special point hamesha is par hota hai?
Solution dikhao
Answer $x^{2} + y^{2} - 2 p x + 2 q y = 0$ ; yeh origin se guzarta hai.
$(x - p)^{2} + (y + q)^{2} = p^{2} + q^{2}$
$x^{2} + y^{2} - 2 p x + 2 q y = 0$
Koi constant term nahi hai, isliye $(0, 0)$ equation ko satisfy karta hai: circle origin se guzarta hai. Yeh samajh mein aata hai, kyunki radius centre se origin tak ke distance ke barabar hai.
3
Short answerEasyNCERT Exercise 10.1 Q6–Q9
Har circle ka centre aur radius nikaalo.
1. $(x - 7)^{2} + (y + 2)^{2} = 20$
  Solution dikhao
  AnswerCentre $(7, - 2)$ aur radius $2 \sqrt{5}$
  (10.1) se seedha padho: $h = 7$ , $k = - 2$ , $r = \sqrt{20} = 2 \sqrt{5} .$
2. $x^{2} + y^{2} + 6 x - 10 y - 2 = 0$
  Solution dikhao
  AnswerCentre $(- 3, 5)$ aur radius 6
  Constant ko doosri side le jao aur 9 aur 25 jodo: $2 + 9 + 25 = 36$ .
  $(x + 3)^{2} + (y - 5)^{2} = 36$
3. $x^{2} + y^{2} + 2 x - 12 y + 12 = 0$
  Solution dikhao
  AnswerCentre $(- 1, 6)$ aur radius 5
  $- 12$ mein 1 aur 36 jodo: right side 25 ban jaati hai.
  $(x + 1)^{2} + (y - 6)^{2} = 25$
4. $3 x^{2} + 3 y^{2} + 2 y = 0$
  Solution dikhao
  AnswerCentre $(0, - \frac{1}{3})$ aur radius $\frac{1}{3}$
  3 se divide karo: $x^{2} + y^{2} + \frac{2}{3} y = 0$ .
  $x^{2} + (y + \frac{1}{3})^{2} = \frac{1}{9}$
4
Long answerMediumNCERT Exercise 10.1 Q10–Q11
Diye gaye do points se guzarne wala woh circle nikaalo jiska centre di gayi line par ho.
1. $(3, 1)$ aur $(- 1, 3)$ se hokar; centre $3 x - y = 2$ par
  Solution dikhao
  Answer $(x - 2)^{2} + (y - 4)^{2} = 10$
  $\begin{aligned} (3 - h)^{2} + (1 - k)^{2} \\ = (1 + h)^{2} + (3 - k)^{2} \end{aligned}$
  Yeh simplify hokar $k = 2 h$ banta hai. $3 h - k = 2$ ke saath: $h = 2$ , $k = 4$ , aur $r^{2} = 1^{2} + 3^{2} = 10.$
2. $(5, 2)$ aur $(- 1, 4)$ se hokar; centre $x + y = 9$ par
  Solution dikhao
  Answer $(x - 3)^{2} + (y - 6)^{2} = 20$
  $\begin{aligned} (5 - h)^{2} + (2 - k)^{2} \\ = (1 + h)^{2} + (4 - k)^{2} \end{aligned}$
  Yeh simplify hokar $k = 3 h - 3$ banta hai. $h + k = 9$ ke saath: $h = 3$ , $k = 6$ , aur $r^{2} = 2^{2} + 4^{2} = 20.$
Short answerMediumNCERT Exercise 10.1 Q12
Radius 13 wale ek circle ka centre $y$ -axis par hai aur woh $(5, 1)$ se guzarta hai. Aise saare circles nikaalo.
Solution dikhao
Answer $x^{2} + (y - 13)^{2} = 169$ ya $x^{2} + (y + 11)^{2} = 169$
$25 + (1 - k)^{2} = 169$
To $(1 - k)^{2} = 144$ , jisse $k = 13$ ya $k = - 11$ milta hai. Dono circles sahi hain: $x^{2} + (y - 13)^{2} = 169$ aur $x^{2} + (y + 11)^{2} = 169.$
6
Short answerMediumNCERT Exercise 10.1 Q13
Ek circle origin se guzarta hai aur axes ko dobara $A$ aur $B$ par kaatta hai.
1. Ise nikaalo jab $A = (6, 0)$ aur $B = (0, - 4)$ hon.
  Solution dikhao
  Answer $x^{2} + y^{2} - 6 x + 4 y = 0$
  $(0, 0)$ se hokar: $c = 0$ . $(6, 0)$ se hokar: $36 + 12 g = 0$ , isliye $g = - 3$ . $(0, - 4)$ se hokar: $16 - 8 f = 0$ , isliye $f = 2$ .
  $x^{2} + y^{2} - 6 x + 4 y = 0$
2. Ise general case mein nikaalo jab $A = (p, 0)$ aur $B = (0, q)$ hon.
  Solution dikhao
  Answer $x^{2} + y^{2} - p x - q y = 0$
  Wahi steps $c = 0$ , $2 g = - p$ aur $2 f = - q$ dete hain, isliye circle $x^{2} + y^{2} - p x - q y = 0.$ hai. Angle $A O B$ ek right angle hai, isliye $A B$ ek diameter hai.
Short answerEasyNCERT Exercise 10.1 Q14
$(- 1, 3)$ par centre wala ek circle $(2, - 1)$ se guzarta hai. Uski equation likho.
Solution dikhao
Answer $(x + 1)^{2} + (y - 3)^{2} = 25$
$r^{2} = 3^{2} + 4^{2} = 25$
$(x + 1)^{2} + (y - 3)^{2} = 25$
Quick answerEasyNCERT Exercise 10.1 Q15
Circle $x^{2} + y^{2} = 16$ ke compare mein point $(3.5, - 2.5)$ kahan hai?
andarcircle parbahar
Answer dekho
Answer (C) bahar
${3.5}^{2} + {2.5}^{2} = 12.25 + 6.25 = 18.5 > 16$ , isliye point centre se 4 se zyada door hai: bahar.
Quick answerMedium
Circle $x^{2} + y^{2} - 8 x + 6 y + k = 0$ ka radius 6 hai. $k$ nikaalo.
Answer dekho
Answer -11
Squares complete karne par $(x - 4)^{2} + (y + 3)^{2} = 25 - k$ . $25 - k = 36$ rakhne par $k = - 11$ milta hai.

10.4 Parabola

Boundary ki taraf uchhali gayi cricket ball, garden hose se nikalti paani ki dhaar, suspension bridge ka cable: har ek (lagbhag) ek hi tarah ka curve follow karta hai. Iske geometric description ke liye ek point aur ek line chahiye.

Definition

Parabola

Ek line $l$ aur uske bahar ek point $F$ lo, dono ek hi plane mein. Parabola us plane ke un points se banta hai jinka $F$ se distance unke $l$ se distance ke barabar ho. Line $l$ directrix hai aur $F$ focus hai.

To ek point $P$ parabola par tab hai jab focus $F$ se uska distance $P F$ , directrix se uske distance $P M$ ke barabar ho, jahan $M$ point $P$ se daale gaye perpendicular ka foot hai (Fig. 10.6).

4/4Yeh ek parabola par padte hain; iska axis $F$ se guzarta hai.

Fig. 10.6Parabola ka har point focus se utna hi door hai jitna directrix se.NCERT Figs 10.13, 10.14 · p. 182

Neeche ke paas ek horizontal dashed directrix aur uske beech ke upar ek focus F. Step 1 mein sirf focus aur directrix hain. Step 2 ek point P jodta hai, jisse F tak ek solid segment aur directrix par M tak ek dashed vertical segment hai, aur dono barabar hain: PF = PM. Step 3 teen aur points jodta hai, har ek ke saath barabar segments ka apna pair. Step 4 in sab se guzarta parabola banata hai, jismein F se guzarta vertical axis hai aur vertex focus aur directrix ke theek beech mein hai.

Focus se guzarne wali aur directrix ke perpendicular line parabola ka axis hai, aur curve iske around symmetric hai. Axis parabola se ek hi point par milta hai, jo vertex hai, aur yeh focus aur directrix ke theek beech mein hota hai.

Try karke dekho

Parabola fold karo

Tumhe chahiye

patle kaagaz ki ek sheet (tracing ya butter paper)
ek pen
ek ruler

Pehle guess karo

Tum ek line ke kai points ko ek dot par fold karoge aur har baar crease banaoge. Creases kya outline karenge?

dot ke around ek circleline se door khulne wala ek curveek straight line

Kaagaz ke neeche wale edge se lagbhag 3 cm upar ek straight line banao.
Line ke beech se lagbhag 4 cm upar ek dot $F$ mark karo.
Kaagaz ko aise fold karo ki line ka koi point theek $F$ par aa jaaye. Achhe se crease karo aur khol do.
Line par faile 20 se 30 alag points ke liye yahi dohrao.

10.4.1 Parabola ki standard equations

Sabse simple equations tab milti hain jab vertex origin par ho aur axis ek coordinate axis ke saath line up ho. Phir bhi parabola chaar taraf face kar sakta hai: right, left, upar ya neeche. Hum pehli derive karte hain aur baaki symmetry se nikaalte hain.

DerivationRight ki taraf khulne wala parabola

Focus $F (a, 0)$ lo jahan $a > 0$ , aur directrix $x = - a$ ; tab dono ke beech ka vertex origin hai. Kisi bhi point $P (x, y)$ ke liye directrix par perpendicular ka foot $M (- a, y)$ hai, isliye $P M = ∣ x + a ∣$ aur $P F = \sqrt{(x - a)^{2} + y^{2}}$ .

$P$ parabola par tabhi hai jab $P F = P M$ . Dono sides non-negative hain, isliye hum bina koi point jode ya khoye square kar sakte hain: $P F^{2} = P M^{2}$ . $(x - a)^{2}$ ko doosri side le jaane par $y^{2}$ akela bachta hai:

\begin{aligned} y^{2} & = (x + a)^{2} \\ - (x - a)^{2} \\ = 4 a x \end{aligned}

Standard parabola

y^{2} = 4 a x, a > 0

(10.3)

Ulta, agar $y^{2} = 4 a x$ hai to $x \geq 0$ aur $P F^{2} = (x - a)^{2} + 4 a x = (x + a)^{2}$ , isliye $P F = x + a = P M$ . Parabola ka har point (10.3) ko satisfy karta hai, aur (10.3) ka har solution parabola par hai.

$y^{2} = 4 a x$ ke kisi bhi point ka $x$ negative nahi hota, kyunki left side kabhi negative nahi ho sakti jab $4 a$ positive hai. Right ki taraf koi limit nahi hai: $y = \pm 2 \sqrt{a x}$ badhta hi rehta hai, isliye arms quadrants I aur IV mein hamesha ke liye khulte jaate hain, aur positive $x$ -axis mein ek doosre ki mirror images hain.

1/4 $y^{2} = 4 a x$ right ki taraf khulta hai.

Fig. 10.7Pehle parabola ko reflect karne se baaki teen standard forms milte hain.NCERT Fig. 10.15 · p. 182

Axes ka ek set. Step 1: parabola y squared equals 4ax right ki taraf khulta hai, focus F(a, 0) aur vertical directrix x = minus a ke saath. Step 2: y squared equals minus 4ax left ki taraf khulta hai, focus F(minus a, 0), directrix x = a. Step 3: x squared equals 4ay upar ki taraf khulta hai, focus F(0, a), horizontal directrix y = minus a. Step 4: x squared equals minus 4ay neeche ki taraf khulta hai, focus F(0, minus a), directrix y = a. Har step mein vertex origin par hai.

$x$ ki jagah $- x$ rakhne se curve $y$ -axis mein reflect hota hai; $x$ aur $y$ ko aapas mein badalne se yeh line $y = x$ mein reflect hota hai. Yeh moves (10.3) ko baaki teen standard forms mein badal dete hain (Table 10.1).

Table 10.1Chaar standard parabolas (

a > 0

, vertex origin par)

Equation	Focus	Directrix	Kidhar khulta hai
$y^{2} = 4 a x$	$(a, 0)$	$x = - a$	right ki taraf
$y^{2} = - 4 a x$	$(- a, 0)$	$x = a$	left ki taraf
$x^{2} = 4 a y$	$(0, a)$	$y = - a$	upar
$x^{2} = - 4 a y$	$(0, - a)$	$y = a$	neeche

10.4.2 Latus rectum

Definition

Latus rectum

Kisi conic ka latus rectum focus se guzarne wali woh chord hai jo axis ke perpendicular ho (ellipse ke liye major axis; hyperbola ke liye transverse axis), aur jiske dono ends curve par hon.

$y^{2} = 4 a x$ ke liye latus rectum line $x = a$ par hota hai. $x = a$ rakhne par $y^{2} = 4 a^{2}$ milta hai, isliye iske ends $L (a, 2 a)$ aur $L^{'} (a, - 2 a)$ hain. Picture se bhi ek proof hai (Fig. 10.8): $L$ parabola par hai, isliye $L F$ , $L$ se directrix tak ke distance ke barabar hai, jo lines $x = - a$ aur $x = a$ ke beech ka gap $2 a$ hai.

3/3To $L L^{'} = 2 a + 2 a = 4 a$ .

Fig. 10.8Latus rectum ka har half focus se directrix tak ke gap

2 a

jitna lamba hai.NCERT Fig. 10.18 · p. 185

Vertex O, focus F aur vertical dashed directrix wala right ki taraf khulta parabola. Step 1 F se guzarti, axis ke perpendicular chord LL prime banata hai. Step 2 L se directrix tak ek dashed horizontal segment jodta hai aur use 2a mark karta hai; segment LF bhi 2a hai, kyunki L parabola par hai. Step 3 L prime F ko bhi 2a mark karta hai aur poori chord LL prime ko 4a likhta hai.

y² = 4ax ka latus rectum

L L^{'} = 4 a

(10.4)

Example 10.5NCERT Example 5 · p. 185

Parabola ko padhna

Parabola $x^{2} = - 12 y$ ka axis batao aur uska focus, directrix aur latus rectum ki length nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.6NCERT Example 6 · p. 186

Focus aur directrix diye hain

Us parabola ki equation likho jiska focus $(0, 4)$ aur directrix $y = - 4$ hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.7NCERT Example 7 · p. 186

Vertex aur focus diye hain

Ek parabola ka vertex $(0, 0)$ par aur focus $(- \frac{5}{2}, 0)$ par hai. Uski equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.8NCERT Example 8 · p. 186

Diye gaye point se hokar

Ek parabola ka vertex origin par hai, axis $x$ -axis ke saath hai, aur yeh $(3, - 6)$ se guzarta hai. Uski equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Parabola

x^{2} = - 8 y

ka focus hai

(- 2, 0)

(0, 2)

(0, - 2)

(0, - 8)

Answer dekho

Answer (C) $(0, - 2)$

Yahan

4 a = 8

, isliye

a = 2

; parabola neeche ki taraf khulta hai, isliye focus

(0, - 2)

hai.

(0, - 8)

chunne ka matlab hai

a

ki jagah

4 a

use karna.

Practice · 6 questions

Practice 10.2: answers ke saath

1
Short answerEasyNCERT Exercise 10.2 Q1–Q6
Har parabola ke liye axis, focus, directrix aur latus rectum ki length batao.
1. $y^{2} = 16 x$
  Solution dikhao
  AnswerFocus $(4, 0)$ ; axis $x$ -axis; directrix $x = - 4$ ; latus rectum 16
  $4 a = 16$ , $a = 4$ ; right ki taraf khulta hai. Focus $(4, 0)$ , axis $x$ -axis, directrix $x = - 4$ , latus rectum 16.
2. $x^{2} = - 10 y$
  Solution dikhao
  AnswerFocus $(0, - \frac{5}{2})$ ; axis $y$ -axis; directrix $y = \frac{5}{2}$ ; latus rectum 10
  $4 a = 10$ , $a = \frac{5}{2}$ ; neeche khulta hai. Focus $(0, - \frac{5}{2})$ , axis $y$ -axis, directrix $y = \frac{5}{2}$ , latus rectum 10.
3. $y^{2} = - 6 x$
  Solution dikhao
  AnswerFocus $(- \frac{3}{2}, 0)$ ; axis $x$ -axis; directrix $x = \frac{3}{2}$ ; latus rectum 6
  $4 a = 6$ , $a = \frac{3}{2}$ ; left ki taraf khulta hai. Focus $(- \frac{3}{2}, 0)$ , axis $x$ -axis, directrix $x = \frac{3}{2}$ , latus rectum 6.
4. $x^{2} = 20 y$
  Solution dikhao
  AnswerFocus $(0, 5)$ ; axis $y$ -axis; directrix $y = - 5$ ; latus rectum 20
  $4 a = 20$ , $a = 5$ ; upar khulta hai. Focus $(0, 5)$ , axis $y$ -axis, directrix $y = - 5$ , latus rectum 20.
2
Short answerEasyNCERT Exercise 10.2 Q7–Q8
Diye gaye focus aur directrix wale parabola ki equation nikaalo.
1. Focus $(0, 7)$ aur directrix $y = - 7$
  Solution dikhao
  Answer $x^{2} = 28 y$
  Vertex origin hai aur focus uske upar hai: $x^{2} = 4 a y$ jahan $a = 7$ , isliye $x^{2} = 28 y .$
2. Focus $(- 4, 0)$ aur directrix $x = 4$
  Solution dikhao
  Answer $y^{2} = - 16 x$
  Vertex origin hai aur focus uske left mein hai: $y^{2} = - 4 a x$ jahan $a = 4$ , isliye $y^{2} = - 16 x .$
3
Short answerEasyNCERT Exercise 10.2 Q9–Q10
Vertex origin par hai. Diye gaye focus se parabola ki equation likho.
1. Focus $(0, - 1)$
  Solution dikhao
  Answer $x^{2} = - 4 y$
  $a = 1$ ke saath neeche khulta hai: $x^{2} = - 4 y .$
2. Focus $(\frac{7}{2}, 0)$
  Solution dikhao
  Answer $y^{2} = 14 x$
  $a = \frac{7}{2}$ ke saath right ki taraf khulta hai: $y^{2} = 4 \cdot \frac{7}{2} x = 14 x .$
4
Short answerMediumNCERT Exercise 10.2 Q11–Q12
Vertex origin par hai. Diye gaye axis aur point se parabola nikaalo.
1. Axis $x$ -axis ke saath; $(- 3, 6)$ se guzarta hai
  Solution dikhao
  Answer $y^{2} = - 12 x$
  $36 = k (- 3) \Rightarrow k = - 12$
  To $y^{2} = - 12 x$ , jo left ki taraf khulta hai, jaisa negative $x$ wale point ke liye zaroori hai.
2. $y$ -axis ke around symmetric; $(6, - 4)$ se guzarta hai
  Solution dikhao
  Answer $x^{2} = - 9 y$
  $36 = k (- 4) \Rightarrow k = - 9$
  To $x^{2} = - 9 y$ , jo neeche ki taraf khulta hai.
Quick answerMedium
Parabola $3 y^{2} = 8 x$ ke focus ka $x$ -coordinate kya hai?
Answer dekho
Answer 0.6667
3 se divide karo: $y^{2} = \frac{8}{3} x$ , isliye $4 a = \frac{8}{3}$ aur $a = \frac{2}{3} \approx 0.667$ . Focus $(\frac{2}{3}, 0)$ hai.
Quick answerMedium
Parabola $y^{2} = 4 a x$ point $(2, - 4)$ se guzarta hai. Uske latus rectum ki length hai
24816
Answer dekho
Answer (C) 8
$16 = 8 a$ se $a = 2$ milta hai, isliye latus rectum $4 a = 8$ hai. Option 2 khud $a$ hai.

10.5 Ellipse

Har planet Sun ke around ek ellipse par chalta hai, aur Sun centre par nahi balki focus naam ke ek special point par hota hai. Yeh Kepler ka first law hai, jo 1609 mein publish hua. Earth ke liye ellipse itna gol hai ki aankh se tum use circle se alag nahi pehchaan paoge.

Definition

Ellipse

Ek plane mein do fixed points $F_{1}$ aur $F_{2}$ lo. Ellipse us plane ke un points $P$ se banta hai jinke liye sum $P F_{1} + P F_{2}$ ki value ek hi rehti hai. $F_{1}$ aur $F_{2}$ iske foci hain (singular: focus).

Try karke dekho

Playground wala ellipse

Tumhe chahiye

lagbhag 3 m gher wala string ka ek loop
chalk ka ek tukda
do dost

Pehle guess karo

Ek baar draw karne ke baad dono dost aur paas khade hote hain aur tum dobara draw karte ho. Shape ka kya hota hai?

Yeh zyada gol ho jaata haiYeh zyada flat ho jaata haiYeh nahi badalta

Do dost lagbhag 1 m ki doori par seedhe khade rehte hain, aur dono loop ko apne pairon ke paas zameen par dabaye rakhte hain.
Chalk ko loop ke andar rakho aur bahar ki taraf kheencho jab tak string tight na ho jaaye aur ek triangle na ban jaaye.
String ko tight rakhte hue chalk ko zameen par poora ek chakkar ghumao.
Doston ke beech 0.5 m ka gap rakh kar dobara draw karo, phir 1.3 m ka gap rakh kar.

Abhi nahi kar sakte? Simulation try karo · Fig. 10.12

1/4Yahan $P F_{1}$ lamba aur $P F_{2}$ chhota hai; dono ka total $2 a$ hai.

Fig. 10.9

P

kaise bhi move kare,

P F_{1} + P F_{2}

same rehta hai: isi se curve ellipse banta hai.NCERT Figs 10.20, 10.21 · p. 187

Ek horizontal line par do foci F1 aur F2. Ek point P, F1 se solid segment aur F2 se dashed segment se juda hai. Drawing ke neeche ek bar do pieces PF1 aur PF2 mein bata hai, jo ek ke baad ek rakhe hain. Steps 1 se 3 P ko curve par teen positions par le jaate hain aur har baar curve ka aur hissa banta hai; aakhri position minor axis ke neeche wale end par hai jahan dono pieces barabar hain; bar ka split khisakta hai lekin uski total length kabhi nahi badalti. Step 4 centre O, do vertices, foci se guzarte major axis aur O se guzarte minor axis ke saath poora ellipse banata hai, aur batata hai ki bar ki length 2a major axis ki length ke barabar hai.

Kuch points aur lengths kisi bhi ellipse ko describe kar dete hain. Foci ke theek beech mein centre hai. Sabse lambi chord, major axis, dono foci se guzarti hai, aur iske ends do vertices hain. Centre se iske square chhoti minor axis chalti hai. Inki lengths $2 a$ aur $2 b$ kehlati hain, aur foci $2 c$ ki doori par hain; isliye semi-axes $a$ aur $b$ hain, aur har focus centre se $c$ door hai.

10.5.1 a, b aur c aapas mein kaise jude hain

Pehli baat, constant sum major axis ki length hai. $F_{2}$ ke aage wale vertex $A$ par $A F_{2} = a - c$ aur $A F_{1} = a + c$ , jo milkar $2 a$ banate hain. Sum har point par same hai, isliye ellipse ke har point $P$ ke liye $P F_{1} + P F_{2} = 2 a$ .

2/4 $B F_{1} = B F_{2} = a$ , isliye $a^{2} = b^{2} + c^{2}$ .

Fig. 10.10

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

ke vertices, foci, latus rectum aur bounding box.NCERT Figs 10.22, 10.23 · p. 188

Origin par centre wala ek ellipse jiska major axis x-axis ke saath hai. Step 1 vertices (a, 0) aur (minus a, 0) aur minor axis ke ends (0, b) aur (0, minus b) mark karta hai. Step 2 foci F1 aur F2 jodta hai aur upar wale point B ko dono se jodta hai; har segment a hai, aur legs b aur c aur hypotenuse a wala right triangle a squared equals b squared plus c squared deta hai. Step 3 F2 se guzarta latus rectum banata hai, jo b squared over a length ke do halves se bana hai. Step 4 lines x = plus ya minus a aur y = plus ya minus b ke beech ke rectangle ko shade karta hai, jise ellipse chhoota hai lekin kabhi bahar nahi jaata.

Ab minor axis ka ek end $B$ lo. Yeh $F_{1} F_{2}$ ke perpendicular bisector par hai, isliye $B F_{1} = B F_{2}$ , aur kyunki dono milkar $2 a$ hain, har ek $a$ ke barabar hai. $B$ , centre aur $F_{2}$ se bane right triangle ke legs $b$ aur $c$ hain aur hypotenuse $a$ hai (Fig. 10.10 ka step 2):

Ellipse

a^{2} = b^{2} + c^{2}

(10.5)

Khaas taur par $c = \sqrt{a^{2} - b^{2}}$ aur $b < a$ : major axis sach mein lambi wali hai.

10.5.2 Eccentricity

Definition

Eccentricity

Ellipse ya hyperbola ki eccentricity $e = \frac{c}{a}$ hai: centre se focus tak ka distance divided by centre se vertex tak ka distance.

Ellipse ke liye $0 \leq c < a$ , isliye $0 \leq e < 1$ , aur har focus centre se $c = a e$ door hai. Eccentricity naapti hai ki ellipse gol hone se kitna door hai. $a$ ko fixed rakho aur $e$ badhao: foci vertices ki taraf khisakte hain aur ellipse flat hota jaata hai, kyunki $b = a \sqrt{1 - e^{2}}$ (Fig. 10.11). $e = 0$ par dono foci centre par hote hain aur ellipse radius $a$ ka circle hai.

3/4Zyada flat: $b = 0.6 a$ .

Fig. 10.11Same

a

, badhta

e

: foci bahar khisakte hain aur ellipse flat hota hai.NCERT 10.5.2 · p. 188

Axes par bana fixed semi-major axis a wala ek ellipse, dono foci mark kiye hue aur ek side panel jo e aur b batata hai. Step 1: e = 0, foci centre par saath hain aur curve b = a wala circle hai. Step 2: e = 0.6, foci alag hote hain aur b = 0.8a. Step 3: e = 0.8, b = 0.6a aur ellipse zyada flat hai. Step 4: e = 0.95, foci vertices ke paas hain aur b lagbhag 0.31a hai, ek bahut flat ellipse.

10.5.3 Ellipse ki standard equations

Centre ko origin par rakho aur foci $F_{1} (- c, 0)$ aur $F_{2} (c, 0)$ ko $x$ -axis par. Ek point $P (x, y)$ ellipse par tabhi hai jab $P F_{1} + P F_{2} = 2 a$ , jahan $a > c$ .

DerivationSum rule se ellipse ki equation

Do square roots ke sum ko square karne ki jagah unke squares ka difference dekho. $y^{2}$ wale terms cancel ho jaate hain:

\begin{aligned} P F_{1}^{2} - P F_{2}^{2} \\ = (x + c)^{2} \\ - (x - c)^{2} \\ = 4 c x \end{aligned}

$P F_{1} + P F_{2} = 2 a$ se divide karne par $P F_{1} - P F_{2} = \frac{2 c x}{a}$ milta hai. Ise $P F_{1} + P F_{2} = 2 a$ mein jodne aur ghatane se har distance alag se milta hai:

Focal distances

\begin{aligned} P F_{1} & = a + \frac{c}{a} x \\ P F_{2} & = a - \frac{c}{a} x \end{aligned}

(10.6)

Ab doosre ko square karo, $P F_{2}^{2} = (x - c)^{2} + y^{2}$ use karke; dono sides ke $- 2 c x$ terms cancel ho jaate hain:

\begin{aligned} x^{2} + c^{2} + y^{2} \\ = a^{2} + \frac{c^{2}}{a^{2}} x^{2} \\ \frac{a^{2} - c^{2}}{a^{2}} x^{2} + y^{2} \\ = a^{2} - c^{2} \end{aligned}

$b^{2} = a^{2} - c^{2}$ likho, jo positive hai kyunki $a > c$ , aur $b^{2}$ se divide karo:

Standard ellipse (a > b)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

(10.7)

Ulta, maano $(x, y)$ (10.7) ko satisfy karta hai. Tab $y^{2} = b^{2} (1 - \frac{x^{2}}{a^{2}})$ , aur expand karne par $P F_{1}^{2} = (x + c)^{2} + y^{2} = (a + \frac{c}{a} x)^{2} .$ milta hai. Kyunki $∣ x ∣ \leq a$ aur $c < a$ , bracket positive hai, isliye $P F_{1} = a + \frac{c}{a} x$ . Isi tarah $P F_{2} = a - \frac{c}{a} x$ . Dono milkar $2 a$ hote hain, isliye point ellipse par hai.

Ellipse kahan tak pahunchta hai? (10.7) ke left wale do terms kabhi negative nahi hote aur milkar 1 hote hain, isliye koi bhi 1 se zyada nahi ho sakta. Isliye $x^{2} \leq a^{2}$ aur $y^{2} \leq b^{2}$ : curve lines $x = \pm a$ aur $y = \pm b$ ke box mein band hai, aur har line ko ek vertex ya minor axis ke ek end par chhoota hai (Fig. 10.10 ka step 4).

$x$ aur $y$ ke roles badal do to khada ellipse milta hai, jiske foci $(0, \pm c)$ hain: $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1.$ Yahi pair ellipse ki standard equations kehlata hai; shift ya tilt kiye gaye ellipse ke liye lambi equation chahiye, jo aage ki classes mein aayegi.

SimulationEllipse ki talaash

Dono axes par minus 6 se 6 tak ek coordinate grid, jismein do fixed foci hain, F1 (minus 4, 0) par aur F2 (4, 0) par, aur ek draggable point P jo (2, 2) se shuru hota hai. P se har focus tak ke distances bane hain aur naape jaate hain, aur P 0.1 ke steps par snap hota hai aur apne coordinates dikhata hai.

Fig. 10.12

P

ko drag karo aur dono distances jodo. Ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1

woh saare points hain jahan inka total 10 hai.

Try karo

$P$ ko $(5, 0)$ par aur phir $(0, 3)$ par le jao. Check karo ki dono jagah $P F_{1} + P F_{2} = 10$ hai.
$(4, 1.8)$ try karo, jo ek latus rectum ka end hai, aur $(- 3, 2.4)$ bhi. Kya total ab bhi 10 hai?
$P$ ko $(3, 2)$ par rakho. Total 10 se zyada hai ya kam? $P$ ellipse ke andar hai ya bahar?

10.5.4 Latus rectum

$F_{2} (c, 0)$ se guzarta latus rectum line $x = c$ par hai. (10.7) mein substitute karke aur $a^{2} - c^{2} = b^{2}$ use karke:

\frac{y^{2}}{b^{2}} = 1 - \frac{c^{2}}{a^{2}} = \frac{b^{2}}{a^{2}}

isliye $y = \pm \frac{b^{2}}{a}$ aur ends $(c, \pm \frac{b^{2}}{a})$ hain. Latus rectum is height ka do guna hai:

Ellipse aur hyperbola

latus rectum = \frac{2 b^{2}}{a}

(10.8)

Example 10.9NCERT Example 9 · p. 192

Ek ellipse ke baare mein sab kuch

Ellipse $\frac{x^{2}}{289} + \frac{y^{2}}{64} = 1$ ke foci aur vertices, dono axes ki lengths, eccentricity aur latus rectum batao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.10NCERT Example 10 · p. 193

Khada hua ellipse

Ellipse $4 x^{2} + y^{2} = 16$ ko poora describe karo: foci, vertices, axis lengths, eccentricity aur latus rectum.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.11NCERT Example 11 · p. 193

Vertices aur foci se

Vertices $(0, \pm 15)$ aur foci $(0, \pm 9)$ wala ellipse nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.12NCERT Example 12 · p. 194

Major axis aur foci se

Ek ellipse ki major axis ki length 14 hai aur uske foci $(\pm 2 \sqrt{6}, 0)$ hain. Uski equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.13NCERT Example 13 · p. 194

Do points se hokar

Standard form wale ek ellipse ki major axis $x$ -axis ke saath hai aur yeh $(2, 2)$ aur $(- 3, 1)$ se guzarta hai. Uski equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1

ke foci hain

(\pm 3, 0)

(0, \pm \sqrt{41})

(\pm 4, 0)

(0, \pm 3)

Answer dekho

Answer (D) $(0, \pm 3)$

Bada denominator, 25,

y^{2}

ke neeche hai, isliye major axis

y

-axis hai,

a = 5

b = 4

aur

c = \sqrt{25 - 16} = 3

: foci

(0, \pm 3)

. Denominators jodne se

\sqrt{41}

milta hai, jo hyperbola ka rule hai aur yahan galat hai.

Practice · 9 questions

Practice 10.3: answers ke saath

1
Long answerEasyNCERT Exercise 10.3 Q1–Q6
Har ellipse ke foci, vertices, major aur minor axes ki lengths, eccentricity aur latus rectum nikaalo.
1. $\frac{x^{2}}{49} + \frac{y^{2}}{24} = 1$
  Solution dikhao
  AnswerFoci $(\pm 5, 0)$ , vertices $(\pm 7, 0)$ ; axes 14 aur $4 \sqrt{6}$ ; $e = \frac{5}{7}$ ; latus rectum $\frac{48}{7}$
  $a = 7$ , $b = 2 \sqrt{6}$ , $x$ -axis ke saath.
  $c = \sqrt{49 - 24} = 5$
  Isliye $e = \frac{5}{7}$ aur latus rectum $\frac{2 \cdot 24}{7} = \frac{48}{7} .$ hai.
2. $\frac{x^{2}}{20} + \frac{y^{2}}{36} = 1$
  Solution dikhao
  AnswerFoci $(0, \pm 4)$ , vertices $(0, \pm 6)$ ; axes 12 aur $4 \sqrt{5}$ ; $e = \frac{2}{3}$ ; latus rectum $\frac{20}{3}$
  $a = 6$ , $y$ -axis ke saath, $b = 2 \sqrt{5}$ .
  $c = \sqrt{36 - 20} = 4$
  Isliye $e = \frac{4}{6} = \frac{2}{3}$ aur latus rectum $\frac{40}{6} = \frac{20}{3} .$ hai.
3. $\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1$
  Solution dikhao
  AnswerFoci $(0, \pm 2)$ , vertices $(0, \pm 4)$ ; axes 8 aur $4 \sqrt{3}$ ; $e = \frac{1}{2}$ ; latus rectum 6
  $a = 4$ , $y$ -axis ke saath, $b = 2 \sqrt{3}$ .
  $c = \sqrt{16 - 12} = 2$
  Isliye $e = \frac{1}{2}$ aur latus rectum $\frac{24}{4} = 6.$ hai.
2
Long answerMediumNCERT Exercise 10.3 Q7–Q9
Har ellipse ko standard form mein likho, phir uske foci, vertices, axes, eccentricity aur latus rectum nikaalo.
1. $16 x^{2} + 25 y^{2} = 400$
  Solution dikhao
  Answer $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ : foci $(\pm 3, 0)$ , vertices $(\pm 5, 0)$ ; axes 10 aur 8; $e = \frac{3}{5}$ ; latus rectum $\frac{32}{5}$
  $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
  $a = 5$ , $b = 4$ , $c = 3$ , $x$ -axis ke saath, isliye $e = \frac{3}{5}$ aur latus rectum $\frac{32}{5} .$ hai.
2. $x^{2} + 4 y^{2} = 4$
  Solution dikhao
  Answer $\frac{x^{2}}{4} + y^{2} = 1$ : foci $(\pm \sqrt{3}, 0)$ , vertices $(\pm 2, 0)$ ; axes 4 aur 2; $e = \frac{\sqrt{3}}{2}$ ; latus rectum 1
  $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
  $a = 2$ , $b = 1$ , $c = \sqrt{3}$ , isliye $e = \frac{\sqrt{3}}{2}$ aur latus rectum $\frac{2 \cdot 1}{2} = 1.$ hai.
3. $9 x^{2} + 5 y^{2} = 45$
  Solution dikhao
  Answer $\frac{x^{2}}{5} + \frac{y^{2}}{9} = 1$ : foci $(0, \pm 2)$ , vertices $(0, \pm 3)$ ; axes 6 aur $2 \sqrt{5}$ ; $e = \frac{2}{3}$ ; latus rectum $\frac{10}{3}$
  $\frac{x^{2}}{5} + \frac{y^{2}}{9} = 1$
  $a = 3$ , $y$ -axis ke saath, $b = \sqrt{5}$ , $c = 2$ , isliye $e = \frac{2}{3}$ aur latus rectum $\frac{10}{3} .$ hai.
3
Short answerEasyNCERT Exercise 10.3 Q10–Q12
Diye gaye vertices aur foci wala ellipse nikaalo.
1. Vertices $(\pm 10, 0)$ aur foci $(\pm 6, 0)$
  Solution dikhao
  Answer $\frac{x^{2}}{100} + \frac{y^{2}}{64} = 1$
  $b^{2} = 100 - 36 = 64$
2. Vertices $(0, \pm 7)$ aur foci $(0, \pm \sqrt{13})$
  Solution dikhao
  Answer $\frac{x^{2}}{36} + \frac{y^{2}}{49} = 1$
  $b^{2} = 49 - 13 = 36$
  Major axis $y$ -axis hai, isliye $a^{2} = 49$ $y^{2}$ ke neeche aata hai.
3. Vertices $(\pm 9, 0)$ aur foci $(\pm 5, 0)$
  Solution dikhao
  Answer $\frac{x^{2}}{81} + \frac{y^{2}}{56} = 1$
  $b^{2} = 81 - 25 = 56$
4
Short answerEasyNCERT Exercise 10.3 Q13–Q14
Axes ke ends se ellipse nikaalo.
1. Major axis ke ends $(\pm 7, 0)$ ; minor axis ke ends $(0, \pm 2)$
  Solution dikhao
  Answer $\frac{x^{2}}{49} + \frac{y^{2}}{4} = 1$
  $a = 7$ $x$ -axis par aur $b = 2$ .
2. Major axis ke ends $(0, \pm \sqrt{10})$ ; minor axis ke ends $(\pm 2, 0)$
  Solution dikhao
  Answer $\frac{x^{2}}{4} + \frac{y^{2}}{10} = 1$
  $a^{2} = 10$ $y^{2}$ ke neeche aur $b^{2} = 4$ $x^{2}$ ke neeche.
5
Short answerMediumNCERT Exercise 10.3 Q15–Q16
Ek axis ki length aur foci se ellipse nikaalo.
1. Major axis ki length 18; foci $(\pm 4, 0)$
  Solution dikhao
  Answer $\frac{x^{2}}{81} + \frac{y^{2}}{65} = 1$
  $a = 9$ aur $c = 4$ .
  $b^{2} = 81 - 16 = 65$
2. Minor axis ki length 10; foci $(0, \pm 12)$
  Solution dikhao
  Answer $\frac{x^{2}}{25} + \frac{y^{2}}{169} = 1$
  $b = 5$ , $c = 12$ , major axis $y$ -axis ke saath.
  $a^{2} = 25 + 144 = 169$
6
Short answerMediumNCERT Exercise 10.3 Q17–Q18
Diye gaye data se origin par centre wala ellipse nikaalo.
1. Foci $(0, \pm 5)$ aur $a = 6$
  Solution dikhao
  Answer $\frac{x^{2}}{11} + \frac{y^{2}}{36} = 1$
  $b^{2} = 36 - 25 = 11$
  Foci $y$ -axis par hain, isliye $a^{2} = 36$ $y^{2}$ ke neeche aata hai.
2. $b = 2$ , $c = 3$ , foci $x$ -axis par
  Solution dikhao
  Answer $\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1$
  $a^{2} = 4 + 9 = 13$
7
Long answerHardNCERT Exercise 10.3 Q19–Q20
Diye gaye do points se guzarta standard ellipse nikaalo.
1. Centre origin par, major axis $y$ -axis par, $(2, 3)$ aur $(1, 5)$ se hokar
  Solution dikhao
  Answer $16 x^{2} + 3 y^{2} = 91$
  $4 A + 9 B = 1, A + 25 B = 1$
  Tab $A = 1 - 25 B$ , isliye $4 - 91 B = 1$ : $B = \frac{3}{91}$ aur $A = \frac{16}{91}$ . Yahan $\frac{1}{B} = \frac{91}{3} > \frac{1}{A} = \frac{91}{16}$ , isliye major axis sach mein vertical hai.
  $16 x^{2} + 3 y^{2} = 91$
2. Major axis $x$ -axis par, $(3, 2)$ aur $(- 1, 3)$ se hokar
  Solution dikhao
  Answer $5 x^{2} + 8 y^{2} = 77$
  $9 A + 4 B = 1, A + 9 B = 1$
  Tab $A = 1 - 9 B$ , isliye $9 - 77 B = 1$ : $B = \frac{8}{77}$ aur $A = \frac{5}{77}$ , aur jaisa chahiye $\frac{1}{A} > \frac{1}{B}$ .
  $5 x^{2} + 8 y^{2} = 77$
Quick answerMedium
Ek ellipse ki minor axis uske foci ke beech ke distance jitni lambi hai. Uski eccentricity hai
$\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ $\sqrt{2}$
Answer dekho
Answer (B) $\frac{1}{\sqrt{2}}$
$2 b = 2 c$ se $b = c$ milta hai, isliye $a^{2} = 2 c^{2}$ aur $e = \frac{c}{a} = \frac{1}{\sqrt{2}}$ . Option $\sqrt{2}$ 1 se zyada hai, jo ellipse ke liye impossible hai.
Quick answerMedium
Assertion (A): Ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ ke foci $y$ -axis par hain. Reason (R): Standard ellipse ki major axis bade denominator wale variable ke saath hoti hai, aur foci major axis par hote hain.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
Yahan $16 > 9$ , isliye major axis $y$ -axis hai aur foci $(0, \pm \sqrt{7})$ hain. R hi theek iska reason hai.

10.6 Hyperbola

Ek doosre se bahut door do radio stations ek hi pal mein ek pulse bhejte hain. Ek ship ko ek pulse doosre se 0.2 ms pehle sunai deti hai. Radio ek millisecond mein 300 km chalta hai, isliye ship pehle station se doosre ke mukable 60 km zyada paas hai. Distances ke isi difference wale saare points ek curve par hain, jo hyperbola hai; stations ka doosra pair ek doosra curve deta hai, aur ship wahan hai jahan dono kaatte hain. LORAN jaise navigation systems isi tarah kaam karte the.

Definition

Hyperbola

Ek plane mein do fixed points $F_{1}$ aur $F_{2}$ lo. Hyperbola us plane ke un points $P$ se banta hai jinke liye $P F_{1}$ aur $P F_{2}$ ka difference, yaani bada minus chhota, ek hi value rakhta hai. $F_{1}$ aur $F_{2}$ iske foci hain.

1/4 $F_{2}$ ke paas: $P F_{1} - P F_{2} = 2 a$ .

Fig. 10.13Dono branches ke har point par bada distance minus chhota distance same rehta hai, yaani

2 a

.NCERT Figs 10.27, 10.28 · p. 196

Ek horizontal line par do foci F1 aur F2, curve ki halki outline ke saath. Ek point P, F1 se solid segment aur F2 se dashed segment se juda hai, aur neeche do bars PF1 aur PF2 dikhate hain, jahan lambe bar ki extra length box mein hai aur uspar 2a likha hai. Steps 1 aur 2 P ko right branch par do jagah rakhte hain, F2 ke paas, doosri jagah vertex ke kareeb; bars badalte hain lekin box wala difference 2a hi rehta hai. Step 3 P ko left branch par rakhta hai, F1 ke paas, jahan ab PF2 lamba bar hai, phir se 2a zyada lamba. Step 4 dono branches ko centre, do vertices A aur A prime, foci se guzarte transverse axis aur vertical conjugate axis ke saath banata hai.

Socho ki dono branches ek doosre ki taraf face kar rahi hain. Har branch ka woh point jo doosri ke sabse paas hai vertex hai, aur vertices $2 a$ ki doori par hain. Inse guzarne wali line dono foci se bhi guzarti hai, jo aur bahar $2 c$ ki doori par hain; yeh line transverse axis hai, aur $F_{1} F_{2}$ ka midpoint centre hai. Centre se iske right angle par guzarne wali line conjugate axis hai.

Kyunki vertices foci ke beech hain, $c > a$ , aur number $b = \sqrt{c^{2} - a^{2}}$ real hai. Conjugate axis par centre ke around $2 b$ length ka segment bhi conjugate axis kehlata hai; Fig. 10.15 dikhata hai ki $b$ kya naapta hai.

Constant difference phir se $2 a$ hai. Dono foci ko aapas mein badalne se “bada minus chhota” nahi badalta, isliye hyperbola apne conjugate axis ke around symmetric hai, aur uske vertices mirror images hain, har ek centre se $a$ door. $F_{2}$ ke paas wale vertex $A$ par $A F_{1} = c + a$ aur $A F_{2} = c - a$ , jinka fark $2 a$ hai. Isliye hyperbola ke har point $P$ par $∣ P F_{1} - P F_{2} ∣ = 2 a$ .

SimulationConstant difference

Dono axes par minus 8 se 8 tak ek coordinate grid, jismein do fixed foci hain, F1 (minus 4, 0) par aur F2 (4, 0) par, aur ek draggable point P jo (6, 3) se shuru hota hai. P se har focus tak ke distances bane hain aur naape jaate hain, aur P grid points par snap hota hai aur apne coordinates dikhata hai.

Fig. 10.14

P

ko drag karo aur bade distance mein se chhota ghatao. Hyperbola

\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1

woh saare points hain jahan difference 4 hai.

Try karo

$P$ ko $(2, 0)$ par le jao, jo ek vertex hai. Check karo ki distances mein 4 ka fark hai.
$x = \pm 4$ wale woh chaar grid points dhoondho jahan difference 4 hai. (Hint: $y = \pm 6$ try karo.)
$P$ ko $y$ -axis par rakho. Wahan difference kya hai, aur hyperbola ka koi point us axis par kyun nahi hai?

10.6.1 Eccentricity

Ratio $e = \frac{c}{a}$ hyperbola ke liye bhi defined hai. Ab foci vertices ke aage hain, $c > a$ , isliye $e > 1$ , aur har focus ab bhi centre se $a e$ door hai. Bada $e$ matlab chaudi branches, kyunki $\frac{b}{a} = \sqrt{e^{2} - 1}$ .

10.6.2 Hyperbola ki standard equation

Centre ko origin par aur foci ko $F_{1} (- c, 0)$ aur $F_{2} (c, 0)$ par rakho. Right branch par ek point $P (x, y)$ lo, jo $F_{2}$ ke paas wali hai, taaki $P F_{1} - P F_{2} = 2 a$ .

DerivationDifference rule se hyperbola ki equation

Ellipse wala trick yahan bhi kaam karta hai. Squares ka difference ab bhi $P F_{1}^{2} - P F_{2}^{2} = 4 c x$ hai; ise $P F_{1} - P F_{2} = 2 a$ se divide karo:

P F_{1} + P F_{2} = \frac{2 c x}{a}

$P F_{1} - P F_{2} = 2 a$ ke saath milkar yeh do focal distances deta hai:

Right branch

\begin{aligned} P F_{1} & = \frac{c}{a} x + a \\ P F_{2} & = \frac{c}{a} x - a \end{aligned}

(10.9)

Doosre ko square karo, $P F_{2}^{2} = (x - c)^{2} + y^{2}$ ke saath; ek baar phir $- 2 c x$ terms cancel ho jaate hain:

\begin{aligned} x^{2} + c^{2} + y^{2} \\ = \frac{c^{2}}{a^{2}} x^{2} + a^{2} \\ \frac{c^{2} - a^{2}}{a^{2}} x^{2} - y^{2} \\ = c^{2} - a^{2} \end{aligned}

$b^{2} = c^{2} - a^{2}$ ke saath, $b^{2}$ se divide karo:

Standard hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

(10.10)

Left branch par, jahan $P F_{2} - P F_{1} = 2 a$ , dono focal distances ka sign badal jaata hai lekin unke squares ka nahi, isliye wahi equation milti hai.

Ulta, agar $(x, y)$ (10.10) ko satisfy karta hai, to $y^{2} = b^{2} (\frac{x^{2}}{a^{2}} - 1)$ aur expand karne par $P F_{1}^{2} = (\frac{c}{a} x + a)^{2}$ aur $P F_{2}^{2} = (\frac{c}{a} x - a)^{2} .$ milte hain. $x \geq a$ ke liye dono brackets positive hain, isliye $P F_{1} - P F_{2} = 2 a$ ; $x \leq - a$ ke liye dono negative hain, isliye $P F_{2} - P F_{1} = 2 a$ . Dono hi case mein point hyperbola par hai.

(10.10) ko $\frac{x^{2}}{a^{2}} = 1 + \frac{y^{2}}{b^{2}}$ ki tarah padho: right side kam se kam 1 hai, isliye $∣ x ∣$ kabhi $a$ se neeche nahi ja sakta. Isliye vertical strip $- a < x < a$ khaali hai (Fig. 10.15). Branches iske bahar rehti hain, $y$ -axis ko kabhi nahi chhooti, aur bina ruke failti jaati hain.

2/4Foci vertices ke bahar hain, aur $c^{2} = a^{2} + b^{2}$ .

Fig. 10.15

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

ke parts.NCERT Fig. 10.29 · p. 197

Coordinate axes par hyperbola x squared over a squared minus y squared over b squared equals 1. Step 1 vertices (a, 0) aur (minus a, 0) mark karta hai aur x = minus a aur x = a ke beech ki strip ko shade karta hai, jismein curve ka koi point nahi hai. Step 2 vertices ke bahar foci F1 aur F2 jodta hai, aur axes ke saath legs a aur b wala ek right triangle, jiska hypotenuse vertex se point (0, b) tak hai aur uski length c hai, isliye c squared equals a squared plus b squared. Step 3 vertices ke beech 2a length ke transverse axis aur y-axis ke saath 2b length ke conjugate axis ko highlight karta hai. Step 4 F2 se guzarta latus rectum banata hai, jo b squared over a length ke do halves se bana hai.

Hyperbola

c^{2} = a^{2} + b^{2}

(10.11)

Foci $y$ -axis par hon to standard equation $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1.$ hai. Jab $a = b$ ho to hyperbola equilateral kehlata hai: $x^{2} - y^{2} = a^{2}$ , jahan $c = a \sqrt{2}$ . Aise hyperbolas bhi hote hain jinke axes koi aur perpendicular lines hon; woh aage ki classes mein aate hain.

10.6.3 Latus rectum

(10.10) mein $x = c$ rakho: $\frac{y^{2}}{b^{2}} = \frac{c^{2}}{a^{2}} - 1 = \frac{b^{2}}{a^{2}}$ , isliye $y = \pm \frac{b^{2}}{a}$ . Isliye hyperbola ke latus rectum ki length bhi ellipse jitni hi hai, $\frac{2 b^{2}}{a}$ , jo (10.8) deta hai.

Table 10.2Ellipse aur hyperbola ki tulna (dono ka centre

O

par, foci

x

-axis par)

	Ellipse	Hyperbola
Rule	sum $2 a$ hai	difference $2 a$ hai
Equation	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$c^{2}$	$a^{2} - b^{2}$	$a^{2} + b^{2}$
$e$	1 se kam	1 se zyada
Vertices	$(\pm a, 0)$	$(\pm a, 0)$
Foci	$(\pm c, 0)$	$(\pm c, 0)$
Axes	major $2 a$ , minor $2 b$	transverse $2 a$ , conjugate $2 b$
Latus rectum	$\frac{2 b^{2}}{a}$	$\frac{2 b^{2}}{a}$

Example 10.14NCERT Example 14 · p. 200

Do hyperbolas ko padhna

(i) $\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$ aur (ii) $4 y^{2} - x^{2} = 4$ ke foci, vertices, eccentricity aur latus rectum nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.15NCERT Example 15 · p. 201

Foci aur vertices se

Foci $(\pm 4, 0)$ aur vertices $(\pm \sqrt{7}, 0)$ wala hyperbola nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.16NCERT Example 16 · p. 201

Foci aur latus rectum se

Foci $(\pm 10, 0)$ wala woh hyperbola nikaalo jiske latus rectum ki length 9 hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 9 questions

Practice 10.4: answers ke saath

1
Short answerEasyNCERT Exercise 10.4 Q1–Q2
Har hyperbola ke foci, vertices, eccentricity aur latus rectum nikaalo.
1. $\frac{x^{2}}{25} - \frac{y^{2}}{144} = 1$
  Solution dikhao
  AnswerFoci $(\pm 13, 0)$ , vertices $(\pm 5, 0)$ , $e = \frac{13}{5}$ , aur latus rectum $\frac{288}{5}$
  $a = 5$ aur $b = 12$ .
  $c = \sqrt{25 + 144} = 13$
  Isliye $e = \frac{13}{5}$ aur latus rectum $\frac{2 \cdot 144}{5} = \frac{288}{5} .$ hai.
2. $\frac{y^{2}}{16} - \frac{x^{2}}{20} = 1$
  Solution dikhao
  AnswerFoci $(0, \pm 6)$ , vertices $(0, \pm 4)$ , $e = \frac{3}{2}$ , aur latus rectum 10
  $a = 4$ , $y$ -axis ke saath, $b^{2} = 20$ .
  $c = \sqrt{16 + 20} = 6$
  Isliye $e = \frac{6}{4} = \frac{3}{2}$ aur latus rectum $\frac{40}{4} = 10.$ hai.
2
Long answerMediumNCERT Exercise 10.4 Q3–Q6
Har hyperbola ko standard form mein likho aur uske foci, vertices, eccentricity aur latus rectum nikaalo.
1. $4 x^{2} - 9 y^{2} = 36$
  Solution dikhao
  Answer $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ : foci $(\pm \sqrt{13}, 0)$ , vertices $(\pm 3, 0)$ , $e = \frac{\sqrt{13}}{3}$ , aur latus rectum $\frac{8}{3}$
  $a = 3$ , $b = 2$ , $c = \sqrt{13}$ ; aur latus rectum $\frac{2 \cdot 4}{3} = \frac{8}{3} .$
2. $16 y^{2} - 9 x^{2} = 144$
  Solution dikhao
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ : foci $(0, \pm 5)$ , vertices $(0, \pm 3)$ , $e = \frac{5}{3}$ , aur latus rectum $\frac{32}{3}$
  $a = 3$ , $y$ -axis ke saath, $b = 4$ , $c = 5$ ; latus rectum $\frac{32}{3} .$
3. $3 x^{2} - y^{2} = 12$
  Solution dikhao
  Answer $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$ : foci $(\pm 4, 0)$ , vertices $(\pm 2, 0)$ , $e = 2$ , aur latus rectum 12
  $a = 2$ , $b^{2} = 12$ , $c = \sqrt{16} = 4$ ; aur latus rectum $\frac{24}{2} = 12.$
4. $7 y^{2} - 9 x^{2} = 63$
  Solution dikhao
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{7} = 1$ : foci $(0, \pm 4)$ , vertices $(0, \pm 3)$ , $e = \frac{4}{3}$ , aur latus rectum $\frac{14}{3}$
  $a = 3$ , $y$ -axis ke saath, $b^{2} = 7$ , $c = 4$ ; latus rectum $\frac{14}{3} .$
3
Short answerEasyNCERT Exercise 10.4 Q7–Q9
Diye gaye vertices aur foci wala hyperbola nikaalo.
1. Vertices $(\pm 6, 0)$ aur foci $(\pm 10, 0)$
  Solution dikhao
  Answer $\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$
  $b^{2} = 100 - 36 = 64$
2. Vertices $(0, \pm 2)$ aur foci $(0, \pm 3)$
  Solution dikhao
  Answer $\frac{y^{2}}{4} - \frac{x^{2}}{5} = 1$
  $b^{2} = 9 - 4 = 5$
3. Vertices $(0, \pm 12)$ aur foci $(0, \pm 13)$
  Solution dikhao
  Answer $\frac{y^{2}}{144} - \frac{x^{2}}{25} = 1$
  $b^{2} = 169 - 144 = 25$
4
Short answerMediumNCERT Exercise 10.4 Q10–Q11
Foci aur ek axis ki length se hyperbola nikaalo.
1. Foci $(\pm 7, 0)$ ; transverse axis ki length 10
  Solution dikhao
  Answer $\frac{x^{2}}{25} - \frac{y^{2}}{24} = 1$
  $a = 5$ aur $c = 7$ .
  $b^{2} = 49 - 25 = 24$
2. Foci $(0, \pm 10)$ ; conjugate axis ki length 12
  Solution dikhao
  Answer $\frac{y^{2}}{64} - \frac{x^{2}}{36} = 1$
  $b = 6$ aur $c = 10$ .
  $a^{2} = 100 - 36 = 64$
5
Long answerHardNCERT Exercise 10.4 Q12–Q13
Foci aur latus rectum ki length se hyperbola nikaalo.
1. Foci $(\pm 2 \sqrt{7}, 0)$ aur latus rectum 6
  Solution dikhao
  Answer $\frac{x^{2}}{16} - \frac{y^{2}}{12} = 1$
  $\frac{2 b^{2}}{a} = 6$ se $b^{2} = 3 a$ milta hai, aur $c^{2} = 28$ .
  $a^{2} + 3 a - 28 = 0$
  Isliye $(a + 7) (a - 4) = 0$ , $a = 4$ aur $b^{2} = 12$ .
2. Foci $(0, \pm 5)$ aur latus rectum $\frac{32}{3}$
  Solution dikhao
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
  $\frac{2 b^{2}}{a} = \frac{32}{3}$ se $b^{2} = \frac{16 a}{3}$ milta hai, aur $c^{2} = 25$ .
  $3 a^{2} + 16 a - 75 = 0$
  Isliye $(a - 3) (3 a + 25) = 0$ , $a = 3$ aur $b^{2} = 16$ , aur transverse axis $y$ -axis ke saath hai.
Short answerMediumNCERT Exercise 10.4 Q14
Vertices $(\pm 2, 0)$ aur eccentricity $\frac{3}{2}$ wala hyperbola nikaalo.
Solution dikhao
Answer $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
$a = 2$ , isliye $c = a e = 3$ .
$b^{2} = 9 - 4 = 5$
Long answerHardNCERT Exercise 10.4 Q15
Standard form wale ek hyperbola ke foci $(\pm 5, 0)$ hain aur yeh $(6, 2)$ se guzarta hai. Uski equation nikaalo.
Solution dikhao
Answer $\frac{x^{2}}{20} - \frac{y^{2}}{5} = 1$
$A = a^{2}$ lene par point se $\frac{36}{A} - \frac{4}{25 - A} = 1$ milta hai. Fractions hatao:
$A^{2} - 65 A + 900 = 0$
Isliye $A = 20$ ya $A = 45$ . Lekin $a < c = 5$ ke liye $A < 25$ chahiye, isliye $a^{2} = 20$ aur $b^{2} = 5$ .
Quick answerMedium
Har equilateral hyperbola ( $a = b$ ) ki eccentricity hai
$1$ $\sqrt{2}$ $2$ $\frac{1}{\sqrt{2}}$
Answer dekho
Answer (B) $\sqrt{2}$
$c^{2} = a^{2} + a^{2} = 2 a^{2}$ , isliye $e = \frac{c}{a} = \sqrt{2}$ . Value $\frac{1}{\sqrt{2}}$ us ellipse ki hai jismein $b = c$ ho.
Quick answerMedium
Ek hyperbola ki eccentricity 2 hai aur uske foci 16 units ki doori par hain. Uske transverse axis ki length nikaalo.
Answer dekho
Answer 8
$2 c = 16$ se $c = 8$ milta hai; $a = \frac{c}{e} = 4$ , isliye transverse axis $2 a = 8$ hai.

Miscellaneous Examples

In har model mein pehla kaam axes ko aise chunna hai ki curve ki standard equation ho; uske baad questions sirf arithmetic reh jaate hain. Shuru karne se pehle neeche wali lab mein chaaron conics try karo.

InteractiveConic lab

Buttons se circle, parabola, ellipse ya hyperbola chuna jaata hai, aur ek aur button axis ko horizontal aur vertical ke beech badalta hai. Sliders size a (ya radius) aur eccentricity e set karte hain. Graph curve ko uske foci, parabola ke liye uski directrix, vertices aur ek latus rectum ke saath dikhata hai. Ek teesra slider, ya Play button, ek point P ko curve par chalata hai, aur P se foci tak (ya focus aur directrix tak) segments bante hain. Ek panel equation, a, b, c, e, latus rectum aur live distances dikhata hai, jo defining rule confirm karte hain: barabar distances, constant sum ya constant difference.

Fig. 10.16Ek conic chuno,

a

aur

e

badlo, aur

P

ko move karke uska distance rule check karo.

Example 10.17NCERT Example 17 · p. 202

Dish antenna

DTH dish ka receiver focus par hai, dish ke vertex se 36 cm door. Dish 16 cm gehri hai. Rim ke aar-paar yeh kitni chaudi hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.18NCERT Example 18 · p. 202

Fountain ki dhaar

Paani ki ek dhaar ground level par ek nozzle se nikalti hai, sabse zyada 2 m ki height tak jaati hai, aur nozzle se 3 m door girti hai, aur ek parabola banati hai. Nozzle se kin horizontal distances par paani zameen se 1.5 m upar hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4/4 $P$ ki positions $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ par hain.

Fig. 10.17Jaise-jaise ladder fisalti hai, mark kiya point ek ellipse ka hissa banata hai.NCERT Fig. 10.33 · p. 203

Ek vertical wall aur horizontal floor O par milte hain. 5 m ki ladder AB tiki hai, uska foot A floor par aur top B wall par hai, aur us par ek point P, A se 2 m aur B se 3 m door hai. Steps 1 se 3 ladder ko steep, 45 degrees par aur lagbhag flat dikhate hain, har baar P mark hai aur P ki pichhli positions halke dots ki tarah bachi hain. Step 4 P ki positions ko jod kar ellipse ka ek quarter banata hai aur uski equation deta hai, x squared over 9 plus y squared over 4 equals 1.

Example 10.19NCERT Example 19 · p. 203

Fisalti ladder

5 m ki ek ladder fisalti hai, uska foot $A$ floor ( $x$ -axis) par aur top $B$ wall ( $y$ -axis) se tika hai. Ladder par paint ka ek mark $P$ , $A$ se 2 m door hai. Dikhao ki $P$ ek ellipse par chalta hai (Fig. 10.17).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Har equation ko uske curve se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$x^{2} + y^{2} = 9$ : circle
$y^{2} = 9 x$ : parabola
$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ : ellipse
$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ : hyperbola

Plus sign ke saath barabar squared terms circle dete hain; sirf ek squared variable parabola deta hai; plus ke saath alag denominators ellipse dete hain; squares ke beech minus sign hyperbola deta hai.

History se

Cones se coordinates tak

Greek geometers ne sadiyon tak conics ko bina kisi algebra ke solid geometry ki tarah padha. Apollonius ki Conics ne dikhaya ki yeh sab ek hi double cone se, kaatne wale plane ka tilt badal kar, aate hain. Prachin India ke Śulba-sūtras ke rassi-aur-khoonti wale constructions bhi isi tarah pure geometry the.

Bada badlaav seventeenth century mein aaya, jab Pierre de Fermat aur René Descartes ne alag-alag kaam karte hue curves ko coordinates mein equations se describe kiya. Fermat ka version haath se likhi copies mein ghooma aur sirf 1679 mein, unki death ke baad, print hua; yeh ek wajah hai ki Descartes ki La Géométrie (1637) ko is subject ki shuruaat maana jaata hai. Agli dedh sadi mein Clairaut, Cramer, Monge aur Lacroix jaise mathematicians ne distances, slopes aur lines ke woh formulas banaye jo tumne Chapter 9 mein use kiye.

Timeline

History mein conics

c. 350 BCE
Menaechmus
Conic sections ki pehli jaani-maani study, cube ko double karne ki problem par kaam karte hue.
c. 200 BCE
Apollonius
Conics likhte hain aur ellipse, parabola aur hyperbola ko naam dete hain.
1609
Kepler
Planets ellipses par chalte hain aur Sun ek focus par hota hai.
1637
Descartes
La Géométrie curves ko equations mein badal deti hai.
1638
Galileo
Dikhate hain ki phenki gayi cheez parabola follow karti hai.
1679
Fermat
Plane aur solid loci par unka introduction, jo dashakon pehle likha gaya tha, print hota hai.
1687
Newton
Dikhate hain ki inverse-square gravity se bane orbits conics hote hain.

Recap

Summary aur key points

Ek plane double cone ko axis se apne angle $β$ ke hisaab se circle, ellipse, parabola ya hyperbola mein kaatta hai; vertex se guzarne par yeh ek point, ek line ya do lines deta hai (Section 10.2).
Circle centre $(h, k)$ se distance $r$ par points ka set hai: $(x - h)^{2} + (y - k)^{2} = r^{2}$ (Section 10.3).
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ ka centre $(- g, - f)$ aur radius $\sqrt{g^{2} + f^{2} - c}$ hai; ise padhne ke liye squares complete karo ((10.2)).
Parabola un points ka set hai jo ek focus aur ek directrix se barabar door hain (Section 10.4).
$y^{2} = 4 a x$ ka focus $(a, 0)$ aur directrix $x = - a$ hai; squared variable axis batata hai aur sign direction (Section 10.4.1).
Latus rectum axis ke perpendicular focal chord hai; $y^{2} = 4 a x$ ke liye iski length $4 a$ hai (Section 10.4.2).
Ellipse un points ka set hai jinke do foci se distances milkar ek constant $2 a$ banate hain, jo major axis ki length hai (Section 10.5).
Ellipse ke liye $a^{2} = b^{2} + c^{2}$ , aur $e = \frac{c}{a} < 1$ batata hai ki woh kitna flat hai (Section 10.5.1, Section 10.5.2).
$a > b$ ke saath $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ke foci $(\pm c, 0)$ hain; bada denominator major axis batata hai (Section 10.5.3).
Ellipse aur hyperbola dono ke liye latus rectum $\frac{2 b^{2}}{a}$ hai ((10.8)).
Hyperbola un points ka set hai jinke do foci se distances mein ek constant $2 a$ ka fark hai, jo transverse axis ki length hai (Section 10.6).
Hyperbola ke liye $c^{2} = a^{2} + b^{2}$ aur $e > 1$ ; $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ mein positive term transverse axis batata hai (Section 10.6.2).
Reflectors, arches, cables aur sliding rods ko aise axes chun kar model kiya jaata hai jo standard equation dein (Miscellaneous Examples).

Practice · 11 questions

Practice questions answers ke saath (NCERT “Chapter 10 ki miscellaneous exercises”)

NumericalEasyNCERT Miscellaneous Exercise Q1
Ek torch ka reflector 12 cm chauda aur 3 cm gehra parabolic bowl hai. Bulb (focus) kahan lagna chahiye?
Solution dikhao
AnswerAxis par, vertex se 3 cm door.
$36 = 4 a \cdot 3 \Rightarrow a = 3$
Focus $(3, 0)$ hai: axis par, vertex se 3 cm door.
NumericalMediumNCERT Miscellaneous Exercise Q2
Ek gateway vertical axis wale parabola ke shape ka hai: yeh 9 m ooncha hai aur zameen par 12 m tak faila hai. 5 m ki height par yeh kitna chauda hai?
Solution dikhao
Answer8 m
Vertex top par ho to $x^{2} = - 4 a y$ point $(6, - 9)$ se guzarta hai.
$36 = 36 a \Rightarrow a = 1$
Isliye $x^{2} = - 4 y$ . 5 m ki height top se 4 m neeche hai, isliye $y = - 4$ : $x^{2} = 16$ , $x = \pm 4$ aur chaudai 8 m hai.
NumericalHardNCERT Miscellaneous Exercise Q3
Ek footbridge ka 120 m lamba level deck hai, jo vertical rods se ek parabolic cable par latka hai. Dono ends ki rods 32 m lambi hain aur centre wali sirf 8 m. Centre se 30 m door kitni lambi rod chahiye?
Solution dikhao
Answer14 m
Vertex sabse neeche wale point par ho to $x^{2} = 4 a y$ . Ek end par $x = 60$ hai aur cable vertex se $32 - 8 = 24$ m upar hai.
$3600 = 96 a \Rightarrow a = 37.5$
$x = 30$ par: $y = \frac{900}{150} = 6$ , isliye rod $8 + 6 = 14$ m lambi hai.
NumericalMediumNCERT Miscellaneous Exercise Q4
Ek tunnel ka entrance aadha ellipse hai, road level par 10 m chauda aur beech mein 3 m ooncha. Ek side se 2 m andar kitni headroom hai?
Solution dikhao
Answer2.4 m
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$x = 3$ par: $y^{2} = 9 (1 - \frac{9}{25}) = \frac{144}{25}$ , isliye $y = 2.4$ m.
Long answerMediumNCERT Miscellaneous Exercise Q5
10 cm ki ek stick ke ends do coordinate axes par fisalte hain. Stick par $x$ -axis wale end se 4 cm door ek dot $P$ paint kiya gaya hai. Dot jis curve par chalta hai uski equation nikaalo.
Solution dikhao
Answer $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$
Agar rod $x$ -axis se angle $θ$ banati hai, to $P$ , $y$ -axis wale end se 6 cm door hai.
$x = 6 \cos θ, y = 4 \sin θ$
$\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$
Short answerMediumNCERT Miscellaneous Exercise Q6
$y^{2} = - 20 x$ ke vertex ko uske latus rectum ke dono ends se jodo. Jo triangle banta hai uska area kya hai?
Solution dikhao
Answer50 square units
$a = 5$ , isliye ends $(- 5, 10)$ aur $(- 5, - 10)$ hain: base 20, height 5.
$\frac{1}{2} \times 20 \times 5 = 50$
Short answerMediumNCERT Miscellaneous Exercise Q7
Ek robotic lawn-mower do beacons se apne distances ka sum 26 m rakhta hai. Beacons 24 m ki doori par hain. Beacons ko $x$ -axis par aur origin ko unke theek beech mein rakh kar iska raasta nikaalo.
Solution dikhao
Answer $\frac{x^{2}}{169} + \frac{y^{2}}{25} = 1$
$a = 13$ aur $c = 12$ .
$b^{2} = 169 - 144 = 25$
Long answerHardNCERT Miscellaneous Exercise Q8
Ek triangle ka ek vertex $y^{2} = 12 x$ ke vertex par hai aur baaki do parabola par hain, axis ke around symmetric rakhe hue, taaki origin par angle right angle ho. Uski sabse lambi side ki length aur uska area nikaalo.
Solution dikhao
AnswerHypotenuse 24 aur area 144 square units
Origin par right angle, jise axis barabar baantta hai, ka matlab hai ki har arm axis se $45^{\circ}$ banati hai, isliye points $(t, \pm t)$ hain.
$t^{2} = 12 t \Rightarrow t = 12$
Vertices $(12, 12)$ aur $(12, - 12)$ hain: hypotenuse 24 hai, aur area $\frac{1}{2} \times 24 \times 12 = 144$ hai.
Quick answerHard
$k$ ki kin values ke liye $\frac{x^{2}}{9 - k} + \frac{y^{2}}{5 - k} = 1$ ek hyperbola represent karta hai?
$k < 5$ $5 < k < 9$ $k > 9$ $k < 9$
Answer dekho
Answer (B) $5 < k < 9$
Hyperbola ke liye dono denominators ke sign ulte hone chahiye: $9 - k > 0$ aur $5 - k < 0$ , isliye $5 < k < 9$ . $k < 5$ ke liye dono positive hain (ellipse); $k > 9$ ke liye dono negative hain (koi real point nahi).
Quick answerMedium
Assertion (A): $y^{2} = 4 a x$ ke latus rectum ki length $4 a$ hai. Reason (R): Latus rectum ka har end directrix se $2 a$ door hai.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
Har end directrix se $2 a$ door hai, isliye definition se woh focus se bhi $2 a$ door hai; dono halves milkar $4 a$ banate hain. R hi reason hai.
Quick answerHard
Ek ellipse ka latus rectum uski minor axis ka aadha lamba hai. Uski eccentricity decimal mein nikaalo.
Answer dekho
Answer 0.866
$\frac{2 b^{2}}{a} = b$ se $b = \frac{a}{2}$ milta hai, isliye $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{3}{4}$ aur $e = \frac{\sqrt{3}}{2} \approx 0.866$ .

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 10.1 Q1–Q4	Practice 10.1 · Q1
NCERT Exercise 10.1 Q5	Practice 10.1 · Q2
NCERT Exercise 10.1 Q6–Q9	Practice 10.1 · Q3
NCERT Exercise 10.1 Q10–Q11	Practice 10.1 · Q4
NCERT Exercise 10.1 Q12	Practice 10.1 · Q5
NCERT Exercise 10.1 Q13	Practice 10.1 · Q6
NCERT Exercise 10.1 Q14	Practice 10.1 · Q7
NCERT Exercise 10.1 Q15	Practice 10.1 · Q8
NCERT Exercise 10.2 Q1–Q6	Practice 10.2 · Q1
NCERT Exercise 10.2 Q7–Q8	Practice 10.2 · Q2
NCERT Exercise 10.2 Q9–Q10	Practice 10.2 · Q3
NCERT Exercise 10.2 Q11–Q12	Practice 10.2 · Q4
NCERT Exercise 10.3 Q1–Q6	Practice 10.3 · Q1
NCERT Exercise 10.3 Q7–Q9	Practice 10.3 · Q2
NCERT Exercise 10.3 Q10–Q12	Practice 10.3 · Q3
NCERT Exercise 10.3 Q13–Q14	Practice 10.3 · Q4
NCERT Exercise 10.3 Q15–Q16	Practice 10.3 · Q5
NCERT Exercise 10.3 Q17–Q18	Practice 10.3 · Q6
NCERT Exercise 10.3 Q19–Q20	Practice 10.3 · Q7
NCERT Exercise 10.4 Q1–Q2	Practice 10.4 · Q1
NCERT Exercise 10.4 Q3–Q6	Practice 10.4 · Q2
NCERT Exercise 10.4 Q7–Q9	Practice 10.4 · Q3
NCERT Exercise 10.4 Q10–Q11	Practice 10.4 · Q4
NCERT Exercise 10.4 Q12–Q13	Practice 10.4 · Q5
NCERT Exercise 10.4 Q14	Practice 10.4 · Q6
NCERT Exercise 10.4 Q15	Practice 10.4 · Q7
NCERT Miscellaneous Exercise Q1	Chapter 10 ki miscellaneous exercises · Q1
NCERT Miscellaneous Exercise Q2	Chapter 10 ki miscellaneous exercises · Q2
NCERT Miscellaneous Exercise Q3	Chapter 10 ki miscellaneous exercises · Q3
NCERT Miscellaneous Exercise Q4	Chapter 10 ki miscellaneous exercises · Q4
NCERT Miscellaneous Exercise Q5	Chapter 10 ki miscellaneous exercises · Q5
NCERT Miscellaneous Exercise Q6	Chapter 10 ki miscellaneous exercises · Q6
NCERT Miscellaneous Exercise Q7	Chapter 10 ki miscellaneous exercises · Q7
NCERT Miscellaneous Exercise Q8	Chapter 10 ki miscellaneous exercises · Q8

Quick check: MCQs

10.1 Introduction2 questions
10.2.2 Degenerated conic sections3 questions
10.3 Circle3 questions
10.4.2 Latus rectum4 questions
10.5.4 Latus rectum4 questions
10.6.3 Latus rectum4 questions
Miscellaneous Examples3 questions

Glossary · 15

Key terms (English + Hindi)

circleवृत्तvritt: Plane ke un points ka set jo ek fixed point (centre) se ek fixed distance (radius) par hain.Chapter mein pehli baar yahan aaya →
conic sectionशंकु परिच्छेदshanku parichchhed: Woh curve jahan ek plane double cone ko kaatta hai: circle, ellipse, parabola ya hyperbola.Chapter mein pehli baar yahan aaya →

degenerate conic: Ek point, ek line ya lines ka pair, jo cone ke vertex se guzarne wale plane ke kaatne se milta hai.Chapter mein pehli baar yahan aaya →
directrixनियताniyata: Parabola ki definition wali fixed line.Chapter mein pehli baar yahan aaya →

eccentricityउत्केंद्रताutkendrata: Ratio $e = c / a$ ; ellipse ke liye 1 se kam, hyperbola ke liye 1 se zyada.Chapter mein pehli baar yahan aaya →
ellipseदीर्घवृत्तdeerghvritt: Plane ke un points ka set jinke do foci se distances milkar ek constant banate hain.Chapter mein pehli baar yahan aaya →

focusनाभिnaabhi: Conic ko define karne wala ek fixed point: parabola ka ek focus hota hai, ellipse aur hyperbola ke do.Chapter mein pehli baar yahan aaya →

generator: Cone ki surface par uske vertex se guzarti ek straight line; ise axis ke around ghumane se cone banta hai.Chapter mein pehli baar yahan aaya →

hyperbolaअतिपरवलयatiparvalay: Plane ke un points ka set jinke do foci se distances mein ek constant ka fark hai.Chapter mein pehli baar yahan aaya →

latus rectumनाभिलंबnaabhilamb: Focus se guzarti woh chord jo conic ke axis ke perpendicular ho.Chapter mein pehli baar yahan aaya →

major axisदीर्घ अक्षdeergh aksh: Ellipse ki sabse lambi chord, jo dono foci se guzarti hai; minor axis centre se guzarti woh chord hai jo iske perpendicular hai.Chapter mein pehli baar yahan aaya →

nappe: Double cone ke do halves mein se koi bhi ek, jo vertex par milte hain.Chapter mein pehli baar yahan aaya →

parabolaपरवलयparvalay: Plane ke un points ka set jo ek fixed point (focus) aur ek fixed line (directrix) se barabar door hain.Chapter mein pehli baar yahan aaya →

transverse axisअनुप्रस्थ अक्षanuprasth aksh: Hyperbola ka foci se guzarta axis; conjugate axis centre se guzarta hai aur iske perpendicular hai.Chapter mein pehli baar yahan aaya →

vertexशीर्षsheersh: Woh point jahan conic focus ya foci se guzarte apne axis se milta hai.Chapter mein pehli baar yahan aaya →

10.1 Introduction

10.2 Cone ke Sections

10.2.1 Circle, ellipse, parabola aur hyperbola

10.2.2 Degenerated conic sections

10.3 Circle

10.4 Parabola

10.4.1 Parabola ki standard equations

10.4.2 Latus rectum

10.5 Ellipse

10.5.1 a, b aur c aapas mein kaise jude hain

10.5.2 Eccentricity

10.5.3 Ellipse ki standard equations

10.5.4 Latus rectum

10.6 Hyperbola

10.6.1 Eccentricity

10.6.2 Hyperbola ki standard equation

10.6.3 Latus rectum

Miscellaneous Examples

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye