Conic Sections: Class 11 Maths · Chapter 10

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
3 simulations
44 practice questions
23 MCQs

What you will learn in Conic Sections

Describe how a plane cutting a double cone gives a circle, an ellipse, a parabola, a hyperbola or a degenerate conic, depending on its angle.
Write the equation of a circle from its centre and radius, and find the centre and radius from an equation by completing squares.
State the focus–directrix definition of a parabola and the two-focus definitions of the ellipse and hyperbola, and derive their standard equations.
Read the focus, vertices, axes, eccentricity and latus rectum from a standard equation, and build the equation from such data.
Use a² = b² + c² for the ellipse, c² = a² + b² for the hyperbola and e = c/a to move between the parameters.
Model reflectors, arches, cables and sliding rods with conics and solve the resulting problems.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 10, pp. 176–207. Not affiliated with or endorsed by NCERT.

10.1 Introduction

Point a torch straight down at the floor and the patch of light is a round disc. Tilt the torch a little and the disc stretches into an oval. Tilt it further and the far end of the patch races away across the floor and never closes. The beam is a cone of light and the floor is a flat plane, so every outline you saw is a curve cut from a cone by a plane.

2/3Tilted: the patch stretches into an oval.

Fig. 10.1Tilting a torch turns its circle of light into an ellipse, then an open parabola.

Top: a side view of a torch above a level floor, its beam drawn as a pale cone with dashed edges and the lit part of the floor marked in green. Bottom: the same lit patch seen from above. Step 1: the torch points straight down and the patch is a round circle. Step 2: the torch is tilted and the patch stretches into an ellipse. Step 3: the torch is tilted so far that the upper edge of the beam runs parallel to the floor, and the patch never closes: its outline is a parabola.

In Chapter 9 you described straight lines by equations. This chapter does the same for four curves: the circle, the ellipse, the parabola and the hyperbola. Because all four come from slicing a cone, they are called conic sections (शंकु परिच्छेद), or simply conics.

These curves are everywhere once you look. Planets and comets travel on ellipses. A ball thrown across a field follows a parabola (if we ignore air resistance), and parabolic shapes collect signals in a DTH dish and focus the beam of a car headlamp. The waist-shaped sides of many power-station cooling towers are hyperbolas, and hyperbolas also helped ships fix their position by radio before satellite navigation.

Quick check

1 of 2

A torch points straight down at a level floor. The edge of its patch of light is

an ellipsea circlea parabolaa hyperbola

Show answer

Answer (B) a circle

With the beam square to the floor, every edge ray travels the same distance, so the patch is round. Tilting the torch stretches it into an ellipse and then opens it into a parabola.

10.2 Sections of a Cone

A double-napped right circular cone is two hollow cones joined tip to tip, each running on without end (Fig. 10.2). Two lines are enough to build one. Keep a vertical line $l$ fixed. Through a point $V$ on it, draw a line $m$ at an acute angle $α$ to $l$ . Spin $m$ about $l$ like a spoke turning round an axle, holding $α$ steady: the positions of $m$ together make up the surface.

3/3The surface swept out: two nappes meeting at the vertex $V$ .

Fig. 10.2Turning the line

m

about the axis

l

sweeps out a double cone.NCERT Figs 10.1, 10.2 · p. 176

A vertical dashed axis l with a slanted line m crossing it at the point V; the angle alpha between them is marked. Step 1 shows only the two lines. Step 2 adds faint copies of m at several positions and dashed rings at the top and bottom, with an arrow showing m turning about l while alpha stays fixed. Step 3 shades the surface swept out: two cones meeting at V, labelled upper nappe and lower nappe, with V marked as the vertex, l as the axis and m as a generator.

Some names: the meeting point $V$ is the vertex, the fixed line $l$ is the axis, and $m$ in any of its positions is a generator. Above and below $V$ lie the two halves of the surface, the upper and lower nappes.

Now bring in a flat plane. Wherever it slices the cone, the edge of the slice is a conic section, and only two things decide its shape: the angle $β$ the plane makes with the axis (measured against the cone's angle $α$ ), and whether or not the plane goes through $V$ .

10.2.1 Circle, ellipse, parabola and hyperbola

First keep the plane away from the vertex. Start with the plane square to the axis and tip it a little at a time:

$β = 90^{\circ}$ : the plane is at right angles to the axis and the edge of the cut is a circle.
$α < β < 90^{\circ}$ : the plane is tilted but still slices right across one nappe, so the cut is a closed oval, an ellipse.
$β = α$ : the plane is parallel to one generator, so on that side it never meets the cone again and the curve stays open; this is a parabola.
$0 \leq β < α$ : the plane is steeper than the side of the cone and cuts both nappes, giving the two separate branches of a hyperbola.

4/7Steeper than the cone: both nappes are cut, a hyperbola.

Fig. 10.3The angle of the cutting plane decides which conic you get.NCERT Figs 10.4–10.10 · p. 177

Left: a double cone seen from the side, with its axis and the angle alpha marked at the vertex V. A dashed line shows the plane edge-on and a thick green stroke shows where it meets the cone. Right: a panel shows the true shape of the cut, its name and the condition on beta. Step 1: a horizontal plane gives a circle. Step 2: a tilted plane across one nappe gives an ellipse. Step 3: a plane parallel to a generator gives an open parabola. Step 4: a vertical plane cuts both nappes and gives a two-branched hyperbola. Steps 5 to 7 put the plane through V: a horizontal plane meets the cone only at V (a point), a plane along a generator gives one line, and a vertical plane through the axis gives two crossing lines.

See the pattern: the more steeply the plane cuts, the more the curve opens up. The circle and the ellipse are closed and lie on one nappe; the parabola lies on one nappe but is open; the hyperbola is open and needs both nappes.

10.2.2 Degenerated conic sections

Now let the plane pass through the vertex $V$ . The same three ranges of $β$ give squashed versions of the conics, called degenerate conics (steps 5 to 7 of Fig. 10.3):

$α < β \leq 90^{\circ}$ : the plane meets the cone only at $V$ , so the section is a single point.
$β = α$ : the plane lies along one generator, and the section is that straight line. This is the degenerate parabola.
$0 \leq β < α$ : the plane passes through both nappes along two generators, giving a pair of intersecting lines. This is the degenerate hyperbola.

From here on we leave the cone aside. Each curve can be described by a simple rule about distances in a plane, and those rules lead straight to equations.

Quick check

1 of 3

The plane does not pass through the vertex. Match the angle

β

with the curve.

Tap an item on the left, then its match on the right.

Show answer

Answer

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

Tipping the plane away from square opens the curve up: circle, then ellipse, then parabola when the plane is parallel to a generator, then hyperbola once it is steep enough to reach both nappes.

10.3 Circle

Definition

Circle

Fix a point $C$ in a plane and a positive number $r$ . The circle with centre $C$ and radius $r$ is made of all the points of that plane at distance $r$ from $C$ .

To turn the rule into an equation, put the centre at $C (h, k)$ , call the radius $r$ , and let $P (x, y)$ be any point of the plane. $P$ is on the circle exactly when $C P = r$ . The distance formula gives $C P$ from the two legs of a right triangle, and squaring removes the square root (Fig. 10.4).

2/3The legs are $x - h$ across and $y - k$ up.

Fig. 10.4Pythagoras in the triangle under the radius gives the equation of the circle.NCERT Fig. 10.12 · p. 180

Coordinate axes with a circle whose centre C(h, k) is in the first quadrant. Step 1 joins C to a point P(x, y) on the circle with a radius r and states CP = r. Step 2 drops a right triangle under the radius: a horizontal leg of length x minus h and a vertical leg of length y minus k, and the panel gives (x minus h) squared plus (y minus k) squared equals r squared. Step 3 adds three more points on the circle, each joined to C by a radius of the same length r, to show that every point of the circle obeys the same equation.

Circle: centre (h, k), radius r

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

(10.1)

Every point of the circle satisfies (10.1), and every point that satisfies it is at distance $r$ from $C$ , so the equation describes the circle exactly. With the centre at the origin it becomes $x^{2} + y^{2} = r^{2} .$

Expanding (10.1) gives $x^{2} + y^{2} - 2 h x - 2 k y + (h^{2} + k^{2} - r^{2}) = 0$ . So the equation of a circle has equal coefficients of $x^{2}$ and $y^{2}$ and no $x y$ term. Going the other way, an equation of the form below is read by completing squares:

General form

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

(10.2)

where

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

If $g^{2} + f^{2} - c = 0$ the “circle” shrinks to the single point $(- g, - f)$ , and if it is negative no real point satisfies the equation at all. A coefficient other than 1 on $x^{2}$ and $y^{2}$ must be divided out first.

SimulationPoints at distance 5 from C

A coordinate grid from minus 8 to 8 on both axes. A fixed point C sits at (1, minus 2) and a draggable point P starts at (4, 2); the segment CP is drawn with its length shown, and P snaps to grid points as you move it, showing its coordinates.

Fig. 10.5Drag

P

and watch the distance

C P

. The circle is every point where it reads 5.

Try this

$P$ starts at $(4, 2)$ with $C P = 5$ . Find all twelve grid points with $C P = 5$ .
For each point you find, check that $(x - 1)^{2} + (y + 2)^{2} = 25$ .
Put $P$ at $(3, 0)$ . Is $C P$ more or less than 5? Is $P$ inside or outside the circle?

Example 10.1NCERT Example 1 · p. 180

Centre at the origin

A circle centred at the origin passes through $(5, - 12)$ . Write its equation.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.2NCERT Example 2 · p. 180

Centre and radius given

Write the equation of the circle whose centre is $(2, - 5)$ and whose radius is 3, then expand it.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.3NCERT Example 3 · p. 180

Reading a circle by completing squares

Find the centre and radius of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10.4NCERT Example 4 · p. 181

Two points and a line for the centre

A circle goes through $A (1, 3)$ and $B (5, 1)$ , and its centre is on the line $x - 2 y = 2$ . Find the circle.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

What is the radius of the circle

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

Show answer

Answer 5

Complete the squares:

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

, so

r = 5

. Answering 15 forgets the 1 and 9 added while completing the squares.

Practice · 9 questions

Practice 10.1: questions with answers

1
Short answerEasyNCERT Exercise 10.1 Q1–Q4
Write the equation of each circle from its centre and radius, in expanded form.
1. Centre $(0, - 3)$ , radius 3
  Show solution
  Answer $x^{2} + y^{2} + 6 y = 0$
  $x^{2} + (y + 3)^{2} = 9$
  $x^{2} + y^{2} + 6 y = 0$
2. Centre $(4, - 1)$ , radius 5
  Show solution
  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 $(\frac{1}{3}, - \frac{1}{2})$ , radius $\frac{1}{6}$
  Show solution
  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: $x^{2} + y^{2} - \frac{2}{3} x + y + \frac{1}{9} + \frac{1}{4} - \frac{1}{36} = 0$ , and $\frac{1}{9} + \frac{1}{4} - \frac{1}{36} = \frac{1}{3} .$ Multiply by 3:
  $3 x^{2} + 3 y^{2} - 2 x + 3 y + 1 = 0$
4. Centre $(- 2, - 2)$ , radius $\sqrt{2}$
  Show solution
  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
Find the equation of the circle with centre $(p, - q)$ and radius $\sqrt{p^{2} + q^{2}}$ . Which special point always lies on it?
Show solution
Answer $x^{2} + y^{2} - 2 p x + 2 q y = 0$ ; it passes through the origin.
$(x - p)^{2} + (y + q)^{2} = p^{2} + q^{2}$
$x^{2} + y^{2} - 2 p x + 2 q y = 0$
There is no constant term, so $(0, 0)$ satisfies the equation: the circle passes through the origin. This makes sense, because the radius equals the distance from the centre to the origin.
3
Short answerEasyNCERT Exercise 10.1 Q6–Q9
Find the centre and radius of each circle.
1. $(x - 7)^{2} + (y + 2)^{2} = 20$
  Show solution
  AnswerCentre $(7, - 2)$ , radius $2 \sqrt{5}$
  Read it off (10.1): $h = 7$ , $k = - 2$ , $r = \sqrt{20} = 2 \sqrt{5} .$
2. $x^{2} + y^{2} + 6 x - 10 y - 2 = 0$
  Show solution
  AnswerCentre $(- 3, 5)$ , radius 6
  Move the constant across and add 9 and 25: $2 + 9 + 25 = 36$ .
  $(x + 3)^{2} + (y - 5)^{2} = 36$
3. $x^{2} + y^{2} + 2 x - 12 y + 12 = 0$
  Show solution
  AnswerCentre $(- 1, 6)$ , radius 5
  Add 1 and 36 to $- 12$ : the right side becomes 25.
  $(x + 1)^{2} + (y - 6)^{2} = 25$
4. $3 x^{2} + 3 y^{2} + 2 y = 0$
  Show solution
  AnswerCentre $(0, - \frac{1}{3})$ , radius $\frac{1}{3}$
  Divide by 3: $x^{2} + y^{2} + \frac{2}{3} y = 0$ .
  $x^{2} + (y + \frac{1}{3})^{2} = \frac{1}{9}$
4
Long answerMediumNCERT Exercise 10.1 Q10–Q11
Find the circle through the two given points whose centre lies on the given line.
1. Through $(3, 1)$ and $(- 1, 3)$ ; centre on $3 x - y = 2$
  Show solution
  Answer $(x - 2)^{2} + (y - 4)^{2} = 10$
  $\begin{aligned} (3 - h)^{2} + (1 - k)^{2} \\ = (1 + h)^{2} + (3 - k)^{2} \end{aligned}$
  This simplifies to $k = 2 h$ . With $3 h - k = 2$ : $h = 2$ , $k = 4$ , and $r^{2} = 1^{2} + 3^{2} = 10.$
2. Through $(5, 2)$ and $(- 1, 4)$ ; centre on $x + y = 9$
  Show solution
  Answer $(x - 3)^{2} + (y - 6)^{2} = 20$
  $\begin{aligned} (5 - h)^{2} + (2 - k)^{2} \\ = (1 + h)^{2} + (4 - k)^{2} \end{aligned}$
  This simplifies to $k = 3 h - 3$ . With $h + k = 9$ : $h = 3$ , $k = 6$ , and $r^{2} = 2^{2} + 4^{2} = 20.$
Short answerMediumNCERT Exercise 10.1 Q12
A circle of radius 13 has its centre on the $y$ -axis and passes through $(5, 1)$ . Find every such circle.
Show solution
Answer $x^{2} + (y - 13)^{2} = 169$ or $x^{2} + (y + 11)^{2} = 169$
$25 + (1 - k)^{2} = 169$
So $(1 - k)^{2} = 144$ , giving $k = 13$ or $k = - 11$ . Both circles work: $x^{2} + (y - 13)^{2} = 169$ and $x^{2} + (y + 11)^{2} = 169.$
6
Short answerMediumNCERT Exercise 10.1 Q13
A circle passes through the origin and cuts the axes again at $A$ and $B$ .
1. Find it when $A = (6, 0)$ and $B = (0, - 4)$ .
  Show solution
  Answer $x^{2} + y^{2} - 6 x + 4 y = 0$
  Through $(0, 0)$ : $c = 0$ . Through $(6, 0)$ : $36 + 12 g = 0$ , so $g = - 3$ . Through $(0, - 4)$ : $16 - 8 f = 0$ , so $f = 2$ .
  $x^{2} + y^{2} - 6 x + 4 y = 0$
2. Find it in general when $A = (p, 0)$ and $B = (0, q)$ .
  Show solution
  Answer $x^{2} + y^{2} - p x - q y = 0$
  The same steps give $c = 0$ , $2 g = - p$ and $2 f = - q$ , so the circle is $x^{2} + y^{2} - p x - q y = 0.$ The angle $A O B$ is a right angle, so $A B$ is a diameter.
Short answerEasyNCERT Exercise 10.1 Q14
A circle centred at $(- 1, 3)$ goes through $(2, - 1)$ . Write its equation.
Show solution
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
Where does the point $(3.5, - 2.5)$ lie compared with the circle $x^{2} + y^{2} = 16$ ?
insideon the circleoutside
Show answer
Answer (C) outside
${3.5}^{2} + {2.5}^{2} = 12.25 + 6.25 = 18.5 > 16$ , so the point is farther than 4 from the centre: outside.
Quick answerMedium
The circle $x^{2} + y^{2} - 8 x + 6 y + k = 0$ has radius 6. Find $k$ .
Show answer
Answer -11
Completing squares, $(x - 4)^{2} + (y + 3)^{2} = 25 - k$ . Setting $25 - k = 36$ gives $k = - 11$ .

10.4 Parabola

A cricket ball lofted towards the boundary, a jet of water from a garden hose, the cable of a suspension bridge: each follows (very nearly) the same kind of curve. Its geometric description needs one point and one line.

Definition

Parabola

Take a line $l$ and a point $F$ off it, both in one plane. A parabola is made of the points of that plane whose distance from $F$ equals their distance from $l$ . The line $l$ is the directrix and $F$ is the focus.

So a point $P$ is on the parabola when its distance $P F$ from the focus $F$ equals its distance $P M$ from the directrix, where $M$ is the foot of the perpendicular from $P$ (Fig. 10.6).

4/4They lie on a parabola; its axis passes through $F$ .

Fig. 10.6Each point of a parabola is as far from the focus as from the directrix.NCERT Figs 10.13, 10.14 · p. 182

A horizontal dashed directrix near the bottom and a focus F above its middle. Step 1 shows just the focus and the directrix. Step 2 adds a point P with a solid segment to F and a dashed vertical segment down to M on the directrix, the two being equal: PF = PM. Step 3 adds three more points, each with its own pair of equal segments. Step 4 draws the parabola through all of them, with the vertical axis through F and the vertex halfway between the focus and the directrix.

The line through the focus perpendicular to the directrix is the axis of the parabola, and the curve is symmetric about it. The axis meets the parabola at a single point, the vertex, which lies exactly halfway between the focus and the directrix.

Try this

Fold a parabola

You’ll need

a sheet of thin paper (tracing or butter paper)
a pen
a ruler

Predict first

You will fold many points of a line onto one dot and crease each time. What will the creases outline?

a circle round the dota curve that opens away from the linea straight line

Draw a straight line about 3 cm above the bottom edge of the paper.
Mark a dot $F$ about 4 cm above the middle of the line.
Fold the paper so that some point of the line lands exactly on $F$ . Crease firmly and unfold.
Repeat for 20 to 30 different points spread along the line.

10.4.1 Standard equations of parabola

The simplest equations come from putting the vertex at the origin and lining the axis up with a coordinate axis. That still leaves four ways for the parabola to face: right, left, up or down. We derive the first and get the others by symmetry.

DerivationA parabola opening to the right

Take the focus $F (a, 0)$ with $a > 0$ and the directrix $x = - a$ ; the vertex, halfway between them, is then the origin. For any point $P (x, y)$ , the foot of the perpendicular to the directrix is $M (- a, y)$ , so $P M = ∣ x + a ∣$ and $P F = \sqrt{(x - a)^{2} + y^{2}}$ .

$P$ is on the parabola exactly when $P F = P M$ . Both sides are non-negative, so we may square without gaining or losing points: $P F^{2} = P M^{2}$ . Moving $(x - a)^{2}$ across leaves $y^{2}$ on its own:

\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)

Conversely, if $y^{2} = 4 a x$ then $x \geq 0$ and $P F^{2} = (x - a)^{2} + 4 a x = (x + a)^{2}$ , so $P F = x + a = P M$ . Every point of the parabola satisfies (10.3), and every solution of (10.3) is on the parabola.

No point of $y^{2} = 4 a x$ has a negative $x$ , because the left side can never be negative while $4 a$ is positive. To the right there is no limit: $y = \pm 2 \sqrt{a x}$ keeps growing, so the arms open out through quadrants I and IV for ever, mirror images in the positive $x$ -axis.

1/4 $y^{2} = 4 a x$ opens to the right.

Fig. 10.7Reflecting the first parabola gives the other three standard forms.NCERT Fig. 10.15 · p. 182

One set of axes. Step 1: the parabola y squared equals 4ax opening to the right, with focus F(a, 0) and the vertical directrix x = minus a. Step 2: y squared equals minus 4ax opening to the left, focus F(minus a, 0), directrix x = a. Step 3: x squared equals 4ay opening upwards, focus F(0, a), horizontal directrix y = minus a. Step 4: x squared equals minus 4ay opening downwards, focus F(0, minus a), directrix y = a. In every step the vertex is at the origin.

Replacing $x$ by $- x$ reflects the curve in the $y$ -axis; exchanging $x$ and $y$ reflects it in the line $y = x$ . These moves turn (10.3) into the other three standard forms (Table 10.1).

Table 10.1The four standard parabolas (

a > 0

, vertex at the origin)

Equation	Focus	Directrix	Opens
$y^{2} = 4 a x$	$(a, 0)$	$x = - a$	right
$y^{2} = - 4 a x$	$(- a, 0)$	$x = a$	left
$x^{2} = 4 a y$	$(0, a)$	$y = - a$	up
$x^{2} = - 4 a y$	$(0, - a)$	$y = a$	down

10.4.2 Latus rectum

Definition

Latus rectum

The latus rectum of a conic is the chord through a focus drawn perpendicular to the axis (for an ellipse, the major axis; for a hyperbola, the transverse axis), with both ends on the curve.

For $y^{2} = 4 a x$ the latus rectum lies on the line $x = a$ . Putting $x = a$ gives $y^{2} = 4 a^{2}$ , so its ends are $L (a, 2 a)$ and $L^{'} (a, - 2 a)$ . There is also a proof by picture (Fig. 10.8): $L$ is on the parabola, so $L F$ equals the distance from $L$ to the directrix, which is the gap $2 a$ between the lines $x = - a$ and $x = a$ .

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

Fig. 10.8Each half of the latus rectum is as long as the gap

2 a

from the focus to the directrix.NCERT Fig. 10.18 · p. 185

A parabola opening to the right with vertex O, focus F and a vertical dashed directrix. Step 1 draws the chord LL prime through F perpendicular to the axis. Step 2 adds a dashed horizontal segment from L to the directrix and marks it 2a; the segment LF is also 2a, since L is on the parabola. Step 3 marks L prime F as 2a as well and labels the whole chord LL prime as 4a.

Latus rectum of y² = 4ax

L L^{'} = 4 a

(10.4)

Example 10.5NCERT Example 5 · p. 185

Reading a parabola

For the parabola $x^{2} = - 12 y$ , name its axis and find its focus, its directrix and how long its latus rectum is.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.6NCERT Example 6 · p. 186

Focus and directrix given

Write the equation of the parabola whose focus is $(0, 4)$ and whose directrix is $y = - 4$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10.7NCERT Example 7 · p. 186

Vertex and focus given

A parabola has its vertex at $(0, 0)$ and its focus at $(- \frac{5}{2}, 0)$ . Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.8NCERT Example 8 · p. 186

Through a given point

A parabola has its vertex at the origin, its axis along the $x$ -axis, and passes through $(3, - 6)$ . Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The focus of the parabola

x^{2} = - 8 y

(- 2, 0)

(0, 2)

(0, - 2)

(0, - 8)

Show answer

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

Here

4 a = 8

, so

a = 2

; the parabola opens downwards, so the focus is

(0, - 2)

. Choosing

(0, - 8)

uses

4 a

instead of

a

Practice · 6 questions

Practice 10.2: questions with answers

1
Short answerEasyNCERT Exercise 10.2 Q1–Q6
For each parabola, give the axis, the focus, the directrix and how long the latus rectum is.
1. $y^{2} = 16 x$
  Show solution
  AnswerFocus $(4, 0)$ ; $x$ -axis; $x = - 4$ ; 16
  $4 a = 16$ , $a = 4$ ; opens right. Focus $(4, 0)$ , axis the $x$ -axis, directrix $x = - 4$ , latus rectum 16.
2. $x^{2} = - 10 y$
  Show solution
  AnswerFocus $(0, - \frac{5}{2})$ ; $y$ -axis; $y = \frac{5}{2}$ ; 10
  $4 a = 10$ , $a = \frac{5}{2}$ ; opens down. Focus $(0, - \frac{5}{2})$ , axis the $y$ -axis, directrix $y = \frac{5}{2}$ , latus rectum 10.
3. $y^{2} = - 6 x$
  Show solution
  AnswerFocus $(- \frac{3}{2}, 0)$ ; $x$ -axis; $x = \frac{3}{2}$ ; 6
  $4 a = 6$ , $a = \frac{3}{2}$ ; opens left. Focus $(- \frac{3}{2}, 0)$ , axis the $x$ -axis, directrix $x = \frac{3}{2}$ , latus rectum 6.
4. $x^{2} = 20 y$
  Show solution
  AnswerFocus $(0, 5)$ ; $y$ -axis; $y = - 5$ ; 20
  $4 a = 20$ , $a = 5$ ; opens up. Focus $(0, 5)$ , axis the $y$ -axis, directrix $y = - 5$ , latus rectum 20.
2
Short answerEasyNCERT Exercise 10.2 Q7–Q8
Find the equation of the parabola with the given focus and directrix.
1. Focus $(0, 7)$ ; directrix $y = - 7$
  Show solution
  Answer $x^{2} = 28 y$
  The vertex is the origin and the focus is above it: $x^{2} = 4 a y$ with $a = 7$ , so $x^{2} = 28 y .$
2. Focus $(- 4, 0)$ ; directrix $x = 4$
  Show solution
  Answer $y^{2} = - 16 x$
  The vertex is the origin and the focus is to its left: $y^{2} = - 4 a x$ with $a = 4$ , so $y^{2} = - 16 x .$
3
Short answerEasyNCERT Exercise 10.2 Q9–Q10
The vertex is at the origin. Use the given focus to write the equation of the parabola.
1. Focus $(0, - 1)$
  Show solution
  Answer $x^{2} = - 4 y$
  Opens down with $a = 1$ : $x^{2} = - 4 y .$
2. Focus $(\frac{7}{2}, 0)$
  Show solution
  Answer $y^{2} = 14 x$
  Opens right with $a = \frac{7}{2}$ : $y^{2} = 4 \cdot \frac{7}{2} x = 14 x .$
4
Short answerMediumNCERT Exercise 10.2 Q11–Q12
The vertex is at the origin. Find the parabola from the given axis and point.
1. Axis along the $x$ -axis; passes through $(- 3, 6)$
  Show solution
  Answer $y^{2} = - 12 x$
  $36 = k (- 3) \Rightarrow k = - 12$
  So $y^{2} = - 12 x$ , which opens to the left, as a point with negative $x$ requires.
2. Symmetric about the $y$ -axis; passes through $(6, - 4)$
  Show solution
  Answer $x^{2} = - 9 y$
  $36 = k (- 4) \Rightarrow k = - 9$
  So $x^{2} = - 9 y$ , opening downwards.
Quick answerMedium
What is the $x$ -coordinate of the focus of the parabola $3 y^{2} = 8 x$ ?
Show answer
Answer 0.6667
Divide by 3: $y^{2} = \frac{8}{3} x$ , so $4 a = \frac{8}{3}$ and $a = \frac{2}{3} \approx 0.667$ . The focus is $(\frac{2}{3}, 0)$ .
Quick answerMedium
The parabola $y^{2} = 4 a x$ passes through $(2, - 4)$ . The length of its latus rectum is
24816
Show answer
Answer (C) 8
$16 = 8 a$ gives $a = 2$ , so the latus rectum is $4 a = 8$ . The option 2 is $a$ itself.

10.5 Ellipse

Every planet moves round the Sun on an ellipse, with the Sun not at the centre but at a special point called a focus. That is Kepler's first law, published in 1609. For the Earth the ellipse is so nearly round that you could not tell it from a circle by eye.

Definition

Ellipse

Take two fixed points $F_{1}$ and $F_{2}$ in a plane. An ellipse is made of the points $P$ of that plane for which the sum $P F_{1} + P F_{2}$ has one and the same value. $F_{1}$ and $F_{2}$ are its foci (singular: focus).

Try this

The playground ellipse

You’ll need

a loop of string about 3 m round
a piece of chalk
two friends

Predict first

After drawing once, the two friends stand closer together and you draw again. What happens to the shape?

It gets rounderIt gets flatterIt does not change

Two friends stand still about 1 m apart, each holding the loop down at their feet.
Put the chalk inside the loop and pull outwards until the string is tight, making a triangle.
Keeping the string tight, walk the chalk all the way round on the ground.
Draw again with the friends 0.5 m apart, then 1.3 m apart.

Can’t do it now? Try the simulation · Fig. 10.12

1/4Here $P F_{1}$ is long and $P F_{2}$ short; their total is $2 a$ .

Fig. 10.9However

P

moves,

P F_{1} + P F_{2}

stays the same: that is what makes the curve an ellipse.NCERT Figs 10.20, 10.21 · p. 187

Two foci F1 and F2 on a horizontal line. A point P is joined to F1 by a solid segment and to F2 by a dashed one. Below the drawing a bar is split into two pieces, PF1 and PF2, laid end to end. Steps 1 to 3 move P to three positions around the curve, tracing more of it each time, the last at the bottom end of the minor axis where the two pieces are equal; the split in the bar moves but its total length never changes. Step 4 draws the whole ellipse with the centre O, the two vertices, the major axis through the foci and the minor axis through O, and notes that the bar's length 2a equals the length of the major axis.

A few points and lengths describe any ellipse. Halfway between the foci is the centre. The longest chord, the major axis, runs through both foci, and its ends are the two vertices. Square to it through the centre runs the shorter minor axis. Their lengths are called $2 a$ and $2 b$ , and the foci are $2 c$ apart; so the semi-axes are $a$ and $b$ , and each focus sits $c$ from the centre.

10.5.1 How a, b and c are related

First, the constant sum is the length of the major axis. At the vertex $A$ beyond $F_{2}$ , $A F_{2} = a - c$ and $A F_{1} = a + c$ , which add to $2 a$ . The sum is the same at every point, so $P F_{1} + P F_{2} = 2 a$ for every point $P$ of the ellipse.

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

Fig. 10.10The vertices, foci, latus rectum and bounding box of

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

.NCERT Figs 10.22, 10.23 · p. 188

An ellipse centred at the origin with its major axis along the x-axis. Step 1 marks the vertices (a, 0) and (minus a, 0) and the ends of the minor axis (0, b) and (0, minus b). Step 2 adds the foci F1 and F2 and joins the top point B to both; each of these segments is a, and the right triangle with legs b and c and hypotenuse a gives a squared equals b squared plus c squared. Step 3 draws the latus rectum through F2, made of two halves of length b squared over a. Step 4 shades the rectangle between the lines x = plus or minus a and y = plus or minus b, which the ellipse touches but never leaves.

Next, take an end $B$ of the minor axis. It lies on the perpendicular bisector of $F_{1} F_{2}$ , so $B F_{1} = B F_{2}$ , and since the two add to $2 a$ , each equals $a$ . The right triangle formed by $B$ , the centre and $F_{2}$ has legs $b$ and $c$ and hypotenuse $a$ (step 2 of Fig. 10.10):

Ellipse

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

(10.5)

In particular $c = \sqrt{a^{2} - b^{2}}$ and $b < a$ : the major axis really is the longer one.

10.5.2 Eccentricity

Definition

Eccentricity

The eccentricity of an ellipse or a hyperbola is $e = \frac{c}{a}$ : the centre-to-focus distance divided by the centre-to-vertex distance.

For an ellipse $0 \leq c < a$ , so $0 \leq e < 1$ , and each focus is $c = a e$ from the centre. The eccentricity measures how far the ellipse is from round. Keep $a$ fixed and increase $e$ : the foci slide out towards the vertices and the ellipse flattens, because $b = a \sqrt{1 - e^{2}}$ (Fig. 10.11). At $e = 0$ both foci sit at the centre and the ellipse is a circle of radius $a$ .

3/4Flatter: $b = 0.6 a$ .

Fig. 10.11Same

a

, growing

e

: the foci move out and the ellipse flattens.NCERT 10.5.2 · p. 188

One ellipse with a fixed semi-major axis a drawn on axes, with both foci marked and a side panel giving e and b. Step 1: e = 0, the foci are together at the centre and the curve is a circle with b = a. Step 2: e = 0.6, the foci move apart and b = 0.8a. Step 3: e = 0.8, b = 0.6a and the ellipse is flatter. Step 4: e = 0.95, the foci are close to the vertices and b is about 0.31a, a very flat ellipse.

10.5.3 Standard equations of an ellipse

Put the centre at the origin with the foci $F_{1} (- c, 0)$ and $F_{2} (c, 0)$ on the $x$ -axis. A point $P (x, y)$ is on the ellipse exactly when $P F_{1} + P F_{2} = 2 a$ , where $a > c$ .

DerivationThe ellipse equation from the sum rule

Instead of squaring a sum of two square roots, look at the difference of their squares. The $y^{2}$ terms cancel:

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

Divide by $P F_{1} + P F_{2} = 2 a$ to get $P F_{1} - P F_{2} = \frac{2 c x}{a}$ . Adding this to $P F_{1} + P F_{2} = 2 a$ , and subtracting it, gives each distance on its own:

Focal distances

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

(10.6)

Now square the second, using $P F_{2}^{2} = (x - c)^{2} + y^{2}$ ; the terms $- 2 c x$ on the two sides cancel:

\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}

Write $b^{2} = a^{2} - c^{2}$ , which is positive because $a > c$ , and divide by $b^{2}$ :

Standard ellipse (a > b)

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

(10.7)

Conversely, let $(x, y)$ satisfy (10.7). Then $y^{2} = b^{2} (1 - \frac{x^{2}}{a^{2}})$ , and expanding gives $P F_{1}^{2} = (x + c)^{2} + y^{2} = (a + \frac{c}{a} x)^{2} .$ Since $∣ x ∣ \leq a$ and $c < a$ , the bracket is positive, so $P F_{1} = a + \frac{c}{a} x$ . In the same way $P F_{2} = a - \frac{c}{a} x$ . The two add to $2 a$ , so the point is on the ellipse.

How far does the ellipse reach? The two terms on the left of (10.7) are never negative and they add to 1, so neither can exceed 1. Hence $x^{2} \leq a^{2}$ and $y^{2} \leq b^{2}$ : the curve is boxed in by the lines $x = \pm a$ and $y = \pm b$ , touching each one at a vertex or at an end of the minor axis (step 4 of Fig. 10.10).

Swap the roles of $x$ and $y$ and you get the upright ellipse, with foci $(0, \pm c)$ : $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1.$ This pair is what we mean by the standard equations of an ellipse; a shifted or tilted ellipse needs a longer equation, met in later classes.

SimulationHunting for the ellipse

A coordinate grid from minus 6 to 6 on both axes with two fixed foci, F1 at (minus 4, 0) and F2 at (4, 0), and a draggable point P that starts at (2, 2). The distances from P to each focus are drawn and measured, and P snaps to steps of 0.1 and shows its coordinates.

Fig. 10.12Drag

P

and add the two distances. The ellipse

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

is every point where they total 10.

Try this

Move $P$ to $(5, 0)$ and then to $(0, 3)$ . Check that $P F_{1} + P F_{2} = 10$ at both.
Try $(4, 1.8)$ , an end of a latus rectum, and $(- 3, 2.4)$ . Is the total still 10?
Put $P$ at $(3, 2)$ . Is the total more or less than 10? Is $P$ inside or outside the ellipse?

10.5.4 Latus rectum

The latus rectum through $F_{2} (c, 0)$ lies on the line $x = c$ . Substituting in (10.7) and using $a^{2} - c^{2} = b^{2}$ :

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

so $y = \pm \frac{b^{2}}{a}$ and the ends are $(c, \pm \frac{b^{2}}{a})$ . The latus rectum is twice this height:

Ellipse and hyperbola

latus rectum = \frac{2 b^{2}}{a}

(10.8)

Example 10.9NCERT Example 9 · p. 192

Everything about one ellipse

For the ellipse $\frac{x^{2}}{289} + \frac{y^{2}}{64} = 1$ , list its foci and vertices, the lengths of both axes, its eccentricity and its latus rectum.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.10NCERT Example 10 · p. 193

An ellipse standing upright

Describe the ellipse $4 x^{2} + y^{2} = 16$ fully: foci, vertices, axis lengths, eccentricity and latus rectum.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.11NCERT Example 11 · p. 193

From vertices and foci

Find the ellipse with vertices $(0, \pm 15)$ and foci $(0, \pm 9)$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10.12NCERT Example 12 · p. 194

From the major axis and the foci

The major axis of an ellipse has length 14 and its foci are $(\pm 2 \sqrt{6}, 0)$ . Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.13NCERT Example 13 · p. 194

Through two points

An ellipse in standard form has its major axis along the $x$ -axis and passes through $(2, 2)$ and $(- 3, 1)$ . Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The foci of the ellipse

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

are

(\pm 3, 0)

(0, \pm \sqrt{41})

(\pm 4, 0)

(0, \pm 3)

Show answer

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

The larger denominator, 25, is under

y^{2}

, so the major axis is the

y

-axis,

a = 5

b = 4

and

c = \sqrt{25 - 16} = 3

: foci

(0, \pm 3)

. Adding the denominators gives

\sqrt{41}

, the hyperbola rule, which is wrong here.

Practice · 9 questions

Practice 10.3: questions with answers

1
Long answerEasyNCERT Exercise 10.3 Q1–Q6
For each ellipse find the foci, the vertices, the lengths of the major and minor axes, the eccentricity and the latus rectum.
1. $\frac{x^{2}}{49} + \frac{y^{2}}{24} = 1$
  Show solution
  AnswerFoci $(\pm 5, 0)$ , vertices $(\pm 7, 0)$ ; axes 14 and $4 \sqrt{6}$ ; $e = \frac{5}{7}$ ; latus rectum $\frac{48}{7}$
  $a = 7$ , $b = 2 \sqrt{6}$ , along the $x$ -axis.
  $c = \sqrt{49 - 24} = 5$
  So $e = \frac{5}{7}$ and the latus rectum is $\frac{2 \cdot 24}{7} = \frac{48}{7} .$
2. $\frac{x^{2}}{20} + \frac{y^{2}}{36} = 1$
  Show solution
  AnswerFoci $(0, \pm 4)$ , vertices $(0, \pm 6)$ ; axes 12 and $4 \sqrt{5}$ ; $e = \frac{2}{3}$ ; latus rectum $\frac{20}{3}$
  $a = 6$ along the $y$ -axis, $b = 2 \sqrt{5}$ .
  $c = \sqrt{36 - 20} = 4$
  So $e = \frac{4}{6} = \frac{2}{3}$ and the latus rectum is $\frac{40}{6} = \frac{20}{3} .$
3. $\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1$
  Show solution
  AnswerFoci $(0, \pm 2)$ , vertices $(0, \pm 4)$ ; axes 8 and $4 \sqrt{3}$ ; $e = \frac{1}{2}$ ; latus rectum 6
  $a = 4$ along the $y$ -axis, $b = 2 \sqrt{3}$ .
  $c = \sqrt{16 - 12} = 2$
  So $e = \frac{1}{2}$ and the latus rectum is $\frac{24}{4} = 6.$
2
Long answerMediumNCERT Exercise 10.3 Q7–Q9
Write each ellipse in standard form, then find its foci, vertices, axes, eccentricity and latus rectum.
1. $16 x^{2} + 25 y^{2} = 400$
  Show solution
  Answer $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ : foci $(\pm 3, 0)$ , vertices $(\pm 5, 0)$ ; axes 10 and 8; $e = \frac{3}{5}$ ; latus rectum $\frac{32}{5}$
  $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
  $a = 5$ , $b = 4$ , $c = 3$ along the $x$ -axis, so $e = \frac{3}{5}$ and the latus rectum is $\frac{32}{5} .$
2. $x^{2} + 4 y^{2} = 4$
  Show solution
  Answer $\frac{x^{2}}{4} + y^{2} = 1$ : foci $(\pm \sqrt{3}, 0)$ , vertices $(\pm 2, 0)$ ; axes 4 and 2; $e = \frac{\sqrt{3}}{2}$ ; latus rectum 1
  $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$
  $a = 2$ , $b = 1$ , $c = \sqrt{3}$ , so $e = \frac{\sqrt{3}}{2}$ and the latus rectum is $\frac{2 \cdot 1}{2} = 1.$
3. $9 x^{2} + 5 y^{2} = 45$
  Show solution
  Answer $\frac{x^{2}}{5} + \frac{y^{2}}{9} = 1$ : foci $(0, \pm 2)$ , vertices $(0, \pm 3)$ ; axes 6 and $2 \sqrt{5}$ ; $e = \frac{2}{3}$ ; latus rectum $\frac{10}{3}$
  $\frac{x^{2}}{5} + \frac{y^{2}}{9} = 1$
  $a = 3$ along the $y$ -axis, $b = \sqrt{5}$ , $c = 2$ , so $e = \frac{2}{3}$ and the latus rectum is $\frac{10}{3} .$
3
Short answerEasyNCERT Exercise 10.3 Q10–Q12
Find the ellipse with the given vertices and foci.
1. Vertices $(\pm 10, 0)$ ; foci $(\pm 6, 0)$
  Show solution
  Answer $\frac{x^{2}}{100} + \frac{y^{2}}{64} = 1$
  $b^{2} = 100 - 36 = 64$
2. Vertices $(0, \pm 7)$ ; foci $(0, \pm \sqrt{13})$
  Show solution
  Answer $\frac{x^{2}}{36} + \frac{y^{2}}{49} = 1$
  $b^{2} = 49 - 13 = 36$
  The major axis is the $y$ -axis, so $a^{2} = 49$ goes under $y^{2}$ .
3. Vertices $(\pm 9, 0)$ ; foci $(\pm 5, 0)$
  Show solution
  Answer $\frac{x^{2}}{81} + \frac{y^{2}}{56} = 1$
  $b^{2} = 81 - 25 = 56$
4
Short answerEasyNCERT Exercise 10.3 Q13–Q14
Find the ellipse from the ends of its axes.
1. Ends of the major axis $(\pm 7, 0)$ ; ends of the minor axis $(0, \pm 2)$
  Show solution
  Answer $\frac{x^{2}}{49} + \frac{y^{2}}{4} = 1$
  $a = 7$ on the $x$ -axis and $b = 2$ .
2. Ends of the major axis $(0, \pm \sqrt{10})$ ; ends of the minor axis $(\pm 2, 0)$
  Show solution
  Answer $\frac{x^{2}}{4} + \frac{y^{2}}{10} = 1$
  $a^{2} = 10$ under $y^{2}$ and $b^{2} = 4$ under $x^{2}$ .
5
Short answerMediumNCERT Exercise 10.3 Q15–Q16
Find the ellipse from an axis length and the foci.
1. Major axis of length 18; foci $(\pm 4, 0)$
  Show solution
  Answer $\frac{x^{2}}{81} + \frac{y^{2}}{65} = 1$
  $a = 9$ , $c = 4$ .
  $b^{2} = 81 - 16 = 65$
2. Minor axis of length 10; foci $(0, \pm 12)$
  Show solution
  Answer $\frac{x^{2}}{25} + \frac{y^{2}}{169} = 1$
  $b = 5$ , $c = 12$ , major axis along the $y$ -axis.
  $a^{2} = 25 + 144 = 169$
6
Short answerMediumNCERT Exercise 10.3 Q17–Q18
Find the ellipse, centred at the origin, from the given data.
1. Foci $(0, \pm 5)$ and $a = 6$
  Show solution
  Answer $\frac{x^{2}}{11} + \frac{y^{2}}{36} = 1$
  $b^{2} = 36 - 25 = 11$
  The foci are on the $y$ -axis, so $a^{2} = 36$ goes under $y^{2}$ .
2. $b = 2$ , $c = 3$ , foci on the $x$ -axis
  Show solution
  Answer $\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1$
  $a^{2} = 4 + 9 = 13$
7
Long answerHardNCERT Exercise 10.3 Q19–Q20
Find the standard ellipse through the two given points.
1. Centre at the origin, major axis on the $y$ -axis, through $(2, 3)$ and $(1, 5)$
  Show solution
  Answer $16 x^{2} + 3 y^{2} = 91$
  $4 A + 9 B = 1, A + 25 B = 1$
  Then $A = 1 - 25 B$ , so $4 - 91 B = 1$ : $B = \frac{3}{91}$ and $A = \frac{16}{91}$ . Here $\frac{1}{B} = \frac{91}{3} > \frac{1}{A} = \frac{91}{16}$ , so the major axis is indeed vertical.
  $16 x^{2} + 3 y^{2} = 91$
2. Major axis on the $x$ -axis, through $(3, 2)$ and $(- 1, 3)$
  Show solution
  Answer $5 x^{2} + 8 y^{2} = 77$
  $9 A + 4 B = 1, A + 9 B = 1$
  Then $A = 1 - 9 B$ , so $9 - 77 B = 1$ : $B = \frac{8}{77}$ and $A = \frac{5}{77}$ , with $\frac{1}{A} > \frac{1}{B}$ as needed.
  $5 x^{2} + 8 y^{2} = 77$
Quick answerMedium
The minor axis of an ellipse is as long as the distance between its foci. Its eccentricity is
$\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ $\sqrt{2}$
Show answer
Answer (B) $\frac{1}{\sqrt{2}}$
$2 b = 2 c$ gives $b = c$ , so $a^{2} = 2 c^{2}$ and $e = \frac{c}{a} = \frac{1}{\sqrt{2}}$ . The option $\sqrt{2}$ is more than 1, impossible for an ellipse.
Quick answerMedium
Assertion (A): The ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ has its foci on the $y$ -axis. Reason (R): The major axis of a standard ellipse lies along the variable with the larger denominator, and the foci lie on the major axis.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
Here $16 > 9$ , so the major axis is the $y$ -axis and the foci are $(0, \pm \sqrt{7})$ . R is exactly the reason.

10.6 Hyperbola

Two radio stations far apart send out a pulse at the same instant. A ship hears one pulse 0.2 ms before the other. Radio travels 300 km in a millisecond, so the ship is 60 km nearer to the first station than to the second. All the points with that same difference of distances lie on one curve, a hyperbola; a second pair of stations gives a second curve, and the ship is where they cross. Navigation systems such as LORAN worked this way.

Definition

Hyperbola

Take two fixed points $F_{1}$ and $F_{2}$ in a plane. A hyperbola is made of the points $P$ of that plane for which the difference between $P F_{1}$ and $P F_{2}$ , the larger minus the smaller, has one and the same value. $F_{1}$ and $F_{2}$ are its foci.

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

Fig. 10.13The larger distance minus the smaller is the same,

2 a

, at every point of both branches.NCERT Figs 10.27, 10.28 · p. 196

Two foci F1 and F2 on a horizontal line, with a faint outline of the curve. A point P is joined to F1 by a solid segment and to F2 by a dashed one, and two bars below show PF1 and PF2, with the extra length of the longer bar boxed and labelled 2a. Steps 1 and 2 put P at two places on the right branch, nearer F2, the second close to the vertex; the bars change but the boxed difference stays 2a. Step 3 puts P on the left branch, nearer F1, where PF2 is now the longer bar, again longer by 2a. Step 4 draws both branches with the centre, the two vertices A and A prime, the transverse axis through the foci and the vertical conjugate axis.

Picture the two branches facing each other. The point of each branch nearest the other is a vertex, and the vertices are $2 a$ apart. The line through them also passes through both foci, which lie farther out, $2 c$ apart; this line is the transverse axis, and the midpoint of $F_{1} F_{2}$ is the centre. The line through the centre at right angles to it is the conjugate axis.

Because the vertices lie between the foci, $c > a$ , and the number $b = \sqrt{c^{2} - a^{2}}$ is real. A segment of length $2 b$ on the conjugate axis, centred at the centre, is also called the conjugate axis; Fig. 10.15 shows what $b$ measures.

The constant difference is again $2 a$ . Swapping the two foci does not change “larger minus smaller”, so the hyperbola is symmetric about its conjugate axis, and its vertices are mirror images, each $a$ from the centre. At the vertex $A$ nearer $F_{2}$ , $A F_{1} = c + a$ and $A F_{2} = c - a$ , which differ by $2 a$ . So $∣ P F_{1} - P F_{2} ∣ = 2 a$ at every point $P$ of the hyperbola.

SimulationConstant difference

A coordinate grid from minus 8 to 8 on both axes with two fixed foci, F1 at (minus 4, 0) and F2 at (4, 0), and a draggable point P that starts at (6, 3). The distances from P to each focus are drawn and measured, and P snaps to grid points and shows its coordinates.

Fig. 10.14Drag

P

and subtract the shorter distance from the longer. The hyperbola

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

is every point where the difference is 4.

Try this

Move $P$ to $(2, 0)$ , a vertex. Check that the distances differ by 4.
Find the four grid points with $x = \pm 4$ where the difference is 4. (Hint: try $y = \pm 6$ .)
Put $P$ on the $y$ -axis. What is the difference there, and why is no point of the hyperbola on that axis?

10.6.1 Eccentricity

The ratio $e = \frac{c}{a}$ is defined for the hyperbola too. Now the foci lie beyond the vertices, $c > a$ , so $e > 1$ , and each focus is still $a e$ from the centre. A bigger $e$ means wider branches, since $\frac{b}{a} = \sqrt{e^{2} - 1}$ .

10.6.2 Standard equation of hyperbola

Put the centre at the origin and the foci at $F_{1} (- c, 0)$ and $F_{2} (c, 0)$ . Take a point $P (x, y)$ on the right branch, the one nearer $F_{2}$ , so that $P F_{1} - P F_{2} = 2 a$ .

DerivationThe hyperbola equation from the difference rule

The trick used for the ellipse works again. The difference of the squares is still $P F_{1}^{2} - P F_{2}^{2} = 4 c x$ ; divide it by $P F_{1} - P F_{2} = 2 a$ :

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

Together with $P F_{1} - P F_{2} = 2 a$ this gives the two focal distances:

Right branch

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

(10.9)

Square the second, with $P F_{2}^{2} = (x - c)^{2} + y^{2}$ ; once more the $- 2 c x$ terms cancel:

\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}

With $b^{2} = c^{2} - a^{2}$ , divide by $b^{2}$ :

Standard hyperbola

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

(10.10)

On the left branch, where $P F_{2} - P F_{1} = 2 a$ , both focal distances change sign but their squares do not, so the same equation results.

Conversely, if $(x, y)$ satisfies (10.10), then $y^{2} = b^{2} (\frac{x^{2}}{a^{2}} - 1)$ and expanding gives $P F_{1}^{2} = (\frac{c}{a} x + a)^{2}$ and $P F_{2}^{2} = (\frac{c}{a} x - a)^{2} .$ For $x \geq a$ both brackets are positive, so $P F_{1} - P F_{2} = 2 a$ ; for $x \leq - a$ both are negative, so $P F_{2} - P F_{1} = 2 a$ . Either way the point is on the hyperbola.

Read (10.10) as $\frac{x^{2}}{a^{2}} = 1 + \frac{y^{2}}{b^{2}}$ : the right side is at least 1, so $∣ x ∣$ can never drop below $a$ . The vertical strip $- a < x < a$ is therefore empty (Fig. 10.15). The branches stay outside it, never touch the $y$ -axis, and spread out without end.

2/4The foci are outside the vertices, with $c^{2} = a^{2} + b^{2}$ .

Fig. 10.15The parts of

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

.NCERT Fig. 10.29 · p. 197

The hyperbola x squared over a squared minus y squared over b squared equals 1 on coordinate axes. Step 1 marks the vertices (a, 0) and (minus a, 0) and shades the strip between x = minus a and x = a, which contains no points of the curve. Step 2 adds the foci F1 and F2 outside the vertices and a right triangle with legs a and b along the axes whose hypotenuse, from the vertex to the point (0, b), has length c, so c squared equals a squared plus b squared. Step 3 highlights the transverse axis of length 2a between the vertices and the conjugate axis of length 2b along the y-axis. Step 4 draws the latus rectum through F2, made of two halves of length b squared over a.

Hyperbola

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

(10.11)

With the foci on the $y$ -axis the standard equation is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1.$ When $a = b$ the hyperbola is called equilateral: $x^{2} - y^{2} = a^{2}$ , with $c = a \sqrt{2}$ . Hyperbolas whose axes are other perpendicular lines exist too; they come in higher classes.

10.6.3 Latus rectum

Put $x = c$ in (10.10): $\frac{y^{2}}{b^{2}} = \frac{c^{2}}{a^{2}} - 1 = \frac{b^{2}}{a^{2}}$ , so $y = \pm \frac{b^{2}}{a}$ . The latus rectum of a hyperbola therefore has the same length as for an ellipse, $\frac{2 b^{2}}{a}$ , given by (10.8).

Table 10.2Ellipse and hyperbola compared (both centred at

O

, foci on the

x

-axis)

	Ellipse	Hyperbola
Rule	sum is $2 a$	difference is $2 a$
Equation	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$c^{2}$	$a^{2} - b^{2}$	$a^{2} + b^{2}$
$e$	less than 1	more than 1
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

Reading two hyperbolas

Find the foci, the vertices, the eccentricity and the latus rectum of (i) $\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$ and (ii) $4 y^{2} - x^{2} = 4$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10.15NCERT Example 15 · p. 201

From foci and vertices

Find the hyperbola with foci $(\pm 4, 0)$ and vertices $(\pm \sqrt{7}, 0)$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10.16NCERT Example 16 · p. 201

From the foci and the latus rectum

Find the hyperbola with foci $(\pm 10, 0)$ whose latus rectum has length 9.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 9 questions

Practice 10.4: questions with answers

1
Short answerEasyNCERT Exercise 10.4 Q1–Q2
For each hyperbola find the foci, the vertices, the eccentricity and the latus rectum.
1. $\frac{x^{2}}{25} - \frac{y^{2}}{144} = 1$
  Show solution
  AnswerFoci $(\pm 13, 0)$ , vertices $(\pm 5, 0)$ , $e = \frac{13}{5}$ , latus rectum $\frac{288}{5}$
  $a = 5$ , $b = 12$ .
  $c = \sqrt{25 + 144} = 13$
  So $e = \frac{13}{5}$ and the latus rectum is $\frac{2 \cdot 144}{5} = \frac{288}{5} .$
2. $\frac{y^{2}}{16} - \frac{x^{2}}{20} = 1$
  Show solution
  AnswerFoci $(0, \pm 6)$ , vertices $(0, \pm 4)$ , $e = \frac{3}{2}$ , latus rectum 10
  $a = 4$ along the $y$ -axis, $b^{2} = 20$ .
  $c = \sqrt{16 + 20} = 6$
  So $e = \frac{6}{4} = \frac{3}{2}$ and the latus rectum is $\frac{40}{4} = 10.$
2
Long answerMediumNCERT Exercise 10.4 Q3–Q6
Write each hyperbola in standard form and find its foci, vertices, eccentricity and latus rectum.
1. $4 x^{2} - 9 y^{2} = 36$
  Show solution
  Answer $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ : foci $(\pm \sqrt{13}, 0)$ , vertices $(\pm 3, 0)$ , $e = \frac{\sqrt{13}}{3}$ , latus rectum $\frac{8}{3}$
  $a = 3$ , $b = 2$ , $c = \sqrt{13}$ ; latus rectum $\frac{2 \cdot 4}{3} = \frac{8}{3} .$
2. $16 y^{2} - 9 x^{2} = 144$
  Show solution
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ : foci $(0, \pm 5)$ , vertices $(0, \pm 3)$ , $e = \frac{5}{3}$ , latus rectum $\frac{32}{3}$
  $a = 3$ along the $y$ -axis, $b = 4$ , $c = 5$ ; latus rectum $\frac{32}{3} .$
3. $3 x^{2} - y^{2} = 12$
  Show solution
  Answer $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$ : foci $(\pm 4, 0)$ , vertices $(\pm 2, 0)$ , $e = 2$ , latus rectum 12
  $a = 2$ , $b^{2} = 12$ , $c = \sqrt{16} = 4$ ; latus rectum $\frac{24}{2} = 12.$
4. $7 y^{2} - 9 x^{2} = 63$
  Show solution
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{7} = 1$ : foci $(0, \pm 4)$ , vertices $(0, \pm 3)$ , $e = \frac{4}{3}$ , latus rectum $\frac{14}{3}$
  $a = 3$ along the $y$ -axis, $b^{2} = 7$ , $c = 4$ ; latus rectum $\frac{14}{3} .$
3
Short answerEasyNCERT Exercise 10.4 Q7–Q9
Find the hyperbola with the given vertices and foci.
1. Vertices $(\pm 6, 0)$ ; foci $(\pm 10, 0)$
  Show solution
  Answer $\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$
  $b^{2} = 100 - 36 = 64$
2. Vertices $(0, \pm 2)$ ; foci $(0, \pm 3)$
  Show solution
  Answer $\frac{y^{2}}{4} - \frac{x^{2}}{5} = 1$
  $b^{2} = 9 - 4 = 5$
3. Vertices $(0, \pm 12)$ ; foci $(0, \pm 13)$
  Show solution
  Answer $\frac{y^{2}}{144} - \frac{x^{2}}{25} = 1$
  $b^{2} = 169 - 144 = 25$
4
Short answerMediumNCERT Exercise 10.4 Q10–Q11
Find the hyperbola from its foci and the length of one axis.
1. Foci $(\pm 7, 0)$ ; transverse axis of length 10
  Show solution
  Answer $\frac{x^{2}}{25} - \frac{y^{2}}{24} = 1$
  $a = 5$ , $c = 7$ .
  $b^{2} = 49 - 25 = 24$
2. Foci $(0, \pm 10)$ ; conjugate axis of length 12
  Show solution
  Answer $\frac{y^{2}}{64} - \frac{x^{2}}{36} = 1$
  $b = 6$ , $c = 10$ .
  $a^{2} = 100 - 36 = 64$
5
Long answerHardNCERT Exercise 10.4 Q12–Q13
Find the hyperbola from its foci and the length of its latus rectum.
1. Foci $(\pm 2 \sqrt{7}, 0)$ ; latus rectum 6
  Show solution
  Answer $\frac{x^{2}}{16} - \frac{y^{2}}{12} = 1$
  $\frac{2 b^{2}}{a} = 6$ gives $b^{2} = 3 a$ , and $c^{2} = 28$ .
  $a^{2} + 3 a - 28 = 0$
  So $(a + 7) (a - 4) = 0$ , $a = 4$ and $b^{2} = 12$ .
2. Foci $(0, \pm 5)$ ; latus rectum $\frac{32}{3}$
  Show solution
  Answer $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
  $\frac{2 b^{2}}{a} = \frac{32}{3}$ gives $b^{2} = \frac{16 a}{3}$ , and $c^{2} = 25$ .
  $3 a^{2} + 16 a - 75 = 0$
  So $(a - 3) (3 a + 25) = 0$ , $a = 3$ and $b^{2} = 16$ , with the transverse axis along the $y$ -axis.
Short answerMediumNCERT Exercise 10.4 Q14
Find the hyperbola with vertices $(\pm 2, 0)$ and eccentricity $\frac{3}{2}$ .
Show solution
Answer $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
$a = 2$ , so $c = a e = 3$ .
$b^{2} = 9 - 4 = 5$
Long answerHardNCERT Exercise 10.4 Q15
A hyperbola in standard form has foci $(\pm 5, 0)$ and passes through $(6, 2)$ . Find its equation.
Show solution
Answer $\frac{x^{2}}{20} - \frac{y^{2}}{5} = 1$
With $A = a^{2}$ , the point gives $\frac{36}{A} - \frac{4}{25 - A} = 1$ . Clearing fractions:
$A^{2} - 65 A + 900 = 0$
So $A = 20$ or $A = 45$ . But $a < c = 5$ needs $A < 25$ , so $a^{2} = 20$ and $b^{2} = 5$ .
Quick answerMedium
The eccentricity of every equilateral hyperbola ( $a = b$ ) is
$1$ $\sqrt{2}$ $2$ $\frac{1}{\sqrt{2}}$
Show answer
Answer (B) $\sqrt{2}$
$c^{2} = a^{2} + a^{2} = 2 a^{2}$ , so $e = \frac{c}{a} = \sqrt{2}$ . The value $\frac{1}{\sqrt{2}}$ belongs to an ellipse with $b = c$ .
Quick answerMedium
A hyperbola has eccentricity 2 and its foci are 16 units apart. Find the length of its transverse axis.
Show answer
Answer 8
$2 c = 16$ gives $c = 8$ ; $a = \frac{c}{e} = 4$ , so the transverse axis is $2 a = 8$ .

Miscellaneous Examples

In each of these models, the first job is to choose the axes so that the curve has a standard equation; after that the questions become arithmetic. The lab below lets you try all four conics before you start.

InteractiveConic lab

Buttons choose a circle, parabola, ellipse or hyperbola, and another button turns the axis between horizontal and vertical. Sliders set the size a (or the radius) and the eccentricity e. The graph shows the curve with its foci, its directrix for a parabola, its vertices and a latus rectum. A third slider, or the Play button, moves a point P along the curve, and the segments from P to the foci (or to the focus and the directrix) are drawn. A panel shows the equation, a, b, c, e, the latus rectum and the live distances, confirming the defining rule: equal distances, a constant sum or a constant difference.

Fig. 10.16Pick a conic, change

a

and

e

, and move

P

to check its distance rule.

Example 10.17NCERT Example 17 · p. 202

A dish antenna

The receiver of a DTH dish sits at the focus, 36 cm from the vertex of the dish. The dish is 16 cm deep. How wide is it across the rim?

Try it yourself first. Then reveal the solution one step at a time.

Example 10.18NCERT Example 18 · p. 202

A fountain jet

A jet of water leaves a nozzle at ground level, rises to a greatest height of 2 m, and lands 3 m from the nozzle, tracing a parabola. At what horizontal distances from the nozzle is the water 1.5 m above the ground?

Try it yourself first. Then reveal the solution one step at a time.

4/4The positions of $P$ lie on $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ .

Fig. 10.17As the ladder slides, the marked point traces part of an ellipse.NCERT Fig. 10.33 · p. 203

A vertical wall and a horizontal floor meet at O. A 5 m ladder AB leans with its foot A on the floor and its top B on the wall, and a point P on it is 2 m from A and 3 m from B. Steps 1 to 3 show the ladder steep, at 45 degrees and nearly flat, with P marked each time and earlier positions of P left as faint dots. Step 4 joins the positions of P into a quarter of an ellipse and gives its equation, x squared over 9 plus y squared over 4 equals 1.

Example 10.19NCERT Example 19 · p. 203

The sliding ladder

A 5 m ladder slides with its foot $A$ on the floor ( $x$ -axis) and its top $B$ against a wall ( $y$ -axis). A paint mark $P$ on the ladder is 2 m from $A$ . Show that $P$ moves on an ellipse (Fig. 10.17).

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Match each equation with its curve.

Tap an item on the left, then its match on the right.

Show answer

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

Equal squared terms with a plus sign give a circle; only one squared variable gives a parabola; unequal denominators with a plus give an ellipse; a minus sign between the squares gives a hyperbola.

From history

From cones to coordinates

Greek geometers studied the conics as solid geometry for centuries, without any algebra. Apollonius's Conics showed that all of them come from one double cone by changing the tilt of the cutting plane. The rope-and-peg constructions of the Śulba-sūtras in ancient India were likewise pure geometry.

The big change came in the seventeenth century, when Pierre de Fermat and René Descartes, working separately, described curves by equations in coordinates. Fermat's version was passed around in handwritten copies and printed only in 1679, after his death, which is partly why Descartes's La Géométrie (1637) is remembered as the start of the subject. Over the next century and a half, mathematicians such as Clairaut, Cramer, Monge and Lacroix shaped the formulas for distances, slopes and lines that you used in Chapter 9.

Timeline

Conics through history

c. 350 BCE
Menaechmus
First known study of conic sections, while attacking the problem of doubling a cube.
c. 200 BCE
Apollonius
Writes Conics and names the ellipse, parabola and hyperbola.
1609
Kepler
Planets move on ellipses with the Sun at one focus.
1637
Descartes
La Géométrie turns curves into equations.
1638
Galileo
Shows that a thrown body follows a parabola.
1679
Fermat
His introduction to plane and solid loci, written decades earlier, is printed.
1687
Newton
Shows that inverse-square gravity gives orbits that are conics.

Recap

Conic Sections: key points and summary

A plane cuts a double cone in a circle, ellipse, parabola or hyperbola, depending on its angle $β$ with the axis; through the vertex it gives a point, a line or two lines (Section 10.2).
A circle is the set of points at distance $r$ from a centre $(h, k)$ : $(x - h)^{2} + (y - k)^{2} = r^{2}$ (Section 10.3).
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ has centre $(- g, - f)$ and radius $\sqrt{g^{2} + f^{2} - c}$ ; complete the squares to read it ((10.2)).
A parabola is the set of points equally far from a focus and a directrix (Section 10.4).
$y^{2} = 4 a x$ has focus $(a, 0)$ and directrix $x = - a$ ; the squared variable gives the axis and the sign the direction (Section 10.4.1).
The latus rectum is the focal chord perpendicular to the axis; for $y^{2} = 4 a x$ its length is $4 a$ (Section 10.4.2).
An ellipse is the set of points whose distances from two foci add to a constant $2 a$ , the length of the major axis (Section 10.5).
For an ellipse $a^{2} = b^{2} + c^{2}$ , and $e = \frac{c}{a} < 1$ measures how flat it is (Section 10.5.1, Section 10.5.2).
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$ has foci $(\pm c, 0)$ ; the larger denominator marks the major axis (Section 10.5.3).
For the ellipse and the hyperbola the latus rectum is $\frac{2 b^{2}}{a}$ ((10.8)).
A hyperbola is the set of points whose distances from two foci differ by a constant $2 a$ , the length of the transverse axis (Section 10.6).
For a hyperbola $c^{2} = a^{2} + b^{2}$ and $e > 1$ ; in $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ the positive term marks the transverse axis (Section 10.6.2).
Reflectors, arches, cables and sliding rods are modelled by choosing axes that give a standard equation (Miscellaneous Examples).

Practice · 11 questions

Practice questions with answers (NCERT “Miscellaneous exercises on Chapter 10”)

NumericalEasyNCERT Miscellaneous Exercise Q1
A torch reflector is a parabolic bowl 12 cm across and 3 cm deep. Where should the bulb (the focus) go?
Show solution
AnswerOn the axis, 3 cm from the vertex.
$36 = 4 a \cdot 3 \Rightarrow a = 3$
The focus is $(3, 0)$ : on the axis, 3 cm from the vertex.
NumericalMediumNCERT Miscellaneous Exercise Q2
A gateway is shaped like a parabola with a vertical axis: it stands 9 m tall and spans 12 m along the ground. How wide is it at a height of 5 m?
Show solution
Answer8 m
With the vertex at the top, $x^{2} = - 4 a y$ passes through $(6, - 9)$ .
$36 = 36 a \Rightarrow a = 1$
So $x^{2} = - 4 y$ . A height of 5 m is 4 m below the top, so $y = - 4$ : $x^{2} = 16$ , $x = \pm 4$ and the width is 8 m.
NumericalHardNCERT Miscellaneous Exercise Q3
A footbridge has a level deck 120 m long, hung from a parabolic cable by vertical rods. The rods at the two ends are 32 m long and the one at the centre only 8 m. What length of rod is needed 30 m from the centre?
Show solution
Answer14 m
With the vertex at the lowest point, $x^{2} = 4 a y$ . At an end, $x = 60$ and the cable is $32 - 8 = 24$ m above the vertex.
$3600 = 96 a \Rightarrow a = 37.5$
At $x = 30$ : $y = \frac{900}{150} = 6$ , so the rod is $8 + 6 = 14$ m long.
NumericalMediumNCERT Miscellaneous Exercise Q4
A tunnel entrance is half an ellipse, 10 m across at road level and 3 m high in the middle. How much headroom is there 2 m in from one side?
Show solution
Answer2.4 m
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
At $x = 3$ : $y^{2} = 9 (1 - \frac{9}{25}) = \frac{144}{25}$ , so $y = 2.4$ m.
Long answerMediumNCERT Miscellaneous Exercise Q5
The ends of a 10 cm stick slide along the two coordinate axes. A dot $P$ is painted on the stick 4 cm from the end on the $x$ -axis. Find the equation of the curve the dot follows.
Show solution
Answer $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$
If the rod makes angle $θ$ with the $x$ -axis, $P$ is 6 cm from the end on the $y$ -axis.
$x = 6 \cos θ, y = 4 \sin θ$
$\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$
Short answerMediumNCERT Miscellaneous Exercise Q6
Join the vertex of $y^{2} = - 20 x$ to both ends of its latus rectum. What is the area of the triangle you get?
Show solution
Answer50 square units
$a = 5$ , so the ends are $(- 5, 10)$ and $(- 5, - 10)$ : base 20, height 5.
$\frac{1}{2} \times 20 \times 5 = 50$
Short answerMediumNCERT Miscellaneous Exercise Q7
A robotic lawn-mower keeps the sum of its distances from two beacons equal to 26 m. The beacons are 24 m apart. Find its path, with the beacons on the $x$ -axis and the origin midway between them.
Show solution
Answer $\frac{x^{2}}{169} + \frac{y^{2}}{25} = 1$
$a = 13$ and $c = 12$ .
$b^{2} = 169 - 144 = 25$
Long answerHardNCERT Miscellaneous Exercise Q8
A triangle has one vertex at the vertex of $y^{2} = 12 x$ and the other two on the parabola, placed symmetrically about the axis so that the angle at the origin is a right angle. Find the length of its longest side and its area.
Show solution
AnswerHypotenuse 24; area 144 square units
A right angle at the origin, split equally by the axis, means each arm makes $45^{\circ}$ with the axis, so the points are $(t, \pm t)$ .
$t^{2} = 12 t \Rightarrow t = 12$
The vertices are $(12, 12)$ and $(12, - 12)$ : the hypotenuse is 24, and the area is $\frac{1}{2} \times 24 \times 12 = 144$ .
Quick answerHard
For which values of $k$ does $\frac{x^{2}}{9 - k} + \frac{y^{2}}{5 - k} = 1$ represent a hyperbola?
$k < 5$ $5 < k < 9$ $k > 9$ $k < 9$
Show answer
Answer (B) $5 < k < 9$
A hyperbola needs the two denominators to have opposite signs: $9 - k > 0$ and $5 - k < 0$ , so $5 < k < 9$ . For $k < 5$ both are positive (an ellipse); for $k > 9$ both are negative (no real points).
Quick answerMedium
Assertion (A): The latus rectum of $y^{2} = 4 a x$ has length $4 a$ . Reason (R): Each end of the latus rectum is $2 a$ from the directrix.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
Each end is $2 a$ from the directrix, so by the definition it is $2 a$ from the focus; the two halves make $4 a$ . R is the reason.
Quick answerHard
The latus rectum of an ellipse is half as long as its minor axis. Find its eccentricity as a decimal.
Show answer
Answer 0.866
$\frac{2 b^{2}}{a} = b$ gives $b = \frac{a}{2}$ , so $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{3}{4}$ and $e = \frac{\sqrt{3}}{2} \approx 0.866$ .

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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	Miscellaneous exercises on Chapter 10 · Q1
NCERT Miscellaneous Exercise Q2	Miscellaneous exercises on Chapter 10 · Q2
NCERT Miscellaneous Exercise Q3	Miscellaneous exercises on Chapter 10 · Q3
NCERT Miscellaneous Exercise Q4	Miscellaneous exercises on Chapter 10 · Q4
NCERT Miscellaneous Exercise Q5	Miscellaneous exercises on Chapter 10 · Q5
NCERT Miscellaneous Exercise Q6	Miscellaneous exercises on Chapter 10 · Q6
NCERT Miscellaneous Exercise Q7	Miscellaneous exercises on Chapter 10 · Q7
NCERT Miscellaneous Exercise Q8	Miscellaneous exercises on Chapter 10 · Q8

Quick check: MCQs on Conic Sections

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: The set of points of a plane at a fixed distance (the radius) from a fixed point (the centre).First seen in the chapter →
conic sectionशंकु परिच्छेदshanku parichchhed: A curve where a plane cuts a double cone: a circle, ellipse, parabola or hyperbola.First seen in the chapter →

degenerate conic: A point, a line or a pair of lines, cut by a plane through the vertex of the cone.First seen in the chapter →
directrixनियताniyata: The fixed line in the definition of a parabola.First seen in the chapter →

eccentricityउत्केंद्रताutkendrata: The ratio $e = c / a$ ; below 1 for an ellipse, above 1 for a hyperbola.First seen in the chapter →
ellipseदीर्घवृत्तdeerghvritt: The set of points of a plane whose distances from two foci add up to a constant.First seen in the chapter →

focusनाभिnaabhi: A fixed point used to define a conic: a parabola has one focus, an ellipse and a hyperbola have two.First seen in the chapter →

generator: A straight line on the surface of a cone through its vertex; turning it about the axis sweeps out the cone.First seen in the chapter →

hyperbolaअतिपरवलयatiparvalay: The set of points of a plane whose distances from two foci differ by a constant.First seen in the chapter →

latus rectumनाभिलंबnaabhilamb: The chord through a focus perpendicular to the axis of the conic.First seen in the chapter →

major axisदीर्घ अक्षdeergh aksh: The longest chord of an ellipse, through both foci; the minor axis is the chord through the centre perpendicular to it.First seen in the chapter →

nappe: Either of the two halves of a double cone that meet at the vertex.First seen in the chapter →

parabolaपरवलयparvalay: The set of points of a plane equally far from a fixed point (the focus) and a fixed line (the directrix).First seen in the chapter →

transverse axisअनुप्रस्थ अक्षanuprasth aksh: The axis of a hyperbola through its foci; the conjugate axis is perpendicular to it through the centre.First seen in the chapter →

vertexशीर्षsheersh: A point where a conic meets its axis through the focus or foci.First seen in the chapter →

10.1 Introduction

10.2 Sections of a Cone

10.2.1 Circle, ellipse, parabola and hyperbola

10.2.2 Degenerated conic sections

10.3 Circle

10.4 Parabola

10.4.1 Standard equations of parabola

10.4.2 Latus rectum

10.5 Ellipse

10.5.1 How a, b and c are related

10.5.2 Eccentricity

10.5.3 Standard equations of an ellipse

10.5.4 Latus rectum

10.6 Hyperbola

10.6.1 Eccentricity

10.6.2 Standard equation of hyperbola

10.6.3 Latus rectum

Miscellaneous Examples

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Quick check: MCQs on Conic Sections

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