Conic Section

 Course Content  Learning Outcomes

 Standard equations of Ellipse and hyperbola. On completion of this unit, students will be able to:
 obtain standard equation of ellipse and
hyperbola.

9.1 Ellipse
Elliptical shapes arise in many situations. Therefore, study of ellipse is very important as it helps to
understand these situations. For example many arches around us have elliptical shapes. The planets
travel in elliptic orbits with the sun as one focal point. The ellipse also has the focal property that a
ray emanating from one focus is reflected to the other. This property makes ellipse more useful in
real world. We first give the definition of ellipse.
An ellipse is the locus of a point which moves in a plane, so that its distance from a fixed point,
called focus, bears a constant ratio to its distance from a fixed line called directrix. The constant
ratio is called the eccentricity and is denoted by e, where e < 1. We first derive the equation of
ellipse in standard form.

9.1.1 Equation of Ellipse in Standard Form

Let S be the focus, ZK be the directrix of the ellipse and SZ be perpendicular to ZK.
Y
K K'
B L1
L
P(x, y) M'
M
A A'
X' Z S N X
O S' Z'

L' L1'
B'
Y'
If A is the point on SZ, then, by definition
SA = e AZ . . . (1)
190 FOUNDATIONS OF MATHEMATICS

where e is the eccentricity of the ellipse and e < 1. Since e < 1, there will be another point A' on ZS
produced, such that
SA' = e A'Z . . . (2)
Here A and A' divide ZS internally and externally in the ratio e : 1, respectively.
Let AA' = 2a and let O be the middle point of AA', then
OA = OA' = a
From (1)
SA = e AZ
or, OA – OS = e(OZ – OA)
or, a – OS = e(OZ – a) . . . (3)
From (2)
SA' = e A'Z
or, OS + OA' = e (OZ + OA')
or, OS + a = e (OZ + a ) . . . (4)
Adding (3) and (4), we get
2a = e  2OZ
a
 OZ =
e
Subtracting (3) from (4) and obtain
2OS = e  2a
 OS = ae
Let O be the origin, then OA' and OB be the x-axis and y-axis respectively. Hence, the coordinates of
a
focus S be (– ae, 0) and the directrix ZM, being parallel to y-axis at a distance OZ = from the
e
a
origin, will have equation as x = – .
e
If P(x, y) is any point on the ellipse and PM the perpendicular on the directrix and PN perpendicular
on axis, then by definition,
PS = e PM  PS = e NZ
or, PS2 = e2 PM2  PS2 = e2 NZ2
or, PS2 = e2 (ON + OZ)2
a 2
or, (x + ae)2 + y2 = e2 x + 
 e
or, x + 2aex + a e + y2 = e2x2 + 2aex + a2
2 2 2

or, x2 (1 – e2) + y2 = a2 (1 – e2)

x2 y2
or 2 + 2 =1 [∵ e < 1]
a a (1 – e2)
x2 y2
 + =1 . . . (5)
a2 b2
 Conic Section UNIT 9 191

where b2 = a2(1 – e2), is the standard form of the equation of the ellipse.
To obtain a sketch of the graph of the ellipse (5), we put x = 0 and y = 0 to get y = ± b and x = ± a.
So that graph intersects the y-axis at the points (0, b) and (0, – b). And the graph cuts the x-axis at
(a, 0) and (– a, 0). Also, we observe from the equation that the graph is symmetric with respect to x
axis and y axis, as the substitution of – x and – y for x and y do not make any change in (3).
NOTE 1. The distance between A(– a, 0) and A'(a, 0) is called the major axis of the ellipse, since it is the longer
axis.
Y
B (0, b)

A (– a, 0 ) A' (a , 0)

S (–c, 0 ) O S' (c, 0) X

X'

B' (0, – b)
Y'
2. The distance between B(0, b) and B'(0, – b) is called the minor axis of the ellipse.
3. The end points of the major axis (± a, 0) are called the vertices and (0, ± b) are called covertices of the
ellipse.
4. The point of intersection of the major and the minor axis is called the center of the ellipse.

9.1.2 Description of Some Terms in Ellipse

1. Second Focus and Second Directrix
Since an ellipse is symmetrical about the axes as the curve contains even powers of x and y
therefore, the points S' and Z' on the x axis are taken as
a
OS' = OS = ae, OZ' = OZ = .
e
then S' will also be a focus, and Z'K', the line through Z' perpendicular to OZ' will be the
corresponding directrix of the ellipse. Thus, an ellipse has two foci (± ae, 0) and two
directrices
a
x=± , as SP = e PM
e
so S'P = e PM'.
2. Latus Rectum
A chord through a focus and at right angles to the major axis is called a latus rectum.
Let LSL' be the length 2l of the latus rectum so that SL = SL' = l Y
L L1
The coordinates of L are (– ae, l).
Since L lies on the ellipse O
S (-ae, 0) S'(ae, 0)
x2 y2 X
+ =1
a2 b2 L'
L1'
a2e2 l 2
or, + 2 =1
a2 b
b2 ... b2 = a2 (1 – e2)  1 – e2 = b2
2
or, l2 = b2 (1 – e2) = b2 ×
a2  a
192 FOUNDATIONS OF MATHEMATICS

b2
or, l =
a
2b2
i.e. 2 l =
a
2b2
Thus, the length of latus rectum (in plural, latera recta) LL' = .
a
The equations of the two latera recta LL' and L1L1' are respectively x = – ae and x = ae.
The coordinates of the extremities of the latera recta are
b2 – b2 b2 – b2
L– ae‚  , L' – ae‚ , L1 ae‚  , L'1 ae‚ .
 a  a   a  a 
3. Eccentricity
x2 y2
The eccentricity of the ellipse 2 + 2 = 1 is given by
a b
b = a (1 – e )
2 2 2

a2 – b2 b2
or, e2 = 2 =1– 2
a a
b2
... e= 1–
a2
4. Focal Distance
Let P(x1, y1) be any point on the ellipse and draw PN perpendicular to x axis. Then, the focal
distance PS and PS' of the point P are:
PS' = e PM' [by definition] PS = e PM
= e (OZ' – ON) = e(NO + OZ)

= e  – x1 = e  x1 + 
a a
e   e
= a – ex1 = a + ex1
 The sum of the focal distances of the point P = PS + PS'
= a – ex1 + a + ex1 = 2a
= Length of the major axis of the ellipse.
Y

M P(x1, y1) M’

X' X
Z S O S' N Z’

a
x=e
-a
x= e

Y'
 Conic Section UNIT 9 193

Remarks:
1. Circle is a particular case of the ellipse: Y Z
The equation of an ellipse is B’ (0, b) b
y=
x2 y2 e
2 + =1 . . . (1)
a b2 S (0, be)
A(-a,0) A’ (a, 0)
such that
O
b2 = a2(1 – e2) . . . (2) S ' (0, -be)

If b = a, then the equation reduces to x2 + y2 = a2, which is a

circle with the centre at the origin and radius equal to a putting B (0, -b)
b = a in (2), we get, Z' -b
y=
e
e = 0.
2. Other Forms of the Ellipse.
If, in the equation
x2 y2
+ = 1, b > a, then
a2 b2
the major axis ( BB') is of length 2b and lies along the y-axis and the minor axis (AA') of the
length 2a and lies along the x-axis.
b a2
The foci are now (0, ± be) and the directrices are y = ± and the eccentricity e = 1–
e b2
The shape of the ellipse is as shown in the figure.

9.1.3 Generalized Equation of Ellipse

The equation of the ellipse with centre C(h, k) and the axes parallel to the axes of coordinates is of
(x – h)2 (y – k)2
the form 2 + = 1.
a b2
Case I: when 0 < b < a
Let X = x – h and Y = y – k Y Y'

X2 Y2 P(x, y)
Then the equation becomes + 2 = 1, which is the standard
a2 b C (h, k)
equation in X and Y coordinates. Therefore, in X and Y X'
M'
coordinates for the given ellipse, we have, the length of major
axis = 2a and the length of minor axis = 2b. Centre at (0, 0) and k
a h
foci (± ae, 0). The equations of directrices are X = ± ,
e M
X
O
2
2b
vertices = (± a, 0) and latus rectum = . Therefore, in the
a
original x, y coordinates, using x = X + h, y = Y + k; centre = (h, k); foci = (ae + h, k) and (– ae + h, k).
a 2b2 b2
The equation of the directrices are x – h = ± , latus rectum = , eccentricity = 1 – 1 – 2 , the
e a a
length of major axis = 2a , and length of minor axis = 2b.
194 FOUNDATIONS OF MATHEMATICS

Case II: when 0 < a < b.

X2 Y2
Let us take X = x – h and Y = y – k then the equation becomes + 2 = 1,
a2 b Y B

which is in standard form in X, Y coordinates, Therefore, in X, Y __

S
coordinates, for the given equation, we have, centre = (0, 0), foci = (0, be)
A A'
and (0, – be) , length of major axis = 2b and length of minor axis = 2a. C
__
b 2a2 S'
The equation of directrices are y = ± , latus rectum = , vertices = (0, b)
e b B'
O X
a2
and (0, – b), eccentricity = 1– .
b2
Therefore, in the original x, y coordinates
Using x = X + h, y = Y + k,
Centre (h, k), foci = (h, be + k) and (h, – be + k); Length of major axis = 2b.
Length of minor axis = 2a
b
The equation of directrices are y – k = ±
e
2a2
Latus rectum =
b
Vertices = (h, b + k) and (h, – b + k)
a2
Eccentricity = 1–
b2
NOTE The equations of an ellipse can be represented by
Ax2 + Cy2 + Dx + Ey + F = 0
in which coefficients of A and C are positive and unequal. This is called the general form of the equation
of an ellipse. Not all equations of this form, however represent ellipses. When the equation (1) is expressed
in the form (x – h)2 + (y – k)2 = r2 , the value of r2 may be zero or negative number. Such cases are called
degenerate cases. In the former case, the locus is the single point (h, k) and in the later case there is no
locus.

Ellipse at a Glance
Length of Equation
Major Minor
Ellipse Center Vertex Focus Eccentricity Latus of
axis axis
rectum Directrix
x2 y2
+ = 1; b2 2b2 a
a2 b2 (0, 0) ( a, 0) ( ae, 0) 2a 2b e= 1– x=
a2 a e
a>b>0
x2 y2
+ = 1; a2 2a2 b
a2 b2 (0, 0) (0, b) (0,  be) 2b 2a e= 1– y=
b2 b e
b>a>0
(x – h)2 (y – k)2
+ = 1; b2 2b2 a
a2 b2 (h, k) (h  a, k) (h  ae, k) 2a 2b e= 1– x=h
a2 a e
a>b>0
(x – h)2 (y – k)2
+ = 1; a2 2a2 b
a2 b2 (h, k) (h, k  b) (h, k  be) 2b 2a e= 1– y=k
b2 b e
b>a>0
 Conic Section UNIT 9 195

 Illustrative Examples
Example 1. Find the lengths of the major and minor axes, coordinates of the foci, vertices and the
eccentricity for the ellipse 9x2 + 16y = 144.
Solution
9x2 + 16y2 = 144
9x2 + 16y2
or, =1
144
x2 y2
or, + =1 . . . (1)
16 9
x2 y2
Comparing (1) with the standard equation of the ellipse + = 1, we get
a2 b2
a2 = 16
or, a = ± 4
b2 = 9
or, b = ± 3
We have
b2 = a2 (1 – e2) [∵ a > b]
or, 9 = 16(1 – e2)

or, e2 = 1 –
9
=
7
=
 7 2
16 16  4 
7
 e=
4
Since the denominator of x2 > denominator of y2, major axis is along the x-axis and
minor axis is along y-axis.
Now
The length of major axis = 2a = 2 × 4 = 8
The length of minor axis = 2b = 2 × 3 = 6
Coordinates of foci are (ae, 0) and (– ae, 0), i.e., ( 7‚ 0) and (– 7‚ 0)
7
Vertices are (a, 0) and (– a, 0), i.e., (4, 0) and (– 4, 0), and the eccentricity = e = .
4
Example 2. Find the eccentricity, distance between the foci, equations of the directrices, the length
of the latus rectum and the coordinates of the ends of the latera recta of the ellipse
25x2 + 16y2 = 400.
Solution
The given equation of the ellipse is
25x2 + 16y2 = 400
x2 y2
or, + =1 . . . (1)
16 25
Since the denominator of y2 is greater than the denominator of x2, the axis of this
x2 y2
ellipse is y-axis. Comparing equation (1) with 2 + 2 = 1, we get
a b
196 FOUNDATIONS OF MATHEMATICS

a2 = 16
or, a = ± 4
and b2 = 25
or, b = ± 5
a2
Now, the eccentricity, e = 1–
b2
16 3
= 1– = .
25 5
3
Distance between two foci = 2be = 2 × 5 × = 6.
5
Equations of the directrices are
b 5 25
y=± =± =±
e 3/5 3
i.e. 3y ± 25 = 0
2a2 2 × 16 32
Length of the latus rectum = = =
b 5 5
Also
3 a2 16
be = 5 × = 3 and =
5 b 5
Therefore, the coordinates of the ends of the latera recta are ± 
a2
 b ‚ ± be
– 16  16  – 16
i.e.  ‚ 3‚  
16
 5   5 ‚ 3‚  5 ‚ – 3,  5 ‚ – 3.
Example 3. Find the latus rectum, the eccentricity and the coordinates of the foci of the ellipse
9x2 + 5y2 – 30y = 0.
Solution
We have
9x2 + 5y2 – 30y = 0
or, 9x2 + 5 (y2 – 6y + 9) = 45
x2 (y – 3)2
or, + =1 . . . (1)
5 9
(x – h)2 (y – k)2
Comparing (1) with + = 1, we get
a2 b2
a2 = 5, b2 = 9, h = 0, k = 3
Since 9 > 5 so the major axis of the ellipse is along the y-axis. Thus,
Centre = (h, k) = (0, 3)
a2
Eccentricity (e) = 1– 2
b
5 2
= 1– =
9 3
2a2
Latus rectum =
b
2 × 5 10 2
= = =3
3 3 3
 Conic Section UNIT 9 197

Foci = (h, k ± be)

= 0‚ 3 ± 3 × 
2
 3
= (0, 3 ± 2)
That is, foci are (0, 5) and (0, 1).
Example 4. Find the lengths of major and minor axes and their equations, latus rectum, eccentricity,
centre, foci, directrices, vertices for the ellipse 3x2 + 4y2 – 12x – 8y + 4 = 0.
Solution
The given equation is
3x2 – 12x + 4y2 – 8y = – 4
or, 3(x2 – 4x + 4) + 4(y2 – 2y + 1) = – 4 + 12 + 4
or, 3(x – 2)2 + 4 (y – 1)2 = 12
(x – 2)2 (y – 1)2
or, + =1 . . . (1)
4 3
Now, putting x – 2 = X and y – 1 = Y, then we get
X2 Y2
2 + =1
2 ( 3)2
X2 Y2
Comparing this with + 2 = 1, we get
a2 b
a2 = 4 i.e., a = 2
and b2 = 3 = ( 3)2 i.e., b = 3
 The length of major axis = 2a = 2(2) = 4
and the length of minor axis = 2b = 2( 3) = 2 3
Since, the denominator of X2 > the denominator of Y2, the equation of the major axis is Y = 0,
i.e., y – 1 = 0
Equation of the minor axis is X = 0 , i.e., x – 2 = 0
2b2 2( 3)2
The length of latus rectum = = =3
a 2
Since b2 = a2 (1 – e2), we have,
( 3)2 = 4(1 – e2)
3
or, 1 – e2 =
4
1
 e=
2
The coordinates of foci (ae, 0) and (– ae, 0) are given by X = ± ae
1
 x–2=±2× =±1
2
and Y = 0 i.e., y – 1 = 0 i.e., y = 1
Therefore, the coordinates of foci are (3, 1) and (1, 1).
198 FOUNDATIONS OF MATHEMATICS

Then the centre of the ellipse is given by X = 0 and Y = 0

 x–2=0
or, x = 2
and y – 1 = 0
or, y = 1
Therefore, the centre is (2, 1).
The equations of the directrices are given by
a
X=±
e
2
or, x – 2 = ± = ± 4
1
2
 x = 6, x = – 2
Aliter
(x – h)2 (y – k)2
Compare (1) with + = 1, we get
a2 b2
a2 = 4 and b2 = 3.
Center C(h, k) = (2, 1)
Vertices A(h  a, k) = (2  2, 1) i.e. (4, 1) or (0, 1)
b2
Eccentricity, (e) = 1– 2
a
3 1
= 1– =
4 2
Foci = (h  ae, k)
= 2  2 × ‚ 1
1
 2 
= (2  1, 1) i.e. (3, 1) or (1, 1)
And the directrix,
a
x=h =22×2
e
 x = 6 and x = 2 are directices.
Example 5. Find the equation of the ellipse passing through the points (– 3, 1) and (– 2, 2). Find
also its eccentricity.
Solution
x2 y2
Let the equation of the ellipse be + =1 . . . (1)
a2 b2
Since the ellipse (1) passes through the points (– 3, 1) and (– 2, 2)
9 1
Therefore, 2 + 2 = 1 . . . (2)
a b
4 4
and 2 + 2 = 1 . . . (3)
a b
Solving (2) and (3), we get,
 Conic Section UNIT 9 199

32 32 32
= 3 or, a2 = , and b2 =
a2 3 5
Substituting these values in (1), we get
x2 y2
+ =1
32 32
3 5
or, 3x + 5y2 = 32
2

Now, if e is the eccentricity of the ellipse, then

b2 = a2 (1 – e2) [∵ a > b]
32 32
or,  = (1 – e2)
5 3
3 2
or, e2 = 1 – =
5 5
2
 e =
5
Example 6. Find the eccentricity of an ellipse if its latus rectum is equal to one-half of its major
axis.
Solution
x2 y2
Let the ellipse be + = 1, where a2 > b2
a2 b2
2b2 1 1
By question, latus rectum, = major axis = × 2a = a
a 2 2
2b2 1
or, =  2a
a 2
or, 2b = a2
2

But, b2 = a2(1 – e2)

 2b2 = 2a2(1 – e2)
 2a2 (1 – e2) = a2
or, 2e2 = 2 – 1 = 1
1
Hence, the eccentricity (e) = .
2
Example 7. An arch is in the form of a semi ellipse. It is 12 m wide at the base and has its height of
5 metre. How wide is the arch at a height of 2 m above the base?
Solution
The adjoining figure shows the given arch. The axis is along the base and the origin is
at the midpoint of the base. Then the ellipse has its principal axis on the x-axis, the
center at the origin, a = 6 and b = 5. Therefore
x2 y2 Y
+ =1, (x', 2 )
62 5 2
is the equation of the ellipse. 2m
Now, (x', 2) is the point on the ellipse, we have A O A' X

Y'
200 FOUNDATIONS OF MATHEMATICS

x' 2 4
+ 2=1
36 5
x' 2 4
or, =1–
36 25
6
 x' = 21 m
5
Example 8. A point moves in a plane in such a way that the sum of its distances from two fixed
points (4, 0) and (– 4, 0) is always 10. Find the locus of P.
Solution Y
Let the point be P(x, y) and let two fixed points
P(x, y)
be S(4, 0) and S'(– 4, 0)
PS + PS' = 10
X' X
S'(–4, 0) O S(4, 0)
i.e. (x + 4)2 + (y – 0)2 + (x – 4)2 + (y – 0)2 = 10
or, (x + 4)2 + y2 = 10 – (x – 4)2 + y2
Y'
Squaring both sides
x2 + 8x + 16 + y2 = 100 – 20 x2 – 8x + 16 + y2 + x2 – 8x + 16 + y2
or, 25 – 4x = 5 x2 – 8x – 16 + y2
Again, squaring both sides
625 – 200x + 16x2 = 25x2 – 200x + 400 + 25y2
or, 9x2 + 25y2 = 225
9x2 + 25y2
or, =1
225
x2 y2
 + = 1, which is an equation of an ellipse.
25 9
Example 9. Show that locus of a point which moves such that its sum of the distances from two
fixed points is constant is ellipse:
Solution
Let S'(c, 0) and S(– c, 0) be the foci on the x-axis, then SS' = 2c and (0, 0) is the
midpoint of the segment SS'. If P(x, y) is any point on the ellipse, then
PS' + PS = 2a (say) . . . (1)
or, (x – c)2 + y2 + (x + c)2 + y2 = 2a
or, (x – c)2 + y2 = 2a – (x + c)2 + y2 Y

or, x2 – 2cx + c2 + y2 = 4a2 – 4a (x + c)2 + y2 + x2 + 2cx + c2 + y2 P(x, y)

or, 4a (x + c)2 + y2 = 4a2 + 4cx
or, a2 [x2 + 2cx + c2 + y2] = a4 + 2a2cx + c2x2 X'
S(– c, 0) O S’(c, 0) X
or, x2 (a2 – c2) + a2y2 = a2(a2 – c2)
x2 y2
or, 2 + 2 2 = 1 . . . (2)
a a –c
Y'
As 2a > 2c, from  PS'S, we have a > c and a2 – c2 > 0
Therefore, from (2), we get
 Conic Section UNIT 9 201

x2 y2
2 + = 1, where b2 = a2 – c2
a b2
Thus, if 2a is the constant distance and an ellipse has its foci at (c, 0) and (– c, 0) then
if b2 = a2 – c2, an equation of the ellipse is
x2 y2
2 + = 1, which is the equation of an ellipse.
a b2

 Exercise 9.1
1. For the following ellipses, find the lengths of major and minor axes, the eccentricity,
coordinates of foci and vertices.
a. 16x2 + 25y2 = 400. b. 3x2 + 2y2 = 6.
c. 9x2 + 16y2 = 144.
2. Find the coordinates of the centre, vertices, the eccentricity, foci and the equations of the
directrices of the following ellipses.
(x – 4 )2 (y + 2)2 (x – 1)2 (y – 4)2
a. + = 1. b. + = 1.
100 64 25 169
(x + 3)2 71
c. + y2 = 1. d. x2 + 5y2 + 3x – 10y – = 0.
9 4
3. Find the equation of the ellipse in standard form, satisfying the following conditions
a. vertices at (± 5, 0), foci at (± 4, 0).
4
b. foci at (0, ± 4), eccentricity .
5

foci at ±
10  2
c.
 3 ‚ 0 and eccentricity 3 .
d. axes along coordinate axes, passing through (4, 3) and (– 1, 4).
e. foci at (± 3, 0), passing through (4, 1).
3
f. eccentricity ; foci on y-axis, centre at origin, passing through (6, 4).
4
1
g. focus at (0, – 5) and eccentricity .
3
2
h. latus rectum is 5 and eccentricity .
3
i. foci at (± 2, 0) and latus rectum is 6.
4. Find the equation of the locus of the set of all points, the sum of whose distances from (3, 0) to
(9, 0) is 12.
5. Find the equation of the ellipse whose
a. major axis is twice the minor axis and which passes through the point (0, 1).
b. latus rectum is equal to half of its major axis and which passes through the point ( 6‚ 1) .
202 FOUNDATIONS OF MATHEMATICS

Answers
3 1
1. a. 10; 8; ; (± 3, 0); (±5, 0) b. 2 3 , 2 2 , ; (0; ± 1), (0, ± 3 )
5 3
7
c. 8; 6; ; ( 7 , 0); (4, 0)
4
3
2. a. (4, – 2); (14, – 2); ( – 6, – 2) ; e = ; (10, – 2); (– 2, – 2); 3x – 62 = 0; 3x + 38 = 0
5
12
b. (1, 4); (1, – 9); (1, 17); e = ; (1, 16); (1, – 8); 12y – 217 = 0; 12y + 121 = 0
13
2
c. ( – 3, 0); (0, 0); (– 6, 0); e = 2 ; (– 3  2 2‚ 0) ; 4x + 12 ± 9 2 = 0
3

d. (–23‚ 1) ; (–23  5‚ 1) ; 25 ; (– 23  2 5‚ 1) ; 2x + 3 – 5 5=0

3. a. 9x2 + 25y2 = 225 b. 25x2 + 9y2 = 225 c. 5x2 + 9y2 = 125

247 2 15 2
d. x + y =1 e. x2 + 2y2 = 18 f. 16x2 + 7y2 = 688
7 247
g. 9x2 + 8y2 = 180 0 h. 20x2 + 36y2 – 405 = 0 i. 3x2 + 4y2 – 48 = 0
4. 3x2 + 4y2 – 36x = 0
5. a. x2 + 4y2 = 4 b. x2 + 2y2 = 8

 Multiple Choice Questions

1
1. The equation to the locus of the point whose distance from (3, 0) is of its distance from the
3
line x = 27 is
x2 y2 x2 y2
a. + =1 b. + =1
9 2 81 72
2
x y2 x2
y2
c. + =1 d. + =1
81 2 81 72
x2 y2
2. The eccentricity of the ellipse + = 1 is
25 16
5
a. 3 b.
3
3 4
c. d.
5 5
x2 y2
3. The equation of + + 1 = 0 represents an ellipse only if
1–m m–3
a. m>1 b. m<3
c. m=0 d. 1<m<3
4. The latus rectum of an ellipse 3x2 + 4y2 = 12 is
4
a. b. 3
3
2 1
c. d.
3 2
 Conic Section UNIT 9 203

5. The equation of the ellipse with its centre at (0, 3), directix parallel to x axis, the major axis 12
1
and eccentricity is
2
x2 (y – 3)2 x2 y2
a. + =1 b. + =1
36 27 36 27
x2
(y – 3)2
(x – 3)2 y2
c. + =1 d. + =1
27 36 27 36
6. The foci of an ellipse 9x2 + 4y2 = 36 are
a. (± 5, 0) b. (0, ± 5)
c. (± 3, 0) d. (± 4, 0)
1
7. The equation of the ellipse whose latus rectum is 3 and eccentricity is is
2
a. x2 + 4y2 = 4 b. x2 + 2y2 = 9
c. 3x2 + 4y2 = 9 d. 2x2 + y2 = 1

Answers
1 2 3 4 5 6 7
b c d b a b b

9.2 Hyperbola
A hyperbola is the locus of a point which moves in a plane so that the ratio of its distances from a
fixed point (called focus) and a fixed line called directrix is a constant. This ratio is called
eccentricity and is denoted by e. For a hyperbola, e > 1.

9.2.1 Equation of Hyperbola in Standard Form

Let S be the focus, MZ the directrix and the eccentricity e ( > 1) of the hyperbola.
Y

B L
M' M P(x, y)

X' X
S' (-ae, 0) A' Z' C(0, 0) Z A S(ae, 0)

B'
L'

Y'

Draw the straight line SZ perpendicular to MZ and divide SZ at A such that

SA = eAZ . . . (1)
Since e > 1, there will be another point A' on the line SZ on the other side of S such that
204 FOUNDATIONS OF MATHEMATICS

SA' = e  A'Z . . . (2)

Let AA' = 2a, and O be the mid point of AA' so that OA = OA' = a. As A and A' lie on the
hyperbola, from (1), we have
SA = eAZ
or, OS – OA = e (OA – OZ)
or, OS – a = e (a – OZ) . . . (3)
From (2), we have
SA' = eA'Z
A'O + OS = e (A'O + OZ)
or, a + OS = e (a + OZ) . . . (4)
From (3) and (4), we have [∵ Adding and subtracting equation (3) and equation (4)]
2OS = 2ae  OS = ae
a
and 2a = 2e OZ OZ =
e
a
Let O be the origin, the coordinate of the focus S is (ae, 0) and the equation of directrix MZ is x = .
e
Let P(x, y) be any point on the hyperbola and R be the foot of the perpendicular drawn from P on the
directrix then by definition.
PS = e PM
or,  PS2 = e2 PM2 = e2 ZR2 [∵ PM = ZR]
or, PS2 = e2 (OR – OZ)2
or,  (x – ae) 2 + y2 = e2 (x – OZ) 2

or,  x2 – 2aex + a2e2 + y2 = e2 x – 

a 2
 e
or,  x2 – 2aex + a2e2 + y2 = e2 x2 – 2 x + 2
a a2
 e e
or,  x2 – 2aex + a2e2 + y2 = e2x2 – 2aex + a2
or,  (e2 – 1) x2 – y2 = a2 (e2 – 1)
x2 y2
or,  2 – 2 2 = 1, a2 (e2 – 1) > 0 as e > 1 . . . (5)
a a (e – 1)
Putting b2 = a2 (e2 – 1) in (5), we get,
x2 y2
2 – 2 =1
a b
which is the standard equation of the hyperbola.
NOTE a. Hyperbola is the curve consisting of two branches. For the right branch we have
PS – PS' = – 2a, while for the left branch we have PS – PS' = 2a
b. Since the difference of any two sides of a triangle is less than the third side, from rt PSS', we have
| PS – PS' | < SS'  2a < 2c i.e. a < c. Also, b2 = c2 – a2 which is positive.
 Conic Section UNIT 9 205

x2 y2
In the standard equation of the hyperbola 2 – 2 = 1. It does not matter whether a > b or a = b or
a b
a < b.
1. The point C(0, 0) is called the centre of the hyperbola.
2. The points of intersection of the hyperbola and x-axis, A(a, 0) and A'(– a, 0) are called the
vertices and A'A is called the transverse axis of the hyperbola. The length of the transverse
axis is 2a. The hyperbola does not meet the y-axis in real point. Let B(0, b) and B'(0, – b) be
two points on the y-axis. Then BB' is called the conjugate axis and its length is 2b. The
equation of transverse axis AA' is y = 0. Equation of conjugate axis BB' is x = 0.
b2
3. The eccentricity e = 1+ where e > 1.
a2
4. S(ae, 0) and S'(– ae, 0) are foci of the hyperbola. Focal distances are SP and S'P and they are
ex – a and ex + a, where P(x, y). |SP – S'P| = |(ex – a) – (ex + a)| = 2a length of transverse axis.
a
5. MZ and M'Z' are the directrices. Equation of directrix MZ is x = and equation of the directrix
e
–a
M'Z' is x = .
e
6. The double ordinate LSL' through the focus is the latus rectum of the hyperbola. Since the
x-coordinate of L is ae, its y coordinate is given by
(ae)2 y2
– 2 =1
a2 b
b2
or, y2 = b2(e2 – 1) = b2  ; as b2 = a2(e2 – 1)
a2
b2
 y = ±
a
2b2
So the coordinates of end points of the latus rectum are ae‚ ±  and the length of latus rectum is
b2
.
 a a

9.2.2 Description of Some Terms in the Hyperbola

Asymptotes
x2 y2 b b
Asymptotes are – = 0 or y = x and y = – x.
a2 b2 a a
These are shown by dotted lines in the above figure. These are also called tangents at infinity. The
b
angle between the two asymptotes is 2 tan–1 .
a
Conjugate Hyperbola
The hyperbola whose transverse and conjugate axes are respectively change into the conjugate and
transverse axes of a given hyperbola is called the conjugate hyperbola (vertical hyperbola) of the
given hyperbola.
The conjugate hyperbola of the hyperbola
Y
S (0, be)

b
y=
e B (0, b)

C (0, 0)
X

-b B' (0, -b)

y=
206 FOUNDATIONS OF MATHEMATICS

x2 y2
2 – 2 = 1 is
a b
y2 x2 x2 y2
2 – 2 = 1 or – =–1
b a a2 b2
In this case the transverse axis is y-axis; vertices (0, b),
(0, – b), foci are (0, be) and (0, – be),
b b
Directrices y = and y = –
e e
a2 a2
Eccentricity e2 = + 1 i.e., e = 1+ .
b2 b2
Rectangular Hyperbola
If the angle between asymptotes is 90°, then the hyperbola is called rectangular hyperbola. In this
case
b 
2 tan–1 =
a 2

or, tan–1   =
b
a 4
 b
or, tan =
4 a
b
or, 1 =
a
or, a = b.
x2 y2
Now, the hyperbola – = 1 becomes x2 – y2 = a2 which is the standard equation of rectangular
a2 b2
hyperbola.
Eccentricity
b2 = a2 (e2 – 1)
or, a2 = a2 (e2 – 1) [∵ a = b]
or, e2 – 1 = 1
or, e2 = 2 
 e= 2
Asymptotes: Asymptotes are given by x2 – y2 = 0 or, y = x and y = – x,
Axes: The transverse and conjugate axes of a rectangular hyperbola are equal in length.
Rectangular hyperbola with asymptotes as axes of coordinates Y
Asymptote


If in x2 – y2 = a2, the axes are rotated by keeping the origin unchanged,
4
O
the equation becomes xy = c2. Asymptote
 Conic Section UNIT 9 207

Axes parallel to the axes of coordinates

As in the case of the ellipse, the hyperbola with centre (h, k) and the axes parallel to the axes of
(x – h)2 (y – k)2
coordinates is of the form – = 1.
a2 b2
Hyperbola at a Glance
x 2
y 2
x2 y2 (x – h)2 (y – k)2 (x – h)2 (y – k)2
Equation of hyperbola – =1 – =–1 – =1 – =–1
a2 b2 a2 b2 a 2
b 2
a2 b2
Center (0, 0) (0, 0) (h, k) (h, k)
Vertices (± a, 0) (0, ± b) (h ± a, k) (h, k ± b)
Foci (± ae, 0) (0, ± be) (h ± ae, k) (h, k ± be)
Equation of transverse axis y=0 x=0 y=k x=h
Equation of conjugate axis x=0 y=0 x=h y=k
Length of transverse axis 2a 2b 2a 2b
Length of conjugate axis 2b 2a 2b 2a
a b a b
Equation of directrices x=± y= ± x=h± y=k±
e e e e
b2 a2 b2 a2
Eccentricity e= 1+ e= 1+ e= 1+ e= 1+
a2 b2 a2 b2
2b2 2a2 2b2 2a2
Latus rectum
a b a b

 Illustrative Examples
Example 1. Find the lengths of the transverse and conjugate axes, coordinates of the foci, vertices,
eccentricity and latus rectum for the hyperbola.
a. 9x2 – 16y2 = 144
y2 x2
b. – =1
36 64
Solution
a. The given equation of the hyperbola is
9x2 – 16y2 = 144
x2 y2
or, – =1 . . . (1)
16 9
x2 y2
Comparing (1) with 2 – 2 = 1, we get
a b
a2 = 16  a = 4
and b2 = 9  b = 3
Now
b2 = a2 (e2 – 1)
b2 9
or, e= 1+ = 1+
a2 16
208 FOUNDATIONS OF MATHEMATICS

5
i.e. e=
4
Now, the length of transverse axis is, 2a = 2 × 4 = 8
The length of conjugate axis is, 2b = 2 × 3 = 6
Coordinates of the foci are (ae, 0) and (– ae, 0)

i.e. 
4×5   4×5 
 4 ‚ 0 and – 4 ‚ 0 i.e. (5, 0) and (– 5, 0)
Vertices are (a, 0) and (– a, 0), i.e. (4, 0) and (– 4, 0)
5 2b2 9 9
Eccentricity, e = , and length of latus rectum = =2× =
4 a 4 2
b. The given equation of the hyperbola
y2 x2
– =1 . . . (1)
36 64
y2 x2
Comparing (1) with 2 – 2 = 1, we get
b a
2
b = 36 i.e., b = 6
and a2 = 64 i.e., a = 8
Transverse axis along x-axis, we have
a2 = b2 (e2 – 1)
or, 64 = 36 (e2 – 1)
64
or, e2 – 1 =
36
64
 or, e2 = +1
36
100
 or, e2 =
36
5
  e=
3
Now, length of transverse axis is, 2b = 2 × 6 = 12
Length of conjugate axis is, 2a = 2(8) = 16
The coordinates of the foci are (0, + be) and (0, – be)
i.e. (0, 10) and (0, – 10)
Vertices are (0, b) and (0, – b) i.e. (0, 3) and (0, – 3)
5
The eccentricity e =
4
2a2 2 × 64 64
and the length of latus rectum = = = .
b 6 3
Example 2. Find the equation of the hyperbola, in which the distance between foci is 16 and
eccentricity is 2 .
Solution
We have, e = 2
 Conic Section UNIT 9 209

In case of hyperbola,
b2 = a2 (e2 – 1)
or, b2 = a2 (2 – 1) = a2
 b = a
Also the distance between the foci, 2ae = 16
i.e. 2a 2 = 16
or, a = 4 2 and b = 4 2
x2 y2
Hence, the required equation of the hyperbola is – =1
a2 b2
x2 y2
i.e. – =1
32 32
 x2 – y2 = 32
Example 3. Find the equation of the hyperbola whose focus is (2, 2) eccentricity 2 and directrix
x + y = 9.
Solution
Let, P(x, y) be any point on the hyperbola whose Y
focus is S(2, 2). Let PM be the perpendicular from
P on the directrix x + y – 9 = 0 P(x, y)
S(2, 2)
Then by definition,
M
SP
=e
PM
or,SP2 = e2 PM2 X
O
or, (x – 2)2 + (y – 2)2 = 22 
x + y – 9 2
x+y=9
 1+1
or, x2 + y2 – 4x – 4y + 8 = 2(x2 + y2 + 81 + 2xy – 18x – 18y)
 x2 + y2 + 4xy – 32x – 32y + 154 = 0
Example 4. Find the centre, eccentricity, foci and directrices of the hyperbola
9x2 – 16y2 – 72x + 96y – 144 = 0
Solution
9(x2 – 8x) – 16(y2 – 64) = 144
or, 9(x2 – 8x + 16) – 16(y2 – 6y + 9) = 144
or, 9(x – 4)2 – 16 (y – 3)2 = 144
(x – 4)2 (y – 3)2
or, – =1 . . . (1)
16 9
Let x – 4 = X and y – 3 = Y
Then (1) can be express as
X2 Y2
– =1 . . . (2)
16 9
For the coordinates of the centre
X = 0 i.e.,x – 4 = 0 i.e., x = 4
210 FOUNDATIONS OF MATHEMATICS

and Y = 0 i.e.,y – 3 = 0 i.e.,y = 3

 The coordinates of the centre are (4, 3)
X2 Y2
Comparing (2) with 2 – 2 = 1, we get
a b
a2 = 16 i.e.,a = 4
and b2 = 9 i.e.,b = 3
Now
b2 = a2 (e2 – 1)
or, 9 = 16 (e2 – 1)
5
 e =
4
Since, the transverse axis lies along the x-axis, the coordinates of the foci are;
X = ± ae
5
or, x – 4 = ± 4 × = ± 5
4
i.e. x = 4 ± 5 and
Y=0
or, y – 3 = 0
or, y = 3
Therefore, foci are; (9, 3) and (– 1, 3).
Equations of the directrices are;
a
X=±
e
4
or, x – 4 = ± 4 ×
5
16
or, x – 4 = ±
5
or, 5x – 20 = ± 16
or, 5x – 36 = 0 and 5x – 4 = 0
Aliter
(x – h)2 (y – k)2
Compare (1) with – =1
a2 b2
where, a2 = 16 i.e., a = 4
and b2 = 9 i.e., b = 3
The centre C(h, k) = (4, 3), i.e. h = 4 and k = 3.
b2 9 5
Eccentricity, e = 1+ 2= 1+ =
a 16 4

Foci, (h  ae, k) = 4  4 × ‚ 3 i.e. (0, 3) or (– 1, 3)

5
 4 
a 4
Directrices, x = h  = 4  4×
e 5
 5x – 36 = 0
and 5x – 4 = 0
 Conic Section UNIT 9 211

Example 5. Find the equation of the locus of all points such that the differences of their distances
from the fixed points (4, 0) and (– 4, 0) is always equal to 2.
Solution
Let S(4, 0) and S'(– 4, 0) be the given two points.
Let P(x, y) be any point on the locus.
By definition
|PS – PS'| = 2
or, (x – 4)2 + y2 – (x + 4)2 + y2 = 2 
or, (x – 4)2
+ y2
= 2 + (x + 4)2 + y2
Squaring both sides, we get
(x – 4)2 + y2 = 4 + 4 (x + 4)2 + y2 + (x + 4)2 + y2
or, 4 (x + 4)2 + y2 = – 16x – 4
Again squaring both sides, we get
(x + 4)2 + y2 = (– 4x – 1)2
or, x2 + 8x + 16 + y2 = 16x2 + 8x + 1
or, 15x2 – y2 = 15
x2 y2
or, – =1
1 15
which is the required equation of the locus (hyperbola).

Example 6. Find the equation of the hyperbola whose foci are (0, ± 10) and which passes through
the point (2, 3).
Solution
Clearly, the transverse axis lies along y-axis. So the required equation of the hyperbola is
y2 x2
– =1 . . . (1)
b2 a2
Now
be = ± 10
or, b2e2 = 10 . . . (2)
Since, (1) passes through the point (2, 3), we have
9 4
– =1 . . . (3)
b2 a2
And, we have
a2 = b2 (e2 – 1)
or, a2 = b2e2 – b2
or, a2 = 10 – b2
i.e. b2 = 10 – a2
Putting the value of b2 in (3), we get
9 4
– =1
10 – a2 a2
212 FOUNDATIONS OF MATHEMATICS

9 4
 =1+ 2
10 – a2 a
 9a2 = 10a2 + 40 – 4a2 – a4
 (a2 + 8) (a2 – 5) = 0
i.e. a2 = 5 or a2 = – 8, not possible and b2 = 10 – 5 = 5
Substituting these values in (1), we get
y2 x2
– =1
5 5
i.e., y2 – x2 – 5 = 0
Example 7. Show that the locus of points in a plane such that the absolute value of the difference
between the distance of a point from two fixed points is constant, is a hyperbola.
Solution
Let XOX' and YOY' be the Y

coordinate axes. Let the two M' M

fixed points be S(c, 0) and P(x, y)
S'(– c, 0), where c > 0. Let
P(x, y) be any point on the R
X' X
locus and let the difference S ' (-c, 0) A' Z' O Z A S(c, 0)
of distances of P from S and
S' be a positive constant, say
2a, where 0 < a < c. Y'
Then, by definition
|PS – PS'| = 2a
or, PS – PS' = ± 2a
or, (x – c)2 + y2 – (x + c)2 + y2 = ± 2a
or, (x + c)2 + y2 = (x – c)2 + y2 ± 2a
or, (x + c)2 + y2 = (x – c)2 + y2 ± 4a (x – c)2 + y2 + 4a2 [Squaring on both sides]
or, 4cx – 4a = ± 4a (x – c) + y
2 2 2

c
or, x – a = ± (x – c)2 + y2
a
c2
or, 2 x2 – 2cx + a2 = (x – c)2 + y2 [Again squaring on both sides]
a
c2
or, 2 x2 – 2cx + a2 = x2 – 2xc + c2 + y2
a

or,  2 – 1 x2 – y2 = c2 – a2
c2
a 
or, (c2 – a2) x2 – a2y2 = a2 (c2 – a2)
x2 y2
or, 2 – 2 =1
a c – a2
x2 y2
or, 2 – 2 = 1,
a b
where b2 = c2 – a2 which is the required equation of the hyperbola.
 Conic Section UNIT 9 213

 Exercise 9.2
1. For the following hyperbolas, find the lengths of transverse and conjugate axes, vertices,
eccentricity and the coordinates of foci
x2 y2 x2 y2
a. – = 1. b. – = – 1.
100 25 9 16
c. 2x2 – 3y2 = 6. d. 9x2 – 16y2 – 18x – 64y – 199 = 0.
2. Show that the equation 9x2 – 16y2 + 18x + 32y – 151 = 0 represents a hyperbola. Find its
eccentricity, centre, foci and directrices.
3. Find the standard equation of the hyperbola with
a. vertices are of (± 6‚ 0) and one of the directrix is x = 4.
b. vertices of (± 5, 0), foci (± 7, 0).
4
c. vertices of (0,±7), eccentricity, e = .
3
d. vertices of (0, ± 8) and passing through (4‚ 8 2) .
e. length of conjugate axis 8 and the distance between the foci 10.
3
f. length of latus rectum is 8 and the eccentricity, e = .
5
Answers
5 5
1. a. 20; 10; (± 10‚ 0); ; (± 5 5‚ 0) b. 8; 6; (0‚ ± 4); ; (0‚ ±5)
2 4
5 5
c. 2 3; 2 2 ; (± 3‚ 0); ‚ (± 5‚ 0) d. 8; 6; ; centre (1‚ – 2); (5, – 2); (– 3, – 2); (6‚ – 2); (– 4‚ – 2)
3 4
5
2. ; (– 1‚ 1); (4‚ 1); (– 6‚ 1); 5x – 11 = 0; 5x + 21 = 0
4
x2 y2
3. a. – =1 b. 24x2 – 25y2 = 600 c. 9x2 – 7y2 + 343 = 0
36 45
y2 x2 x2 y2 x2 y2
d. – =1 e. – =1 f. – =1
64 16 9 16 25 20

 Multiple Choice Questions

1. The angle between the asymptotes of the hyperbola 27x2 – 9y2 = 24 is
 5
a. b.
6 6
 
c. d.
3 2
2. What is the equation of the hyperbola in standard position such that its transverse axis is 4 and
conjugate axis is 5?
x2 4y2 x2 y2
a. – =1 b. – =1
25 25 4 25
2 2
x y x x2
2
c. – =1 d. – =1
25 4 4 25
214 FOUNDATIONS OF MATHEMATICS

3. The equation of the hyperbola with a focus at (7, 0) and a vertex at (5, 0) is
x2 4y2 y2 x2
a. – =–1 b. – =1
4 24 24 25
x2 y2 x2 y2
c. – =1 d. – =1
25 24 24 25
4. The foci of the hyperbola 9x2 – 16y2 = 144 is
a. (± 5, 0) b. (± 4, 0)
c. (0, ± 5) d. (0, ± 4)
7
5. The equation of a hyperbola with a focus at (– 7, 0) and eccentricity is
4
x2 y2 x2 y2
a. – =1 b. – =1
16 3 16 33
2
y x2 x2
y2
c. – =1 d. – =–1
16 33 16 33
x2 y2
6. The eccentricity of the hyperbola – = 1 is
36 64
3 3
a. b.
5 4
5 4
c. d.
3 3
x2 y2
7. Which of the following is not true for – = 1?
a2 b2
a. The curve of the hyperbola is symmetric with respect to both the axes.
b. The eccentricity of the hyperbola is greater than unity.
b
c. y = ± x are called the asymptotes of the hyperbola.
a
a
d. Equation of its directrices are y = ± .
e

Answers
1 2 3 4 5 6 7
c a c a b c d

Conic Section UNIT 9 215
216 FOUNDATIONS OF MATHEMATICS


9.1 Ellipse ...................................................................................................... 189
9.1.1 Equation of Ellipse in Standard Form ......................................... 189
9.1.2 Description of Some Terms in Ellipse ........................................ 191
9.1.3 Generalized Equation of Ellipse .................................................. 193
 Illustrative Examples ............................................................................ 195
 Exercise 9.1 .......................................................................................... 201
 Multiple Choice Questions ................................................................... 202
9.2 Hyperbola ................................................................................................ 203
9.2.1 Equation of Hyperbola in Standard Form ................................... 203
9.2.2 Description of Some Terms in the Hyperbola ............................. 205
 Illustrative Examples ............................................................................ 207
 Exercise 9.2 .......................................................................................... 213
 Multiple Choice Questions ................................................................... 213