5

Conic Section

A section cut off from a circular (not necessary a right circular) cone by a plane in
various way is a conic section. Its shape depends upon the position of the plane.
Consider a double right circular (nappe) cone of semi vertical angle  and let it be
cut by a plane inclined at an angle  to the axis of the cone. We will get different
sections (curves) as follows.
S P  (Semi vertical
Case I: If the plane passes through the vertex
angle)
O(0,0)
V Pair of lines
The curve of intersection is pair of straight lines Vertex
passing through the vertex which are Q R
i) Real and distinct for  < 
axis
ii) Coincident for  =  i.e. plane touches the cone
iii) Imaginary for  > 
Case – II :- If the plane does not pass through the vertex O(0,0)
The curve of intersection is called

i) a circle if  = 2 i.e. plane intersects cone perpendicular to axis

ii) a parabola for  =  i.e. if the plane not passing through vertex is parallel to the
generator PQ
 
iii) an ellipse for  >    2  i.e. if the plane cuts both the generating lines PQ
 
and RS (with the axis greater then semi-vertical angles)
 Mathematics | 286
iv) a hyperbola for  <  i.e. if the plane cuts both the cones (angles between the
axis and the plane is less than the semi vertical-angle)
Thus, we may get the section either as a pair of straight line, a circle, a parabola, an
ellipse or hyperbola depending upon the different positions of the cutting plane. The
curves of intersection are called the conic sections which have been clearly shown in
the following figures.

=  

For circle For ellipse

(=  ) (     ) For parabola For parabola
 ( = ) (  )

Definitions
i) Conic section:- A conic section or conic is the locus of a point (say P) which
moves in such a way that its distance from a fixed point (say S) always bears a
constant ratio to the perpendicular distance from a fixed line all being in the
same plane.
ii) Focus:- The fixed point is called focus of the conic section.
iii) Directrix:- The fixed straight line is called the directrix of the conic section.
I n general every conic has four foci, two of them are real and the other two are
imaginary. Due to two real foci, every conic has two directrix corresponding to
each real focus
iv) Eccentricity:- The constant ratio is called eccentricity of the conic section and
is denoted by e.
On the basis of e, conic sections are classified as follows.
a) for e → , the conic is obtained as straight line
b) for e > 1, the conic is obtained as hyperbola
c) for e = 1, the conic section is obtained as parabola
d) for e < 1, the conic section is obtained as ellipse
e) If e = 0, the conic section is obtained as circle
 Mathematics | 287
v) Axis:- The straight line passing through the focus and perpendicular to the
directrix is called axis of the conic.
vi) Vertex:- The point of intersection of the conic section and the axis are called
vertices of the conic.
vii) Centre:- The point which bisects every chord of conic section through it is
called centre of conic section.
Directisc Latus rectum
viii) Latus-Rectum:- The chord passing through the
focus and perpendicular to the axis is called latus P(x, y)
M
rectum
axis
Here, vertex S
(focus)
PS
e = PM . For parabola e = 1. So, PS = PM

Equation of the parabola in its standard form

Let, xx1 be axis, S(a, 0) be focus, A(0, 0) be vertex and MZ be the directrix of the
parabola. Let P(x, y) be any point on the parabola,
y
PM ⊥ MZ.
L y = 4ax
2

By the definition of parabola AS = AZ.

(–a, y)M
So, co-ordinates of Z is (– a, 0) and M is (– a, y) Z
x' x
Now, PS = PM (By definition) (–a, 0) A S(a, 0)
(0, 0)
 PS2 = PM2
L'
 (x – a)2 +(y – 0)2 + = {x – (–a)}2 + (y – y)2 y'
 x2 – 2ax + a2 + y2 = x2 + 2ax + a2 + 0
 y2 = 4ax, is the equation of parabola in standard form.
Tracing of the parabola y2 – 4ax, a > 0
Given equation can be written as y =  2 ax
a. Origin: The curve passes through origin and the tangent at origin is x = 0.
b. Intersection with the axes: The curve meets the co–ordinate axes only at the
origin.
c. Region: For every negative value of x, the value of y is imaginary, therefore no
part of the curve lies left of y –axis.
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d. Symmetry: For every positive value of x, there are two equal and opposite
value of y.
e. Position occupied: As x → , y → . Therefore, the curve extends to infinity
to the right of axis of y.
Note: Equation of directrix as per above figure is x = – a ⇒ x + a = 0
Some more terms
a. Double ordinate: A chord passing through P and perpendicular to the axis of
parabola is called double ordinate.
b. Latus Rectum: A double ordinate through the focus is called latus rectum.
Hence, latus rectum of a parabola is a chord passing through the focus and
perpendicular to the axis.
Form above figure, when we put x = a in y2 = 4ax, we get, y =  2a
Hence two ends of latus rectum are L (a, 2a) and L1(a, – 2a).
Also, length of latus rectum = 2a + 2a = 4a
Note: [Latus rectum = 2 [length of perpendicular from the focus on directrix]
c. Focal distance of any point : The distance of P(x, y) from the focus S is called
the focal distance of the point P.
 Focal distance (SP) = (x – a2) + (y – 0)2
= (x – a)2 + y2 = (x – a2) + 4ax = (x + a)2 = x + a.
d. Focal chord : A chord of a parabola passing through the focus is called focal
chord.
All the standard forms of parabola
1. y2 = 4ax 2. y2 = – 4ax
y y
L y = 4ax
2
Direction
(–a, y)M M
(x, y)p
Z x
x' x x'
(–a, 0) A S(a, 0) S(– a, 0) A(0,0)
(0, 0)
L' y2 = – 4ax
y' x=a
y'
 Mathematics | 289

3. x2 = 4ay 4. x2 = – 4ay

y y
x2 = 4ay
M y=a
S(0, a )
A(0, 0)
x' x
p(x, y)
x' x p(x, y)
(0, 0)A S
y=–a
M
x2 = – 4ay
y' y'

Regarding the shapes of the curves in the four-standard form, their corresponding
results are given below in the table:

Co–ordinate
Equation Length of Focal
Co–ordinate Equation Symmetry
Equation of latus distance of Opens
of Vertex of focus of the axis (about)
directrix rectum point P(x, y)

y2 = 4ax (0, 0) (a, 0) x = –a y=0 4a x+a Right x – axis

x2 = – 4ax (0, 0) (–a, 0) x=a y=0 4a a–x left x– axis

x2 = – 4ay (0, 0) (0, a) y = –a x=0 4a y+a Up y – axis

x2 = – 4ay (0, 0) (0, –a) y=a x=0 4a a–y down y – axis

Equation of a parabola with its axis parallel to x–axis and vertex at any point (h, k)
Let a be the distance between vertex and focus, axis y = k is parallel to x – axis. By
definition
y
PS = PM
 PS2 = PM2
M
 {x – (h + a)} + (y – k) = {x – (h – a)}
2 2 2
a a y=k
z A S(h + a, k)
 (x – h)2 – 2(x – h) a + a2 + (y – k)2 (h, k)
h– a
= {(x – h) + a}2
 (x – h)2 – 2(x – h) a + a2 + (y – k)2 x
O
= (x – h) + 2 (x – h) a + a
2 2

 (y – k)2 = 4a(x – h) is the required equation.

 Mathematics | 290
Note: Equation of a parabola with its axis parallel to the y–axis and the vertex at
(h, k) is (x – h)2 = 4a(y – k)
Corresponding results of above two equations are as follows
Latus Symmetry
Equation Vertex Focus Directrix Axis Opens
Rectum (about)

(y – k)2 =
(h, k) (h + a, k) x = h – a y = k 4a y=k Right
4a(x – h)
(x –h)2 =
(h, k) (h, k + a) y = k – a x = h 4a x=h Above
4a(y – k)

Worked out Examples

1. Find the vertex, focus, equation of directrix, axis and length of latus rectum for the
following parabolas.
a. y2 = 12x
b. x2 = 12y
c. y2 = 6y – 12x + 45
d. x2 + 2y – 3x + 5 = 0
Solution:-
a. Given y2 = 12x …………… (i) Comparing with y2 = 4ax, we get, a = 3
(i) Vertex = (0, 0)
(ii) Focus = (a, 0) = (3, 0)
(iii) Equation of directrix is, x = – a  x = –3  x+3 = 0
(iv) Axis is x-axis, i.e. y = 0
(v) Latus rectum = |4a| = | 4× 3| = 12
b. Given x2 = 12y …………………. (i)
Comparing (i) with x2 =4ay, we get, a = 3
(i) Vertex = (0, 0)
(ii) Focus = (0, a) = (0, 3)
(iii) Equation of directrix is, y = – a  y = –3  y + 3 = 0
(iv) Latus rectum = |4a| = | 4 × 3| = 12
 Mathematics | 291
c. Given, y = 6y – 12x +45
2

 y2 – 6y + a = – 12x + 45 + 9 → (y – 3)2 = – 12x + 54

 9
 (y – 3)2 = (– 12) x – 2 …… (i)
 
9
Comparing with (y – k)2 = 4a (x – h), we get, h = 2 , k = 3,

4a = – 12  a = – 3
9 
(i) Vertex = (h, k) = 2 3
 
9  3 
(ii) Focus = (h + a, k) = 2 – 3  3 = 2  3
   
9
(iii) Equation of directrix is, x = h – a  x = 2 – (– 3)  2x – 15 = 0

(iv) Length of Latus rectum = |4a| = | 4 × (– 3)| = |– 12| = 12

d. Given parabola is
x2 + 2y – 3x + 5 = 0
 x2 – 3a = – 2y – 5
3 32 32
 x2 – 2. x . 2 + 2 = – 2y – 5 + 2
   
 32  11
 x – 2 = – 2 y + 8  …… (i)
   
Comparing with (x – h)2 = 4a (y – k) we get,
3 – 11 –1
h = 2 , k = 8 , 4a = – 2  a = 2

3 – 11
(i) Vertex = (h, k) =   
2 8 
3 – 11 – 1 3 – 15
(ii) Focus = (h, k + a) = 2  8 2  = 2  8 
   
– 11 – 1
(iii) Equation of directrix is, y = k – a  y = 8 2  8y + 15 = 0
 –1
(iv) Length of Latus rectum = |4a| = 4 ×  2  = |–2| = 2
  
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2. Find the equation of the parabola satisfying the following conditions
a. Focus at (4, 0), vertex at (0 ,0)
b. Focus at (0, -4), vertex (0 ,0)
Solution :
a. Focus = (a, 0) = (4, 0)  a = 4
We have equation of parabola is
y2 = 4ax  y2 = 4 × 4 × x  y2 = 16x
b. Focus = (0, a) = (0, – 4)  a = – 4
We have, equation of parabola is,
x2 = 4ay  x2 = 4 × (– 4) × y  x2 = – 16y
y
3. Find the equation of parabola whose vertex is
(4, 2) and focus is (5, 2)
Solution: Since, y – co–ordinate of the vertex and the
focus are equal, the axis is parallel to x – axis. A S (5, 2)
(4, 2)
Given, (h, k) = (4, 2) = vertex and
focus = (h + a, k) = (5, 2) x
O
Then, h + a = 5  4 + a = 5  a = 1
Now, equation of the parabola is (y – k)2 = 4a (x – h)
 (y – 2)2 = 4 × 1. (x – 4)
 y2 – 4y + 4 = 4x – 16
 y2 – 4y – 4x + 20 = 0 is required parabola.
4. Find the equation of parabola whose vertex is (4, 2) and focus is (4, 3).
Solution: Since, y – co–ordinate of the vertex and the focus are equal, the axis is
parallel to x – axis.
Given, (h, k) = (4, 2) = vertex and
focus = (h, k + a) = (4, 3)
Then, k + a = 3  2 + a = 3  a = 1
Now, equation of the parabola is (x – h)2 = 4a (y – k)
 (x – 2)2 = 4 × 1. (y – 4)
 Mathematics | 293
 x2 – 4x + 4 = 4y – 16
 x2 – 4x – 4y+ 20 = 0 is required parabola.
5. Find the equation of the parabola whose focus is (1, – 1) and vertex is (2, 1). Also
find its axis and latus rectum.
Solution:
Let, M (x1, y1) be any point on the directrix.
x1 + 1 y1 + (– 1) directrix
 2 = 2 and =1
2 p(x, y)
 x1 = 3, y1 = 3
S
 axis meets directrix at (3, 3) axis
A M (x1, y11)
(1, – 1)
1 – (– 1) (2, 1)
Slope of axis (m1) = 2 – 1 = 2

1 -1
Slope of directrix (m2) = m = 2
1

-1
Equation of directrix is y – 3 = 2 (x - 3)  x + 2y – 9 = 0

Let, P(x, y) be any point on the parabola, then

Distance of P from focus = Perpendicular distance of P from directrix
x + 2y - 9
 (x - 1)2 + (y + 1)2 =   2 
 1 + 22 
Squaring,
(x + 2 y - 9 )2
(x – 1)2 + (y + 1)2 = 5
 4x2 + y2 – 4xy + 8x + 46y – 71 = 0
Now,
Slope of axis (m1) = 2
And axis passes through (1, -1), then equation of axis is
y – (-1) = 2 ( x – 1 )  2x – y – 3 = 0
Also,
Latus rectum = 2 [perpendicular length from focus on the directrix]
= 2 [length of the perpendicular from (1, -1) on x + 2y – 9 = 0]
 Mathematics | 294

1 - 2 - 9 
=2 =4 5
 1+4 
6. Find the equation of the parabola in which ends of the latus rectum have the co-
ordinates (-1,5) and (-1,-11) and the vertex is (-5,-3)
Solution:- Since the end points of latus rectum are (-1, 5) and (-1, -11). So, the
equation of latus rectum is x = -1. Hence, latus rectum is parallel to y – axis. So, axis
is parallel to x-axis.
Also, vertex = (h, k) = (-5, -3)
Now, equation of parabola is
(y – k)2 = 4a (x + h)
 (y + 3)2 = 4a (x + 5)…….(i)
Equation (i) also passes through (-1, 5) then
(5 + 3)2 = 4a (-1 + 5)  a = 4
Putting the value of ‘a’ is (i)
(y + 3 )2 = 4 × 4 (x + 5)
 y2 + 6y – 16x – 71 = 0 is the required parabola
Exercise
1. Find vertex, focus, axis, equation of directrix and length of latus rectum of the
following parabolas.
a. y2 = 8x b. y2 – 4y – 3x + 1 = 0-
c. 4x2 + y = 0 d. x2 + y = 6x – 14
2. Find the equation of the parabola whose focus is the point (2, 3) and directrix is
the line x – 4y + 3 = 0. Also find the length of its latus rectum.
3. Find the equation of the parabola if
a. Focus is at (0, – 3) and vertex is at (0, 0).
b. Focus is at (0, – 3) and vertex is at (– 1, – 3).
c. Focus is at (– 6, – 6) and vertex is at (– 2, 2).
Answer
1. a. (0, 0), (2, 0), y = 0, x = – 2, 8
– 1  7
b. (– 1, 2),  4  2 , y = – 2, x = – 4 , 3
 
 Mathematics | 295

 1 1 1
c. (0, 0), 0 – 16 , x = 0, y = 16 , 4
 
 – 21
d. (3, – 5), 3 4  , x = 3, 4y + 19 = 0
 
14
2. 16x2 + y2 + 8xy – 74x – 78y + 212 = 0,
17
3. a. x2 – 12y b. y2 + 6y – 4x + 5 =0
c. (2x – y)2 + 4(26x + 37y – 31) = 0

Multiple Choice Questions

Choose the best answer
1. The parabola y2 = – 4ax with positive 'a' opens
(a) up (b) down (c) left (d) right
2. The directrix of y2 = – 4ay is
a) x = a (b) x = – a (c) y = – a (d) y = a
3. x2 = 4ay is symmetric about
(a) y-axis (b) x-axis (c) both y-axis and x-axis (d) none
4. Focus of the parabola x2 + 10y = 0 is
3   3
(a) (0, 0) (b) 2 0 (c) 0 2  (d) none
   
Answers
1. c 2. d 3. a 4. b

Activities and project work

1) A prepare a concrete material to show parabola by using thread and nail in
wooden panel.
2) Prepare a project work to show the application of parabolic curve in everyday
life.
 Mathematics | 296

Application of Parabola
Parabola is in a projectile thrown shape. So, it has many application. Parabolic
reflectors have the property that the light rays or sound waves coming parallel to its
axis converge at the focus and then it reflects them parallel to the axis. Due to this
property, parabolic reflectors are used in cars, automobiles, loudspeakers, solar
cookers, telescopes etc.
If the road way of a suspension bridge is loaded uniformly per horizontal meters, the
suspension cable hangs in the form of arcs which closely approximate to the
parabolic arcs. Therefore, parabolic arcs are used in suspension cable bridge
construction.

Example-1
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:-
Let LAM be the parabolic reflector such that LM is its diameter
L
and AN is its depth. It is given that AN = 5 cm and LM = 20 cm
10 cm
 LN = 10 cm A x
5 cm
Taking A as origin, AX along x-axis and a line through A 10 cm
perpendicular to AX as y=axis, let the equation of the reflector be
M
y2 = 4ax ………… (i)
The point L has co-ordinates (5, 10) and lies on (i)
 10 = 4a × 5  a = 5.
So, the equation of the reflector is y2 = 20x.
Its focus is at (5, 0) i.e. at the point N.
Hence, the focus is at the mid-point of the given diameter.

Example-2
The focus of a parabolic mirror is at distance 6 cm from its vertex. If the mirror is 20
cm deep, find the distance LM as given in figure below,
Solution: Let the axis of the mirror be along the positive direction of x-axis and the
vertex A be the origin.
L(20, LN)

A 20 CM
x
S(6,0) N(20,0)

M
 Mathematics | 297
Since, the focus is at a distance of 6 cm from the vertex. Then, the co-ordinates of the
focus are (6, 0). Therefore, the equation of the parabolic section is, y2 = 24x [Putting
a = 6 in y2 = 4ax]
Since, L(20, LN) lies on this parabola,
(LN)2 = 24 × 20  LN = 4 30
 LM = 2LN = 8 30 cm.

••