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Ghonse Maths Academy,
A’nagar

Conic Section

Double napped right circular cone :

Conic Sections :

i) Circle

ii) Parabola

iii) Ellipse iv) Hyperbola

A plane cutting the double napped right circular
cone in different positions will produce sections
shown in the table given below :

Conic section - Parabola
XI - (Sci
...

2
They are obtained by taking sections of a
double napped right circular cone by a plane shown
in figures
...

Conic sections have wide range of applications
5
such as planetary motion, in designs of telescopes
6
and antennas, reflection in flash light, automobile
headlights, construction of bridges, navigation,
*
projectiles etc
...

Plane is parallel to
the generator
Plane is parallel to the axis

Section
a point
a pair of lines
a circle
an ellipse
a parabola
a hyperbola

Focus Directrix Property of Conic :
It is the property of conic used to derive standard
equations of parabola, ellipse and hyperbola
...

“The locus of point which moves so that the ratio
of its distance from a fixed point to its distance
from a fixed line is constant
...

Fixed point is called as focus of conic, fixed line is
called as directrix of conic, the constant ratio is
called eccentricity of conic
...


By focus directrix property for any conic
SP
 e , Where e is eccentricity of conic
...

iii) If e > 1, conic is called as hyperbola
...
Axis of Conic : The line about which given conic
is symmetric is called axis of conic
...
Focal chord : The chord passing through focus
of conic is called as focal chord
...
Latus Rectum (L
...
) : The focal chord that is
perpendicular to axis of conic is called as ‘Latus
Rectum’ of conic
...
R
...
Centre of a Conic : The point which bisect every chord of the conic passing through it, is called
the centre of the conic
...
Double ordinate : A chord passing through any
point on the conic and perpendicular to the axis is
called double ordinate
...

i
...
SP = PM
...
Vertex of Conic : The point of intersection of
conic and its axis is called vertex of conic
...
Focal distance : The distance of a point on a
conic section from its focus is called the focal distance of the point
...

 Equation of directrix is x = -a
consider point P x1 , y1 is any point on parabola
...

l (PM)
= l(PN) + l(MN)
= x1 + a
By focus directrix property for any conic
...
(i)
 SP = PM
By distance fromula,

b x  a g  b y  0g
2

1

1

2

 x1  a

Squaring on both sides
...






Tracing of parabola : y 2  4ax, a  0
1) Symmetry : Equation of the parabola can be written as y  2 ax i
...
for every value of x, there
are two values of y which are negatives of each
Standard equation of parabola :
other
...
S is
axis
...

nary therefore no part of the curve lies to the left
Draw SZ  directrix d
...

is mid point of SZ
...

the tangent at the vertex is Y-axis
...

x  a
x
y
5)
Extent of parabola : As x   , y  
...


b g

b g

1 1

X'

Z

ba,0g

O

b g

S a,0

X

*

d

Focal Distance of a point :
Let P x1 , y1 be a point on the parabola

b

g

y 2  4ax with focus at S(a, 0) and directrix d
Y'

Conic section - Parabola
XI - (Sci
...
consider LL' is the L
...
of
parabola y 2  4ax
...

 x - co-ordinate of L and L' is a
...

Y

ba,2Lag

M
X'

Z
a,0

b g

O

b g

y  4ax
2

X

S a,0
L'
ag
ba,2

d
Y'






Put x = a in y 2  4ax
y 2  4a 2
y  2a
End points of L
...
for standard parabola

b g

b

y 2  4ax are, L  a,2a and L'  a,2a
Length of L
...
is

b g ba  a g  b2a  2 a g
 0  b 4a g
 b 4a g  4a

 LL' 

2

g

2

 Length of L
...
= 4a

* Other forms of parabola :
I) Parabola y 2  4ax :
Y

X'

baL,2ag
S

O

ba,0g

ba,L'2ag

b g

Z a,0

Equation of parabola is y 2  4ax
Axis of parabola is X-axis, its equation is y=0
...
R
...
R
...
R
...
R
...
R
...

Focal distance of point P is
...
from (i)

X

Y

Z b0, bg

yb

O

X'

b2b, bg L'

X

S b0,bg

Lb2b, bg

x 2  4 by
Y'

i) Equation of parabola is x 2  4 by
ii) Axis of parabola is y-axis, its equation is x= 0
iii) Vertex of parabola is origin O(0, 0)
iv) Focus of parabola is S (0, - b)
v) Equation of directrix is y = b
vi) End points of L
...
are L(2b, -b)& L' 2 b, b
vii) Equation of L
...
is y = -b
viii) Length of L
...
= 4b

b

d

g

Y'

Conic section - Parabola
XI - (Sci
...

3) Find the equation of the parabola with focus at
(4,0) and the equation of directrix is x  4  0
...

a,
2a
,

a,
2a
,
2b,
b
,
2b,

b
,



 
 

Ends of
5) Find the equation of the parabola whose vertex is
L
...

O(0,0) and focus is at (- 4, 0)
...
Also find the
out
Y  axis
Y  axis X  axis
X  axis
focus
...
Also find the equation of
P
y  4ax
the directrix
...

X'
X
O
R
S
having focus at (5/3,0) and directrix 3x + 5=0
...
The line
the origin, axis along the X-axis and passing
passing through R and perpendicular to axis is
through the point (i) (1, -4) (ii) (2, 3)
...

11) Find the equation of a parabola having vertex at
the origin, Y-axis as the axis and passing through
* Parametric equations of parabola y2 = 4ax :
the point (3, - 9)
...

symmetric about theY-axis
...

ordinate equal to 3 times the abscissa
...
e
...

the parabola y 2  12 x, whose parameter is -2/3
...

15) Find the cartesian co-ordinates of the points on
Note :
the parabola y 2  8x , whose parameters are,
2
2
i) If P t 1  at 1 ,2at 1 and Q t 2  at 2 ,2at 2 are
i) 3
ii) -2 iii) 2/5
iv) 4/5
the end points of the focal chord of a parabola 16) Find the cartesian co-ordinates of the points on
y 2  4ax, then t1 t 2  1
...

Ht t K

i

2

Exercise :
1) Find the co-ordinates of the focus, equation of
the directrix, length of latus-rectum and the co
ordinates of endpoints of latus-rectum of each of
following parabolas
...


* GHONSE MATHS ACADEMY -Regular Notes XI * GHONSE MATHS ACADEMY *

Parabola
Focus
Directrix
Axis
Vertex
L
...


17) For the parabola y 2  24 x , find the parameter
of each of the points, i) (6, -12) ii) (3/2, 6)
18) For the parabola 5y2 = 32x, find the parameter of
each of the points, i) (10,-8)
ii) (5/2, 4)
19) Find the focal distance of a point P of the parabola y 2  8x, if the ordinate of P is 8
...

21) Find the co-ordinates of a point of the parabola
y 2  4 x, having focal distance 5
...

= 40 cm
Also find the length of latus rectum and the  The distance PQ = 40 cm
equation of its axis
...
Also draw its rough sketch
...

24) Find the co-ordinates of vertex, focus and equa- Solution :
tion of the directrix of the parabola y 2  4 x  4 y
...

4b = 12
 b=3
26) Find the co-ordinates of the vertex, focus and
equation of the directrix of the parabola
4 y 2  12 x  12 y  39  0
...
If the mirror is 25 cm deep, find the distance
PQ
...

As the focus is at a distance of 4cm from
the vertex
...
i
...
a = 4
 The equation of the parabolic section is of
the form y 2  4ax
put a = 4
 y 2  16x
It is given that mirror is 25cm deep
...


* GHONSE MATHS ACADEMY -Regular Notes XI * GHONSE MATHS ACADEMY *

x

Ghonse Maths Academy,
A’nagar

Coordinates of its focus is
S (0,b), i
...

Ends of the latus rectum are
L(2b,b) and L1  2b, b 
i
...
L(6,3) and L1  6,3
 l  LL1   6  6  12 and l  OS   3
...
units
...

Solution :

[5]



Ghonse Maths Academy,
A’nagar

Let the axis of the parabolic reflector be
along positive X-axis and vertex O be the
origin
...

Let AB be the diameter of the parabolic reflector,
where l (AB) = 20 cm
...

Since the reflector is 5cm deep, l(OC) = 5
 A= (5,10) and B(5, - 10)
Since A(5,10) lies on y 2  4ax


10 

2

 4a  5

 100 = 20a
 a=5
 focus is S (a,0) i
...
S (5,0)

* GHONSE MATHS ACADEMY -Regular Notes XI * GHONSE MATHS ACADEMY *

Ghonse Maths Academy,
A’nagar

As the point P(100,25) lies on this parabola,
 100   4b  25 
 b = 100
 the eqution of parabola is
x 2  400 y
Let QR be the length of vertical supporting cable
which is 30 m away from the centre
...

 l(OS) = 30 m
Let QS be h
...
e 900 = 400h
900
 2
...
25 + 5
= 7
...
If the cable is 5 m
above the roadway at the centre of the bridge,
find the length of the vertical supporting cable
* Tangent to a Parabola :
30m from the centre
...

of the parabolic cable and the axis is vertical
...
Then
Consider the coordinate axes as in the figure
...

x 2  4by
...

Conic section - Parabola
XI - (Sci
...
e
...
(i)

 yt  2at 2  x  at 2
 yt  x  at 2

y 2  4ax
Differentiating w
...
t x
dy
 4a
 2y
dx
dy 2a
 dx  y
dy 2a
at P   x1 , y1  , dx  y
1

*

 the equation of the tangent at P is
y  y1  m  x  x1 
2a
i
...
y  y1  y  x  x1 
1
2
 yy1  y 1  2ax  2ax1

at P  t    at 2 , 2at 
...

OR
Find the equation of the tangent to the parabola

y 2  4ax in terms of its slope and write the
coordinates of the point of contact
...
e
...
from (i)
i
...
yy1  2ax  2ax1
 yy1  2a  x  x1 

is the equation of the tangent to y 2  4ax
at P   x1 , y1 
...
r
...
e
...


* GHONSE MATHS ACADEMY -Regular Notes XI * GHONSE MATHS ACADEMY *

is the equation of the tangent to y 2  4ax

It is slope of tangent to parabola at P   x1 , y1 

2

1
x  at 2 

t

Consider the line y = mx + c be tangent
to the parabola y 2  4ax at the point P(x1 , y1 ),
i
...
mx - y + c = 0

...
e
...
e
...
(ii)
Since equation i) and ii) represent the same tangent
line,  comparing the corresponding coefficients
of
equation (i) and (ii) we get,

2a  y1 2ax1


m
1
c
i
...


2a 2ax1 2a  y1


,
m
c
m
1

 x1 

c
2a
and y1 
m
m

 c 2a 
P  , 
m m 

[7]

Ghonse Maths Academy,
A’nagar

Ghonse Maths Academy,
A’nagar
in its plane
...


Now, point P lies on parabola y2 = 4ax
2

c
 2a 
   4a 
m
m

m1  m 2 

4a 2 4ac

m2
m
a
c 
m

If the tangents are perpendicular then

Put value of c in equation y = mx + c

a
is the equation of tangent to the
m

parabola y 2  4ax in terms of its slope m

 c 2a 
,  i
...

m m 

and point of contact is 

 a 2a 
 2, 
...

Hence find the locus of the point from which the
tangents are perpendicular to each other
...

Let m be the slope of the tangent drawn from P
 its equation is
a
y  mx 
m
 P   x1 , y1  lies on the tangent
a
 y1  mx1 
m
2
 y1m  x1m  a

 x1m  y1m  a  0
...

Hence, in genral, two tangents can be drawn to
the parabola y 2  4ax from any point P   x1 , y1 
Conic section - Parabola
XI - (Sci
...


Examples :
1) Find the equation of the tangent to the parabola
y 2  9x at the point  4, 6 
...

3) Find the equation of the tangent to the parabola
y 2  12x from the point  2,5 
...

5) Show that the tangents drawn from the point

 4, 9 to the parabola y 2  16x are perpen-

dicular to each other
...

2
7) Show that the line y = x + 2 touches the parabola

y 2  24x having slope

y 2  8x
...


8) If the tangents drawn from the point  6,9  to the
parabola y 2  kx are perpendicular to each other,,
find k
...

10) Find the equation of common tangent to parabola
y 2  8x and x 2  8y
...
If
PQ = 4, prove that the equation of the locus of

[8]

Ghonse Maths Academy,
A’nagar

Ghonse Maths Academy,
A’nagar
the point of intersection of two tangents is

 t1  t 2 

y  8 x  2
...

The equations of tangents at A  t1  and B  t 2  are
and yt 2  x  at 22

i
...
yt1  x  2t12
...
(ii)
Let tangents at A and B meet the tangent at vetrex
at P and Q respectively
...

 To find P, Put x = 0 in (i)
2
 yt1  2t 1

 y  2t1  P   0, 2t1 
To find Q, Put x = 0 in (ii)
 Q   0, 2t 2 
It is given that PQ = 4

 0  0    2t1  2t 2 
2

2

4

 2t1  2t 2  4

i
...
t1  t 2  2
...

To find R, solving (i) and (ii)
...
(iv)
put in equation (i),

2  t1  t 2  t1  x  2t12

* GHONSE MATHS ACADEMY -Regular Notes XI * GHONSE MATHS ACADEMY *

2
 y  8  x  2
is the eqution of required locus
...
e
...

Solution : Equation of parabola is
y 2  18x,

Comparing with y 2  4ax ,
9
4a = 18  a 
2
Equation of tangent to parabola y 2  4ax in terms
of slope m is
a
9
y  mx 
i
...
y  mx 
m
2m
Cnsider tangents are drawn from P  x1 , y1  ,
9
2m
2
i
...
2x1m  2y1m  9  0
This is a quadratic equation in m
...

 2y1  y1
 m1  m 2    2x   x

1 
1
But it is given that m1  m 2  3
y1
 x  3
1
 3x1  y1  0
Locus of point P is 3x + y = 0
...
(v)
 x  2t1t 2
To find the equation of locus of R,
We have to eliminate t1 and t 2 from the equations (iii), (iv) and (v)
We know that ,
Conic section - Parabola
XI - (Sci

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