1. PARABOLA
A conic section is a section cut off from a right circular cone by a plane in various ways. The
shape of the section depends upon the position of the cutting plane.
1. If the plane passes through the axis of the cone, the curve of intersection will be a
pair of straight lines.
2. If the plane is perpendicular to the axis of the cone, the curve of intersection will be a
circle.
3. If the plane is parallel to a generator of the cone (Say, PQ), the curve of intersection
will be a parabola.
the cone at an angle  (0
4. If the plane cuts the axis of
<  < /2), the curve of intersection will be an
ellipse.
V
A
P Q
B
O
Plane
P Q
O
V
Generator
Parabola
Plane
P Q
O
V
Ellipse
Plane
P Q
O
V
Plane
Circle
Circular
base

2. 5. If the plane is parallel to the axis of the cone, the curve of intersection will be a
hyperbola.
1.1 DEFINITION OF CONIC
A conic section or conic is the locus of a point, which moves such that its distance from a
fixed point is in a constant ratio to its distance from a fixed straight line, not passing through
the fixed point.
(i) The fixed point is called the focus.
(ii) The fixed straight line is called the directrix.
(iii) The constant ratio is called the eccentricity of the conic and is denoted by e.
(iv) When the eccentricity is unity; i.e., e = 1, the conic is called a parabola; when e < 1,
the conic is called an ellipse; and when e > 1, the conic is called a hyperbola.
(v) The line of symmetry of the conic section is called its axis.
(vi) A point of intersection of a conic with its axis is called vertex.
(vii) The chord of a conic which passes through the focus and perpendicular to the axis is
called the latus rectum.
1.2 GENERAL EQUATION OF CONIC
Let S (, ) be the focus and Ax + By + C = 0 be the equation
of the directrix QN of the conic section.
Let P (x, y) be any point on it and let PN  QN.
If ‘e’ be the eccentricity of the conic, then by definition,
PN
PS
= e  PS2 = e2 . PN2 ……(1)
 (x – )2 + (y – )2 = e2
2
2
2
B
A
C
By
Ax











which on simplification takes the form
Q
N
P (x, y)
S (, )
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ……(2)
where a, b, c, f, g and h are constants. (2) being the locus of P, is the equation of the conic.
1.3 RECOGNITION OF CONICS
The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents
A pair of straight line If  = 0 where  = abc + 2fgh - af2 - bg2 - ch2 , 2
h ab

Hyperbola
Axis

3. A circle if   0, h = 0, a = b
A parabola if   0, ab - h2 = 0
An ellipse if   0, ab - h2 > 0
An hyperbolaif   0, ab – h2 < 0
Example -1: Find the equation of the parabola whose focus is (3, -4) and directrix is x - y + 5 = 0.
Solution: Let P(x, y) be any point on the parabola. Then
   
1
1
5
y
x
4
y
3
x
2
2






  (x - 3)2
+ (y + 4)2
=
 
2
5
y
x
2


 x2
+ y2
+ 2xy - 22x + 26y + 25 = 0  (x + y)2
= 22x - 26y – 25.
Example -2: If (0, 4) and (0, 2) are respectively the vertex and focus of a parabola,
then its equation is
(A) x2 + 8y = 32 (B) y2 + 8x = 32
(C) x2 - 8y = 32 (D) y2 - 8x = 32
Solution: AS = 2 = a. Vertex (0, 4) lies on y-axis. Hence
the parabola X2
= -4aY is a downward parabola
as focus is below the vertex.
Or (x –0)2
= -4  2(y –4)
Or x2
+ 8y = 32
Hence (A) is the correct answer.
A (0, 4)
S (0, 2)
X
Y’
Example -3: Co-ordinates of the focus of the parabola x2 –4x –8y –4 = 0 are
(A) (0, 2) (B) (2, 1)
(C) (1, 2) (D) (-2, -1)
Solution: (x –2)2
= 8(y + 1)
Focus:- X = 0
 x –2 = 0  x = 2
Similarly Y = a  y + 1 = 2  y = 1
 Focus is (2, 1)
Hence (B) is the correct answer.
Example -4: If the point (2, 3) is the focus and x = 2y + 6 is the directrix of a parabola, find
(I) The equation of the axis (ii) The co-ordinates of the vertex
(iii) Length of the latus rectum (iv) Equation of the latus rectum
Solution: (i) We know that the axis of a parabola is the line through the focus and
perpendicular to the directrix.
The equation of any line passing through the focus (2, 3) is
y – 3 = m(x-2)  y – mx = 3 –2m.
If the line be perpendicular to the directrix x – 2y = 6 we have, m + 2=0  m = –2
Hence the equation of the axis is y – 3 = –2(x-2)  2x + y = 7
(ii) The co-ordinates of the point of intersection (say) A of the directrix
x – 2y = 6 and the axis 2x + y = 7 are obtained by solving the two equations; thus
they are (4, -1). Since the vertex O is the middle point of A (4, –1) and the focus S (2,
3); the co-ordinates of the vertex are ,
2
1
3
,
2
2
4





 

i.e. (3, 1).
(iii) Since OS = 5
)
3
1
(
)
2
3
( 2
2



 , the length of the latus rectum
= 4OS = 5
4 .

4. (iv) Since the latus rectum is the line through the focus parallel to the directrix, its
equation is x-2y+c=0, where c is given by 2 – 2.3 + c = 0, i.e. c = 4
1.4 STANDARD EQUATION OF A PARABOLA:
Let S be the focus, ZM the directrix and P the moving point.
Draw SZ perpendicular from S on the directrix. Then SZ is
the axis of the parabola. Now the middle point of SZ, say A,
will lie on the locus of P, i.e., AS = AZ. Take A as the origin,
the x-axis along AS, and the y- axis along the perpendicular
to AS at A, as in the figure.
Let AS = a, so that ZA is also a. Let (x, y) be the coordinates
of the moving point P. Then MP = ZN = ZA + AN = a + x.
But by definition MP = PS  MP2
= PS2
So that, (a + x)2
= (x – a)2
+ y2
.
Hence, the equation of parabola is y2
= 4ax.
Z
M
S
X
N
P
A
Y
2.1 DIFFERENT FORMS OF THE PARABOLA
Form y2
= 4ax Y2
= –4ax x2
= 4ay x2
= –4ay
Shape of the
parabola
X
Y
S
L
L
A
N
P(x,y)
(0,0)
(a,0)
x a
 
2
y 4ax

 x
x = a
A
(0, 0)
P (x, y)
Y
2
y 4ax
 
x
y = -b
S (0, a)
L' L
A
(0, 0)
P (x, y)
Y
2
x 4ay

x
y =a
S (0, -a)
L' L
A(0, 0)
P (x, y)
Y
2
x 4ay
 
Vertex (0, 0) (0, 0) (0, 0) (0, 0)
Focus (a, 0) (–a, 0) (0, a) (0, –a)
Equation of
Directrix
x = –a x = a y = –a y = a
Equation of axis y = 0 y = 0 x = 0 x = 0
Tangent at the
vertex
x = 0 x = 0 y = 0 y = 0
Parametric
Equation
x = at2
, y = 2at x = –at2
, y = 2at x = 2at, y = at2
x = 2at, y = –at2
Focal Distance
of (x1, y1)
x1 + a –x1 + a y1 + a –y1 + a
Equation of Tangent
(a) At the point
(x1, y1)
yy1 = 2a(x + x1) yy1 = –2a(x + x1) xx1 = 2a(y + y1) xx1 = –2a(y + y1)
(b) in terms of m y = mx + a/m y = mx – a/m x = my + a/m x = my – a/m
(c) in terms of t yt = x + at2
yt = –x + at2
xt = y + at2
xt = –y + at2
Equation of Normal
(a) At the point
(x1, y1)
y – y1 =
a
2
y1

(x – x1) y – y1 =
a
2
y1
(x – x1) x–x1 =
a
2
x1

(y–y1) x – x1 =
a
2
x1
(y – y1)
(b) in terms of m y = mx – 2am – am3
y = mx +2am + am3
x= my –2am – am3
x = my +2am + am3
(c) in terms of t y=–tx+2at+at3
y = tx + 2at + at3
x=–ty+ 2at+ at3
x = ty + 2at + at3

5. Latus Rectum of the Parabola:
Let the given parabola be y2
= 4ax.
In the figure LSL is the latus rectum.
Also by definition, LSL =2   a
4
a
.
a
4  = double ordinate
through the focus S.
y2
= 4ax
L
S(a, 0)
L
x

A
y
Notes:
 Any chord of the parabola y2
= 4ax which is perpendicular to its axis is called the double
ordinate.
 Two parabolas are said to be equal when their latus recta are equal.
Illustration 2: Find the equation of parabola whose focus is (1, - 1), and whose vertex
is (2, 1). Also find equation of its axis and latus rectum.
Solution: As we know that vertex is mid point of focus and point of intersection of
directrix with the axis so point P(h, k) is 3
h
2
2
1
h




1
1 3
2
k
k

  
 P is (3, 3)
slope of axis = 2
1
2
)
1
(
1




 equation of axis = y – 1 = 2 (x – 2) 
2x – y = 3
P O S
 equation of directrix y – 3 = -
2
1
(x – 3)  x + 2y = 9
 equation of parabola is (x – 1)2 + (y + 1)2 =
5
)
9
y
2
x
( 2


equation of latus roctum
y + 1 =
2
1
(x – 1)  x + 2y + 1 = 0
Parametric Equations of a Parabola:
If the coordinates of any point (x, y) on a curve can be expressed as functions of a variable t, given
by x =  (t), y =  (t) ……(1)
Then the equations in (1) are said to be parametric equations of the curve where ‘t’ is called the
parameter.
Clearly x = at2
, y = 2at satisfy the equation y2
= 4ax for all real values of t. Hence the parametric
equations of the parabola y2
= 4ax are x = at2
, y = 2at, where t is the parameter.
Also, (at2
, 2at) is a point on the parabola y2
= 4ax for all real values of t. This point is also described
as the point ‘t’ on the parabola.
Example -6: The tangents at the points (a 2
1
t , 2at1), (a 2
2
t , 2at2) on the parabola y2 = 4ax
are at right angles if
(A) t1t2 = -1 (B) t1t2 = 1
(C) t1t2 = 2 (D) t1t2 = -2
Solution: 1
t
1
.
t
1
2
1

  t1t2 = -1
Hence (A) is the correct answer.

6. Example -7: If P(a 2
1
t , 2at1) and Q (a 2
2
t , 2at2) are two variable points on the curve y2 =
4ax and PQ subtends a right angle at the vertex, then t1t2 is equal to
(A) –1 (B) –2
(C) –3 (D) –4
Solution: Slope of OP =
1
t
2
Slope of OQ =
2
t
2
Given OP  OQ 
1
t
2
.
2
t
2
= –1  t1t2 = –4
Hence (D) is the correct answer.
Focal Chord:
Any chord to the parabola y2
= 4ax which passes through the
focus is called a focal chord of the parabola y2
= 4ax.
Let y2
= 4ax be the equation of a parabola and (at2
, 2at) a
point P on it. Suppose the coordinates of the other extremity
Q of the focal chord through P are (at1
2
, 2at1).
Then, PS and SQ, where S is the focus (a, 0) have the same
slopes.

a
at
0
at
2
a
at
0
at
2
2
1
1
2





 tt1
2
– t = t1 t2
- t1  (tt1 + 1)(t1 – t) = 0.
Hence t1 = -1/t, i.e. the point Q is (a/t2
, -2a/t).
y2
= 4ax
Q(at1
2
, 2at1)
S(a, 0)
P(at2
, 2at)
x
A
y
i.e. the extremities of a focal chord of the parabola y2
= 4ax may be taken as the points t and -1/t.
Example -8: Prove that the circle with any focal chord of the parabola y2 = 4ax as its
diameter always touches its directrix.
Solution: Let AB be a focal chord. If A is (at2
, 2at), then B is 






t
a
2
,
t
a
2
.
Equation of the circle with AB as diameter is
(x - at2
) 





 2
t
a
x + (y - 2at) 






t
a
2
y = 0
For x = -a, this gives
 
2
2
2
2
t
t
1
a 
+ y2
- 2ay 






t
1
t - 4a2
= 0  a2
2
t
1
t 





 + y2
- 2ay(t - 1/t) = 0
 [y - a(t - 1/t)]2
= 0, which has equal roots.
 x + a = 0 is a tangent to the circle with diameter AB.
Example -9: In the parabola y2 = 4ax, the length of the chord passing through the
vertex and inclined to the axis at an angle /4 is
(A) 4a 2 (B)
2
a
4
(C) 2a 2 (D) none of these

7. Solution:
4
tan
at
at
2
2

  t = 2
 P = (4a, 4a)
 OP = 4 2 a.
Hence (A) is the correct answer.
X
N
P(at
2
, 2at)
O
Y
/4
Focal Distance of any Point:
The focal distance of any point P (x, y) on the
parabola y2
= 4ax is the distance between the point
P and the focus S, i.e. PS.
Thus the focal distance
= PS = PM =ZN = ZA + AN = a + x
Z
M
S
X
N
P
A
Y
Position of a point relative to the Parabola:
Consider the parabola : y2
= 4ax. If (x1 , y1) is a given point
and y2
1 -4ax1 = 0, then the point lies on the parabola. But
when y1
2
- 4ax1  0, we draw the ordinate PM meeting the
curve in L. Then P will lie outside the parabola if
PM > LM, i.e., PM2
– LM2
> 0
Now, PM2
= y1
2
and LM2
= 4ax1 by virtue of the coordinates
of L satisfying the equation of the parabola. Substituting
these values in equation of parabola, the condition for P to
lie outside the parabola becomes y1
2
- 4ax1 > 0.
P(x1, y1)
M
L
X

A
y
Interior
Exterior
Similarly, the condition for P to lie inside the parabola is y1
2
-4ax1 < 0.
Example -10:The coordinates of a point on the parabola y2 = 8x, whose focal distance
is 4, are
(A) 





 2
,
2
1
(B) (1, 2 2 )
(C) (2, 4) (D) none of these
Solution: Focal distance of a point P (x, y) on y2
= 4ax is (x + a).
 4 = x + 2  x = 2
y2
= 8  2 = 16  y =  4
Hence (C) is the correct answer.
1. GENERAL EQUATION OF A PARABOLA
Let (h, k) be the focus S and lx + my + n = 0 the equation
of the directrix ZM of a parabola. Let (x,y) be the
coordinates of any point P on the parabola. Then the
relation, PS = distance of P from ZM, gives
(x – h)2 + (y – k)2 = (lx + my + n)2 / (l2 + m2)
2
(mx ly) 2gx 2fy d 0
     
This is the general equation of a parabola.
O
Z
M
P(x, y)
S(h, k)
X
Y

8. Note: From the general equation of the parabola it is clear that the second-degree terms in
the equation of a parabola form a perfect square. The converse is also true, i.e. if in
an equation of the second degree, the second-degree terms form a perfect square
then the equation represents a parabola, unless it represents two parallel straight
lines.
Special case:
Let the vertex be (, ) and the axis be parallel to the x-axis. Then
the equation of parabola is given by (y - )2 = 4a (x – ) which is
equivalent to x = Ay2 + By + C
Similarly, when the axis is parallel to the y-axis and vertex be(, ),
the equation of parabola is given by (x - )2 = 4a (y – ) which is
equivalent to y = A’x2 + B’x + C’
Illustration 7: Find the vertex, axis, directrix, tangent at the vertex and the length of
the latus rectum of the parabola 2y2 + 3y - 4x - 3 = 0.
Solution: The given equation can be re-written as 














32
33
x
2
4
3
y
2
which is of the form Y2 = 4aX.
Hence the vertex is 







4
3
,
32
33
The axis is 0
4
3
y 
  y = -3/4
The directrix is X + a = 0
 x +
32
33
+
2
1
= 0  x = 
32
49
The tangent at the vertex is 0
32
33
x 
  x = -
32
33
Length of the latus rectum = 4a = 2.
INTERSECTION OF A STRAIGHT LINE WITH THE PARABOLA
Points of Intersection of a Straight Line with the Parabola:
Points of intersection of y2
= 4ax and y = mx + c are given by (mx + c)2
= 4ax
i.e. m2
x2
+ 2x(mc - 2a) + c2
= 0 ……(1)
Since (1) is a quadratic equation, the straight line meets the parabola in two points (real, coincident,
or imaginary). The roots of (1) are real or imaginary according as
{2(mc – 2a)}2
 4m2
c2
is positive or negative, i.e. according as –amc +a2
is positive or negative, i.e.
according as mc is less than or greater than a.
Note:
 When m is very small, one of the roots of equation (1) is very large; when m is equal to zero,
this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets
the curve in one point at a finite distance and in another point at an infinite distance from the
X
Y
( , )
 
X
Y
( , )
 

9. vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one
point.
Length of the Chord:
As in the preceding article, the abscissae of the points common to the straight line y = mx + c and
the parabola y2
= 4ax are given by the equation m2
x2
+ ( 2mc – 4a) x + c2
= 0.
If (x1, y1) and (x2, y2) are the points of intersection, then
(x1 - x2)2
= (x1 + x2)2
- 4x1 x2
=
   
4
m
mc
a
16a
2
m
2
4c
4
m
2
2a
mc
4 



and (y1 - y2) = m(x1 - x2)
Hence, the required length
=        
mc
a
a
m
1
m
4 2
2








 2
1
2
2
2
1
2
2
1 x
x
m
1
x
x
y
y
TANGENT TO A PARABOLA
Tangent at the Point (x1, y1):
Let the equation of the parabola be y2
= 4ax.
Hence, value of
dx
dy
at P(x1, y1) is
2a
y1
and the equation of the tangent at P is
y - y1 =
1
y
2a
(x - x1) i.e. yy1 = 2a(x - x1) + y1
2
 yy1 = 2a(x + x1)
Tangent in Terms of m:
Suppose that the equation of a tangent to the parabola y2
= 4ax ……(1)
is y = mx + c ……(2)
The abscissae of the points of intersection of (1) and (2) are given by the equation
(mx + c)2
= 4ax. But the condition that the straight line (ii) should touch the parabola is that it should
meet the parabola in coincident points
 (mc - 2a)2
= m2
c2
……(3)
 c = a /m.
Hence, y = mx + a/m is a tangent to the parabola y2
= 4ax, whatever be the value of m.
Equation (mx +c)2
= 4ax now becomes (mx - a/m)2
= 0
 x = 2
m
a
and y2
= 4ax  y =
m
2a
.
Thus the point of contact of the tangent y = mx + a/m is 





m
2a
,
m
a
2
.
Tangent at the Point ‘t’:
Let the equation of the parabola is y2
= 4ax.
The equation of the tangent at (x1, y1) to this parabola is yy1 = 2a(x + x1).
If the point (x1, y1)  (at2
, 2at)
Equation of tangent becomes y.2at = 2a(x + at2
)  yt = x + at2
.
Note:
 The point of intersection of the tangents at ‘t1’ and ‘t2’ to the parabola y2
= 4ax is
(at1t2, a(t1 + t2)).
Example -11: Two tangents are drawn from the point (-2, -1) to the parabola y2 =
4x. If  is the angle between these tangents, then tan  equals
(A) 3 (B) 1/3

10. (C) 2 (D) 1/2
Solution: Here a = 1. Any tangent is y = mx +
m
1
.
It passes through (-2, -1)
 2m2
–m –1 = 0
 m = 1,
2
1
  tan  = 3
2
1
1
2
1
1



Hence (A) is the correct answer.
Example -12: If the line 2x + 3y = 1 touches the parabola y2
= 4ax, find the length of the latus rectum.
Solution: Equation of any tangent to y2
= 4ax is
m
a
mx
y 
  m2
x – my + a = 0
Comparing it with the given tangent 2x + 3y – 1 = 0, we find
1
a
3
m
2
m2



 
9
2
3
m
a
,
3
2
m 




Hence the length of the latus rectum = 4a =
9
8
ignoring the negative sign for length.
Example -13: The angle between tangents to the parabola y2 = 4ax at the point
where it intersects with the line x – y –a = 0 is
(A)
3

(B)
4

(C)
6

(D)
2

Solution: The given chord id a focal chord as it passes through focus (a, 0) and we know that
tangents at the extremities of a focal chord meet at right angles on the directrix.
Hence (D) is the correct answer.
Example -14: On the parabola y2 = 4ax, three points E, F, G are taken so that
their ordinates are in G.P. Prove that the tangents at E and G intersect
on the abscissa of F.
Solution: Let the points E, F, G be (at1
2
, 2at1), (at2
2
, 2at2), (at3
2
, 2at3) respectively.
Since the ordinates of these points are in G.P., t2
2
= t1t3.
Tangents at E and G are t1y = x + at1
2
and t3y = x + at3
2
.
Eliminating y from these equation, we get
x = at1t3 = at2
2
, which is the abscissa of F.
Equation of the Tangents from an External Point:
Let y2
= 4ax be the equation of a parabola and (x1, y1) an external point P. Then, equations of the
tangents from (x1, y1) to the given parabola are given by
SS1 = T2
, where S = y2
- 4ax, S1 = y1
2
- 4ax1, T = yy1 - 2a(x + x1)
Chord of Contact:
Equation to the chord of contact of the tangents drawn from a point (x1, y1), to the parabola
y2
= 4ax is T= 0, i.e. yy1 - 2a(x+x1) =0.
Equation of a Chord with Midpoint (x1, y1):

11. The equation of the chord of the parabola y2
= 4ax with mid point (x1, y1) is T= S1
i.e. yy1-2a(x+x1)= y1
2
-4ax1 Or yy1 - 2ax = y1
2
- 2ax1 .
Example -15: Find the equation of the chord of the parabola y2
= 12x which is bisected at the point (5,
–7).
Solution: Here (x1, y1) = (5 –7) and y2
= 12x, a = 3
The equation of the chord is S1 = T
or 2
1
y - 4ax1 = yy1 – 2a (x + x1)
or (–7)2
– 12.5 = y(–7) – 6(x+5)
 6x + 7y + 19 = 0.
NORMAL TO THE PARABOLA
Normal at the Point (x1, y1 ):
The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1). Since the slope of
tangent = 2a/y1 , slope of normal is -y1/ 2a . Also it passes through (x1, y1).
Hence its equation is y - y1 =  
1
1
x
x
2a
y

 . . . . . (i)
Normal in Terms of m:
In equation (i), put m
2a
y1

 so that y1 = -2am and x1 = 2
2
1
am
4a
y
 , then the equation becomes y =
mx - 2am - am3
. . . . . (ii)
where m is a parameter. Equation (ii) is the normal at the point (am2
, -2am) of the parabola.
Notes:
 If this normal passes through a point (h, k), then k = mh – 2am - am3
.
For a given parabola and a given point (h, k) , this cubic in m has three roots say m1,
m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes
are m1, m2, m3 . For the cubic, we have
m1+ m2 + m3 = 0
m1 m2 +m2 m3 +m3 m1 = (2a-h) /a
m1 m2 m3 = - k/a
If we have an extra condition about the normals drawn from a point (h, k) to a given parabola
y2
=4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we
can get the locus of (h, k).
 Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the
normals from a given point is zero.
Normal at the Point ‘t’:
Equation of the normal to y2
= 4ax at the point (x1, y1) is y – y1 = –
a
2
y1
(x – x1)
If (x1, y1)  (at2
, 2at)
Equation of normal becomes
y – 2at = –
a
2
at
2
(x – at2
)  y = –tx + 2at + at3
.
Notes:
 If normal at the point t1 meets the parabola again at the point t2, then t2 = -t1 –2/t1.
 Point of intersection of the normals to the parabola y2
= 4 ax at (at1
2
, 2at1) and
(at2
2
, 2at2) is (2a + a(t1
2
+ t2
2
+ t1t2),– at1t2(t1+ t2)) .

12. Example -16: Find the locus of the foot of the perpendicular drawn from the
vertex on a tangent to the parabola y2 = 4ax.
Solution: Any tangent of slope m to the parabola y2
= 4ax is
y = mx + a/m . . . (1)
The perpendicular from the vertex on (1) is
x + my = 0 . . . . (2)
By eliminating m between (1) and (2) we obtain the required locus as xy2
+ x3
+ ay2
= 0
Example -17: If the normal at the point (at1
2
, 2at1) meets the parabola y2
= 4ax again at the point (at2
2
,
2at2), prove that t2 = –
1
1
t
2
t 
Solution: The equation of the normal at (at1
2
, 2at1) is
y = –t1x + 2at1 + at1
3
.
Since the normal passes through the point (at2
2
, 2at2),
we have, 2at2 = -t1. at2
2
+2at1 + at1
3
 2a (t1 – t2) = at1 t2
2
– at1
3
= at1 )
t
t
( 2
1
2
2  = at1(t2 – t1) (t2 + t1)
 2 = –t1 (t2 + t1), as t1  t2
 t2 + t1 =
1
t
2
  t2 = –t1
1
t
2

Example -18: The normal at any point P to a parabola y2 = 4ax meets its axis at
G. Q is another point on the parabola such that QG is perpendicular to
the axis of the parabola. Prove that QG2 – PG2 = constant.
Solution: Let P be the point (at2
, 2at)
Equation of normal at P is
y = –tx + 2at + at3
……(1)
Equation (1) meets the x-axis at G whose
coordinates are (2a + at2
, 0)
 PG2
= 4a2
+ 4a2
t2
……(2)
Since QG is perpendicular to the axis, the
abscissa of Q is the same as that of G and its
ordinate is QG.
 Point Q is (2a + at2
, QG)
But Q lies on the parabola y2
= 4ax.
y
2
= 4ax
G
P(at
2
, 2at)
(0, 0)
Q
QG2
= 4a(2a + at2
) = 8a2
+ 4a2
t2
……(3)
 QG2
– PG2
= 4a2
= constant by (2) and (3).
Subtangent and Subnormal:

13. Let the tangent and normal at any point
P (x1, y1) on the parabola y2
= 4ax meet the axis
in T and G respectively. Then PT is called the
length of the tangent at P and PG is called the
length of the normal at P.
NT is called the subtangent and NG the
subnormal at P. The coordinates of T and G can
be easily found by putting x = 0 in the equations
of the tangent and normal at P. It is evident from
figure that
Subtangent = NT = twice the abscissa of P
Subnormal = NG = 2a = semilatus rectum
y
2
= 4ax
G
P(x1, y1)
A
Q
X
T (–x1, 0) S
(a, 0)
(x1, 0)
N (x1 + 2a, 0)
X
yy1 = 2a(x + x1)
Y
Y

Example -19: If P is a point on the parabola y2 = 4ax such that the subtangent
and subnormal at P are equal, find the coordinates of P.
Solution: Let P (x, y) be the required point.
Length of subtangent = twice the abscisse of P = 2x
Length of subnormal = 2a
Since subtangent = subnormal, 2x = 2a  x = a  y2
= 4a2
 y =  2a
 The required points are (a, 2a) and (a, –2a).
PROPERTIES OF THE PARABOLA
P
y
2
= 4ax
S (Focus) x
A
T
M
y
K
Z
M
P
N G
(i) The tangent at any point P on a parabola bisects the angle between the focal
chord through P and the perpendicular from P on the directrix.
In the given figure MPT = TPS :
Similarly the normal at any point on a parabola bisects the angle between the focal chord
and the line parallel to the axis through that point.
(ii) The portion of a tangent to a parabola cut off between the directrix and the
curve subtends a right angle at the focus.
In the given figure SP is perpendicular to SK i.e. KSP = 900
.
(iii) Tangents at the extremities of any focal chord intersect at right angles on the
directrix.
(iv) Any tangent to a parabola and the perpendicular on it from the focus meet on
the tangent at the vertex.

14. Pole and Polar
To find the equation of the polar of the point (x1, y1) with respect to the parabola y2
= 4ax.
Let Q and R be the points in which any chord drawn through the point P, whose coordinates are (x1,
y1), meets the parabola.
Let the tangents at Q and R meet in the point whose coordiantes are (h, k).
R’
T’
P
Q
R
T
(h, k)
(x1, y1)
R’
T’
P
Q’
R
T
(h, k)
(x1, y1)
Q
We require the locus of (h, k).
Since QR is the chord of contact of tangents from (h, k) its equation is ky = 2a (x + h)
Since this straight line passes through the point (x1, y1) we have
ky1 = 2a (x 1+ h) ….(i)
Since the relation (i) is true, it follows that the point (h, k) always lies on the straight line
yy1 = 2a(x + x1) ….(ii)
Hence (ii) is the equation to the polar of (x1, y1).
Example -20:Prove that the locus of poles of focal chord of the parabola y2 = 4ax is
the directrix.
Solution: Let (h, k) be the pole. Then the equation of the chord is ky = 2a(x + h).
Since it is a focal chord, it passes through the focus (a, 0).
 2a(a + h) = 0 or a + h = 0
Hence locus of pole (h, k) is
x + a = 0, which is the directrix.

15. ASSIGNMENT
1. The focus of the parabola x2 – 8x + 2y + 7 = 0 is
(A)
1
0, -
2
æ ö
ç ÷
è ø
(B)
9
4,
2
æ ö
ç ÷
è ø
(C) (4, 4) (D)
9
-4, -
2
æ ö
ç ÷
è ø
Solution: Given parabola is (x – 4)2 = –2 






2
9
y , focus is given by x – 4 = 0,
y –
2
9
=
2
1
  focus is (4, 4).
Hence (C) is the correct answer.
2. For the parabola y2 = 4ax, the ratio of the subtangent to the abscissa is
(A) 1 : 1 (B) 2 : 1
(C) x : y (D) x2 : y2
Solution:
x
y
y
x
gent
tan
Sub

 & 2yy = 4a  y =
y
a
2
from y2 = 4ax

ax
2
ax
4
x
gent
tan
Sub
 =
1
2
.
Hence (B) is the correct answer.
3. The length of subnormal to the parabola y2 = 4ax at any point is equal to
(A) a 2 (B) 2 2 a
(C) a/ 2 (D) 2a
Solution: Length of subnormal at point (x1, y1) = y1 a
2
y
a
2
.
y
dx
dy
1
1 
 .
Hence (D) is the correct answer.
4. The length of the latus rectum of the parabola x = ay2 + by + c is
(A)
4
a
(B)
3
a
(C)
a
1
(D)
a
4
1
Solution: x = a c
a
4
b
a
4
b
y
a
b
y
2
2
2
2












or, a c
a
4
b
x
a
2
b
y
2
2









 or, 






 









a
4
ac
4
b
x
a
1
a
2
b
y
2
2
 the latus rectum =
a
1
Hence (C) is the correct answer.
5. The number of normals drawn from a point (3, 0) to the parabola y2 = 4x is /are
(A) 1 (B) 2
(C) 3 (D) none of these
Solution: y = mx – 2m – m3, this passes through (3, 0) hence

16. 0 = 3m – 2m – m3  m3 – m = 0  m = 0, m =  1
Hence three normals can be drawn.
Hence (C) is the correct answer.
6. If a normal chord to the parabola y2 = 4x is drawn at (1, 2), then the chord meets the
parabola again at:
(A) (9, 6) (B) (9, –6)
(B) (6, 6) (D) none of these
Solution: For point (1, 2) value of t = 1,
the value of t2 for other end of the normal is t2 = –1 –
1
2
= –3.
So point is (at2, 2at) = (9, –6)
Hence (B) is the correct answer.
7. The area of the triangle formed by the lines joining the vertex of the parabola
x2 = 12y to the ends of its latus rectum is
(A) 12 sq. units (B) 16 sq. units
(C) 18 sq. units (D) 24 sq. units
Solution: Required area is given by  =
2
1
(12x3) = 18 sq. units
Hence (C) is the correct answer.
8. If the line x+ y –1 = 0 touches the parabola y2 = kx , then the value of k is
(A) 4 (B) –4
(C) 2 (D) –2
Solution: Any tangent to y2 = kx is ;
y = mx +k/4m
Comparing it with the given line y = 1 – x,
we get, m = -1 and k/4m = 1  k = -4
Alternative Solution:
If x+ y –1 = 0 touches y2 = kx, then y2 = k(1-y) would have equal roots
 k2 + 4k = 0
 k = 0 or -4 but k  0,
hence k = -4
Hence (B) is the correct answer.
9. The coordinates of the point on the parabola y = x2 + 7x +2, which is nearest to
the straight line y = 3x – 3 are
(A) ( -2, -8) (B) ( 1, 10)
(C) ( 2, 20) (D) ( -1, -4)
Solution: Any point on the parabola is (x , x2+ 7x + 2)
Its distance from the line y = 3x –3 is given by
P =
 
10
5
x
4
x
1
9
3
2
x
7
x
x
3 2
2








=
10
5
x
4
x2


(as x2 +4x + 5 > 0 for all x  R)
dx
dP
= 0  x = -2
The required point  ( -2, -8)
Hence (A) is the correct answer.

17. 10. The point (1, 2) is one extremity of focal chord of parabola y2 = 4x. The length of
this focal chord is
(A) 2 (B) 4
(C) 6 (D) none of these
Solution: The parabola y2 = 4x. Here a =1 and
focus is (1, 0).
The focal chord is ASB. This is clearly
latus rectum of parabola, its value = 4.
Hence (B) is the correct answer.
S(1, 0)
A(1, 2)
X
y
B
11. If AFB is a focal chord of the parabola y2 = 4ax and AF = 4, FB = 5, then the latus-
rectum of the parabola is equal to
(A)
9
80
(B)
80
9
(C) 9 (D) 80
Solution: FA = 4, FB = 5
We know that
FB
1
AF
1
a
1


 a =
9
20
 4a =
9
80
.
Hence (A) is the correct answer.
A
B
F
(a, 0)
O
12. Vertex of the parabola whose parametric equation is x = 2
t - t + 1, y = 2
t + t + 1; t R,
is
(A) (1, 1) (B) (2, 2)
(C) (1/2, 1/2) (D) (3, 3)
Ans. (A)
Solution. x = 2
t - t + 1, y = 2
t + t + 1
 x + y = 2( 2
t + 1) and y – x = 2t

x y
2

= 1 +
2
y x
2

 
 
 
 2
(y x)
 = 2(x + y) – 4
 2
(y x)
 = 2 (x + y – 2).
Vertex will be the point where lines y – x = 0 and x + y – 2 = 0 meet, i.e., the point
(1,1).
13. If the vertex of the parabola y = x2 – 8x + c lies on x-axis, then the value of c is
(A) –16 (B) –4
(C) 16 (D) 4
Solution: The parabola is y = (x – 4)2 + c – 16. So the vertex is (4, c – 16). As vertex is
on x-axis, c = 16.
14. If (2, 0) is the vertex and y-axis the directrix of a parabola, then its focus is
(A) (2, 0) (B) (-2, 0)
(C) (4, 0) (D) (-4, 0)
Solution: a = distance of vertex from focus = distance of vertex from directrix = 2.
So, focus is at (4, 0).

18. 15. If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the
minimum value taken by c is
(A) 12 (B) 8
(C) 4 (D) 6
Solution: The tangent of slope m must be of the form y = m(x + 2) +
m
a
.
So, 2m +
m
2
= c  c = 2 






m
1
m  2  2. So cmin = 4.
16. The length of the latus rectum of the parabola whose focus is (3, 3) and directrix is
3x – 4y – 2 = 0 is
(A) 2 (B) 3
(C) 1 (D) 4
Solution: Semilatus rectum =
5
|
2
3
4
3
3
| 



= 1, so latus rectum = 2.
17. The length of the subtangent to the parabola y2 = 16x at the point whose abscissa is
4, is
(A) 2 (B) 4
(C) 8 (D) none of these
Solution: The length of subtangent = 2x1 = 2  4 = 8.
18. If the line x + y = 1 touches the parabola y2 –y + x = 0, then the co-ordinates of the
point of contact are
(A) (1, 1) (B) 





2
1
,
2
1
(C) (0, 1) (D) (1, 0)
Solution: The tangent at (x1, y1) is 2yy1 – y – y1 + x + x1 = 0 comparing it with x + y – 1 =
0, we get
(x1, y1)  (0, 1).
19. If the chord y = mx + c subtends a right angle at the vertex of the parabola
y2 = 4ax, then the value of c is
(A) – 4am (B) 4am
(C) –2am (D) 2am
Solution: y2 – 4ax 




 
c
mx
y
= 0
Coefficient of x2 + Coefficient of y2 = 0  c = –4am.
20. Minimum distance between the curve 2
y = 4x and 2
x + 2
y -12x + 31 = 0 is equal to
(A) 21 (B) 26 5

(C) 21 5
 (D) 28 5

Ans. (C)
Solution. Shortest distance will take place along the common normal.

19. (6,0)
y
x
A
O
C
Equation of normal for 2
y = 4x at (a 2
t , 2t) is
y = -tx + t + 3
t
If it passes through (6, 0), then
3
t - 5t = 0
 t = 0, t = 5
 .
 A  (5, 2 5 ), C  (5, -2 5 ).
Now, PA = PC = 1 20
 = 21, OP = 6
 Minimum distance = ( 21 5
 )units.
21. If (x1, y1) and (x2, y2) are the ends of a focal chord of the parabola y2 = 4ax, then
x1x2+ y1y2 =
(A) –3a2 (B) 3a2
(C) -4a2 (D) 4a2
Solution: x1x2 + y1y2 = a2 2
2
2
1t
t + 4a2t1t2 = a2 – 4a2 = –3a2.
22. If y = 2x + 3 is a tangent to the parabola y2 = 24x, then its distance from the parallel
normal is
(A) 5 5 (B) 10 5
(C) 15 5 (D) none of these
Solution: distance =
4
1
48
24
3
m
1
am
am
2
m
a
2
3







= 15 5 .
23. If the segment intercepted by the parabola y2 = 4ax with the line lx + my + n = 0
subtends a right angle at the vertex, then
(A) 4am + n = 0 (B) 4am + 4al + n = 0
(C) 4al + n = 0 (D) al + n = 0
Solution: y2 + 4ax 




 
n
my
lx
= 0,
now coefficient of x2 + coefficient of y2 = 0  4al + n = 0.
24. The length of a focal chord of the parabola y2 = 4ax making an angle  with the axis
of the parabola is
(A) 4a cosec2  (B) 4a sec2 
(C) a cosec2  (D) none of these
Solution: length = a|t1 – t2| 4
)
t
t
( 2
2
1 

Now, t1t2 = –1 and
2
1 t
t
2

= tan 

20. So, t1 + t2 = 2 cot , |t1 – t2| = )
1
(
4
)
cot
2
( 2


 = 2 |cosec |
So, required length = a  2|cosec |  2|cosec | = 4a cosec2 .
25. Equation of parabola having it’s vertex at A(1,0) and focus at S(3, 0) is
(A) 2
y = 4 (x - 1) (B) 2
y = 4 (x + 1)
(C) 2
y = 8 (x - 1) (D) 2
y = 8 (x + 1)
Ans. (C)
Solution. Clearly, axis of parabola is the x-axis. Corresponding value of a = 3 – 1 = 2.
Thus, equation of parabola is
2
y = 8(x – 1).
26. Equation of parabola having it’s focus at S(2, 0) and one extremity of it’s latus rectum
as (2, 2) is
(A) 2
y = 4 (3 - x) (B) 2
y = 4 (1 - x)
(C) 2
y = 8 (3 - x) (D) 2
y = 8 (3 - x)
Ans. (A)
Solution. Clearly, the other extremity of latus rectum is (2, 2). It’s axis is x-axis.
Corresponding value of a =
2 0
2

= 1
Hence it’s vertex is (1, 0) or (3, 0)
Thus it’s equation is
2
y = 4(x – 1) or 2
y = 4(x – 3).
27. Equation of parabola having the extremities of it’s latus rectum as (3, 4) and (4, 3) is
(A)
2 2 2
7 7 x y 6
x y
2 2 2
 
     
   
     
     
(B)
2 2 2
7 7 x y 8
x y
2 2 2
 
     
   
     
     
(C)
2 2 2
7 7 x y 4
x y
2 2 2
 
     
   
     
     
(D) None of these
Ans. (B,A)
Solution. Focus is
7 7
,
2 2
 
 
 
and it’s axis is the line y = x. Corresponding value of ‘a’
=
1 2
( 1 1)
4 4
  . Let the equation of it’s directrix be y + x +  = 0.

| 3 4 | 2
2.
4
2
  

  = -6, -8
Thus equation of parabola is
2 2 2
7 7 (x y 6)
x y
2 2 2
 
   
   
   
   
or
2 2 2
7 7 (x y 8)
x x
2 2 2
 
   
   
   
   
28. AB is focal chord of 2
x - 2x + y – 2 = 0 whose focus is ‘S’. If AS = 1 then BS is equal
to

21. (A) 1
1
4
4 1

(B) 1
1
4 1

(C) 1
1
2
4 1

(D) None of these
Ans. (B)
Solution. 2
x – 2x + y – 2 = 0
 2
x - 2x + 1 = 3 - y
 2
(x 1)
 = -(y – 3)
Length of it’s latus rectum is 1 unit.
Since AS,
1
2
, BS are in H.P., therefore
1
2
=
2.AS.BS
AS BS

 BS = 1
1
I
(4 1)

29. Locus of mid-point of chords of the parabola 2
y = 4ax that pass through the point
(3a, a) is
(A) 2
y + 2ax + ay – 6 2
a = 0 (B) 2
y + 2ax - ay + 6 2
a = 0
(C) 2
y - 2ax + ay + 6 2
a = 0 (D) 2
y - 2ax - ay + 6 2
a = 0
Ans. (D)
Solution. Let the mid point of chord be P(h, k) then it’s equation is
T = 1
S
i.e., yk – 2a(x + h) = 2
k - 4ah
It must pass through (3a, a), hence
ak – 2a(3a + h) = 2
k - 4ah
Thus locus of P is
2
y - 2ax – ay + 6 2
a = 0
30. Locus of trisection point of any double ordinate of 2
y = 4ax is
(A) 3 2
y = 4ax (B) 2
y = 6ax
(C) 9 2
y = 4ax (D) None of these
Ans. (C)
Solution. Let AB be a double ordinate, where
A  (a 2
t , 2at), B  (a 2
t , - 2at).
If P(h, k) be it’s trisection point, then
3h = 2a 2
t + a 2
t , 3k = 4at – 2at
 2
t =
h
a
, t =
3k
2a
Thus locus is,
2
2
9k h
4a a

i.e., 9 2
y = 4ax.
31. If the line joining the points 2
1 1
A(at ,at ) and 2
2 2
B(at ,at ) passes through C(0, b), then
(A) 1 2 1 2
b(t t ) 2at t
  (B) 1 2 1 2
2b(t t ) at t
 

22. (C) 1 2 1 2
b(t t ) at t
  (D) None of these
Ans. (A)
Solution. Equation of line joining A, B is
y( 1 2
t t
 ) = 2(x + a 1 2
t t )
It will pass through (0, b), if
b( 1 2
t t
 ) = 2a 1 2
t t
 1 2
t t
 =
2a
b 1 2
t t .
32. If the lines (y – b) = 1
m (x + a) and (y – b) = 2
m (x + a) are the tangents of 2
y = 4ax,
then
(A) 1 2
m m 0
  (B) 1 2
m m 1

(C) 1 2
m m 1
  (D) 1 2
m m 1
 
Ans. (C)
Solution. Clearly, both the lines pass through (-a, b) which is a point lying on the directrix of
the parabola.
Thus, 1 2
m m = -1.
Because tangents drawn from any point on the directrix are always mutually
perpendicular.
33. Tangents drawn to parabola 2
y = 4ax at the points A and B intersect at C. Ordinate
of A, C and B forms
(A) a A.P. (B) a G.P.
(C) a H.P. (D) None of these
Ans. (A)
Solution. If A  (a 2
1
t , 2a 1
t ), B  (a 2
2
t , 2a 2
t )
 C  (a 1 2
t t , a( 1 2
t t
 ))
Clearly the ordinates of A,C, B are in A.P.
34. Tangents drawn to parabola 2
y = 4ax at the point A and B intersect at C. If ‘S’ be the
focus of the parabola then, SA, SC and SB forms
(A) a A.P. (B) a G.P.
(C) a H.P. (D) None of these
Ans. (B)
Solution. If A  (a 2
1
t , 2a 1
t ), B  (a 2
2
t , 2a 2
t ),
 C  (a 1 2
t t , a( 1 2
t t
 ))
Now, SA = a + a 2
1
t , SB = a + a 2
2
t ,
SC = 2 2 2
1 2 1 2
(at t a) a (t t )
  
 SC = a 2 2
1 2 1 2
(t t 1) (t t )
  
= a 2 2 2 2
1 2 1 2
t t 1 t t
  
= a 2 2
1 2
(1 t )(1 t )
 
Clearly, 2
SC = SA.SB
35. Locus of trisection point of any arbitrary double ordinate or the parabola 2
x 4by
 is

23. (A) 2
9x by
 (B) 2
3x 2by

(C) 2
9x 4by
 (D) 2
9x 2by

Ans. (C)
Solution. Let A  (2bt, b 2
t ), B  (-2bt, b 2
t ) be the extremities on the double ordinate AB. If
C(h, k) be it’s trisection point, then
3h = 4bt – 2bt, 3k = 2b 2
t + b 2
t .
 t =
3h
2b
, 2
t =
k
b

k
b
=
2
2
9h
4b
Thus locus of C is 9 2
x = 4by
36. Equation of common tangent of 2
y = 4ax and 2
x 4by
 is
(A) x + y – a = 0 (B) x - y – a = 0
(C) x - y + a = 0 (D) x + y + a = 0
Ans. (D)
Solution. Let y = mx +
a
m
be a tangent of 2
y = 4ax. It will touch 2
x = 4ay, provided
2
x = 4a
a
mx
m
 

 
 
has equal roots.
 16 2 2
a m = -16
2
a
m
 m = -1.
Thus common tangent is
y + x + a = 0.
37. If the line ax + by + c = 0 is a tangent to the parabola 2
y - 4y – 8x + 32 = 0, then
(A) 2
4b = a(7a + 2c + 4b) (B) 2
4b = a(7a + c - 4b)
(C) 2
4b = a(7a + 2c - b) (D) 2
4b = a(7a + c - b)
Ans. (A)
Solution. Line will touch the parabola provided 2
y – 4y + 32 =
8( by c)
a
 
has equal roots.
 4 2
b = 7 2
a + 2a(c + 2b).
38. The straight line y = m(x –a) will meet the parabola 2
y = 4ax in two distinct real
points if
(A) m R
 (B) m [ 1,1]
 
(C) m [ ,1] [1
, )
    (D) m R ~ {0}

Ans. (D)
Solution. y = m(x – a) passes through the focus (a, 0) of the parabola. Thus for this to be
focal chord m  R ~ {0}.
39. Length of the shortest normal chord of the parabola 2
y = 4ax is
(A) a 27 (B) 3a 3
(C) 2a 27 (D) None of these
Ans. (C)
Solution. Let AB be a normal chord where A  (a 2
1
t , 2a 1
t ), B  (a 2
2
t , 2a 2
t ).

24. We have, 2 1
1
2
t t
t
  .
2 2 2 2 2 2 2
1 2 1 2
AB [a (t t )] 4a (t t )
   
= 2 2 2
1 2 1 2
a (t t ) [(t t ) 4]
  
=
2
2
1 1 2
1 1
2 4
a t t 4
t t
   
  
   
   
=
2 2 3
1
4
1
16a (1 t )
t


2
1
d(AB )
dt
= 16
4 2 2 2 3 3
2 1 1 1 1 1
8
1
t [3(1 t ) 2t ] (1 t ) .4t
a
t
 
  
 
 
=
2 2 2
2
1
1
5
1
a .32(1 t )
(t 2)
t


1
t = 2 is indeed the point of minima of A 2
B .
Thus, 3 / 2
mini
4a
AB (1 2)
2
  = 2a 27 units.
40. Mutually perpendicular tangents TA and TB are drawn to 2
y = 4ax, minimum length
of AB is equal to
(A) 4a (B) 6a
(C) 8a (D) 2a
Ans. (A)
Solution. Chord of contact of mutually perpendicular tangents is always a focal chord. Thus
minimum length of AB is 4a.
41. If three distinct and real normals can be drawn to 2
y = 8x from the point (a, 0), then
(A) a > 2 (B) a > 4
(C) a (2,4)
 (D) None of these
Ans. (B)
Solution. Equation of normal in terms of m is
y = mx – 4m – 2 3
m
If it passes through (a, 0), then
am – 4m - 2 3
m = 0
 m(a - 4 - 2 2
m ) = 0
 m = 0, 2
m =
a 4
2

.
For three distinct normals,
a – 4 > 0
 a > 4.
42. PA and PB are tangents drawn to 2
y = 4ax from arbitrary point P. If the angle
between tangents is / 4
 , then locus of point P is
(A) 2 2 2
y x a 6ax
   (B) 2 2 2
y x a 6ax
  
(C) 2 2 2
y x a 6ax
   (D) 2 2 2
y x a 6ax
  

25. Ans. (A)
Solution. Let P  (h, k)
Tangent in terms of m is
y = mx +
a
m
 2
m x – my + a = 0.
If it passes through (h, k), then
2
m h – mk + a = 0.
Since, angle between these tangents is
4

,
tan
4

= 1 2
1 2
| m m |
|1 m m |


 2 2
1 2 1 2
(m m ) (1 m m )
  
 2 2
1 2 1 2 1 2
(m m ) 4m m (1 m m )
   

2
2
2
k 4a a
1
h h h
 
  
 
 
Thus required locus is,
2
y = 4ax + (a + x)2.
43. The parabola 2
y = 4x and the circle 2 2 2
(x 6) y r
   will have no common tangent if
‘r’ is equal to
(A) r 20
 (B) r 20

(C) r 18
 (D) r ( 20, 28)

Ans. (A)
Solution. Any normal of parabola is
y = -tx + 2t + 3
t
If it pass through (6, 0), then
-6t + 2t + 3
t = 0
 t = 0, 2
t = 4, A  (4, 4).
y
x
A
O (6,0)
Thus for no common tangent
AC > 4 16
 > r
 r > 20
44. Parabola 2
y = 4x and the circle having it’s centre at (6, 5) intersect at right angle.
Possible point of intersection of these curves can be
(A) (9, 6) (B) (2, 8)
(C) (1, 2) (D) (3,2 3)

26. Ans. (A)
Solution. Let the possible point be ( 2
t , 2t)
Equation of tangent at this point is
yt = x + 2
t
It must pass through (6, 5)
 2
t - 5t + 6 = 0
 t = 2, 3
Thus possible points are (4, 4), (9, 6).
45. Normal PO, PA and PB (‘O’ being the origin) are drawn to 2
y = 4x from P(h, 0). If
AOB /2
   , then area of quadrilateral OAPB is equal to
(A) 12 sq. units (B) 24 sq. units
(C) 6 sq. units (D) 18 sq. units
Ans. (B)
Solution. Let A  ( 2
t , 2t), B  ( 2
t , -2t)
OA OB
2 2
m , m
t t
  
2
t 4
 
 t = 2
y
x
A
O P
B
Equation of normal AP is
y = -2x + 4 + 8
 P  (6, 0)
Thus area of quadrilateral
OAPB =
1
2
(OP)(AB)
=
1
2
.6.8 = 24 sq. units.
46. Locus of the midpoint of any focal chord of 2
y = 4ax is
(A) 2
y = a(x – 2a) (B) 2
y = 2a(x – 2a)
(C) 2
y = 2a(x – a) (D) None of these
Ans. (C)
Solution. Let the midpoint be P(h, k)
Equation of this chord is
T = 1
S .
i.e., yk – 2a(x + h) = 2
k - 4ah
It must pass through (a, 0)
 2a(a + h) = 2
k - 4ah
Thus required locus is
2
y = 2ax – 2 2
a

27. 47. Locus of the midpoint of any normal chord of 2
y = 4ax is
(A)
2 2
2 2
4a y
x a 2
y 2a
 
  
 
 
(B)
2 2
2 2
4a y
x a 2
y 2a
 
  
 
 
(C)
2 2
2 2
4a y
x a 2
y 2a
 
  
 
 
(D)
2 2
2 2
4a y
x a 2
y 2a
 
  
 
 
Ans. (B)
Solution. Let AB be a normal chord where
A  (a 2
1
t ,2 1
t ), B  (a 2
2
t ,2 2
t ),
If it’s midpoint is P(h, k), then
2h = a( 2
1
t + 2
2
t )
= a[ 2
1 2
(t t )
 - 2 1 2
t t ]
and 2k = 2a 1 2
(t t )

We also have 2
t = - 1
t -
1
2
t
 1 2
t t
 =
1
2
t
and 1 2
t t = - 2
1
t - 2.
 1
t = -
2a
k
and h = a 2
1 2
1
2
t 2
t
 
 
 
 
Thus required locus is
x = a
2 2
2 2
4a y
2
y 2a
 
 
 
 
.
48. The locus of the centre of a circle that passes through P(a, b) and touch the line y =
mx + c (it is given that b ma c
  ) is
(A) a straight Circle (B) circle
(C) parabola (D) hyperbola
Ans. (C)
Solution. Let O(h, k) be the centre of circle
Clearly distance of O from P and the line y = mx + c will be equal
Thus locus of P will be a parabola.
49. Slope of the normal chord of 2
y = 8x that gets bisected at (8, 2) is
(A) 1 (B) -1
(C) 2 (D) -2
Ans. (C)
Solution. Let AB be the normal chord where
A  (2 2
1
t , 4 1
t ), B  (2 2
2
t , 4 2
t ),
It’s slope =
1 2
2
t t

We also have
2 1
1
2
t t
t
  
and 16 = 2( 2 2
1 2
t t
 ), 4 = 4 1 2
(t t )

 1 2
t t
 = 1

28. Thus slope is 2.
50. Radius of the circle that passes through origin and touches the parabola 2
y = 4ax at
the point (a, 2a) is
(A)
5
a
2
(B) 2 2a
(C) 5 2a (D) 3 2a
Ans. (A)
Solution. Equation of tangent of parabola at (a, 2a) is
y.2a = 2a(x + a),
i.e., y – x – a = 0.
Equation of circle touching the parabola at (a, 2a) is
2 2
(x a) (y 2a)
     (y – x – a) = 0
Since, it passes through (0, 0), therefore
2 2
a 4a
   (-a) = 0
  = 5a.
Thus required circle is
2 2
x y
 - 7ax + ay = 0
It’s radius =
2
2
49 a 5
a a.
4 4 2
 
51. The length of latus rectum of the parabola, whose focus is (2 cos2  , a cos2  )
and directrix is the line y = a, is
(A) |4a 2
cos  | (B) |4a 2
sin  |
(C) |4a cos2  | (D) |a cos2  |
Ans. (B)
Solution. Distance of focus from directrix is
|a cos2 – a|
Thus length of latus rectum is |4a 2
sin  |.
52. If the line y = mx + c is a tangent to 2
y = 4mx, then distance of this tangent from the
parallel normal is
(A) 2
1 m
 (B) 2
2 m

(C) 2 3 / 2
(1 m )
 (D) 2 3 / 2
(2 m )

Ans. (C)
Solution. Equation of parallel normal is
y = mx - 2 2 4
m m

 Required distance =
2 4
2
| c 2m m |
1 m
 

Here, c = 1
Thus required distance = (1 + 2
m )3/2
53. Angle between the tangents drawn to 2
y = 4x at the points where it is intersected by
the line y = x – 1 is equal to
(A)
4

(B)
3


29. (C)
6

(D)
2

Ans. (D)
Solution. The line y = x – 1 passes through (1, 0). that means it is a focal chord. Hence the
required angle is /2.
54. Tangents PA and PB are drawn to parabola 2
y = 4ax. If slope of bisector of the angle
PAB is 3 , then locus of point ‘p’ is
(A) 2 2 2
y x 3a 2ax
   (B) 2 2 2
y 3x 3a 2ax
  
(C) 2 2 2
y x 3a 10ax
   (D) 2 2 2
y 3x 3a 10ax
  
Ans. (D)
Solution. Slope of tangents will be tan
3

 
 
 
 
, tan
3

 
 
 
 
,
Let A  (a 2
1
t , 2a 1
t ), B  (a 2
2
t , 2a 2
t ),
 P 1 2 1 2
[at t ,a(t t )]
 
We have
1
t = cot
3

 
 
 
 
, 2
t = cot
3

 
 
 
 
,
 1 2
1 1
.cot 1 .cot 1
3 3
t ,t
1 1
cot cot
3 3
   
 
   
 1 2
1 2
3 t t 3
cot
1 3t 1 3t
 
  
 
 2 1 1 2 2 1 2 1
3 3t t 3t t t 3 3t t 3t
      
 1 2 1 2
2 3 2 3t t 2(t t )
  
 2 2 2
2 2 1 2 1 2 1 2
3(1 t t ) (t t ) (t t ) 4t t
     
Thus locus of P is
2 2 2
y 3x 3a 10ax
  
55. Parabolas 2
1
y 4a(x c )
  and 2
2
x 4a(y c )
  where 1
c and 2
c are variable, are such
that they touch each other. Locus of their point of contact is
(A) xy = 2 2
a (B) xy = 4 2
a
(C) xy = 2
a (D) None of these
Ans. (B)
Solution. Let P(x, y) be the point of contact.
At ‘P’ both of them must have same slope.
We have, 2y
dy
dx
= 4a, 2x = 4a
dy
dx
.
Eliminating
dy
dx
, we get
xy = 4 2
a
56. The circle 2 2
x y
 + 2gx + m 2fy + c = 0, cuts the parabola 2
x = 4ay at points
i i i
A (x ,y ), i = 1, 2, 3, 4; then

30. (A) i
y
 = 0 (B) i
y
 = -4(f + 2a)
(C) i
x
 = - 4(g + 2a) (D) i
x
 = - 2(g + 2a)
Ans. (B)
Solution. Putting y =
2
x
4a
in the equation of circle, we get
4 2
2
2
x fx
x 2gx c 0
16a 2a
    
 4 2 2 2 2
x x (16a 8af) 32a gx 16a c 0
    
 1 2
x x
 = 8a(f + 2a),
2 2
1 2 3 1 2 3 4
x x x 32a g, x x x x 16a c
    
 2 2
i 1 1 1 2
1 1
y x [( x ) 2 x x ]
4a 4a
      
= -4(f + 2a)
57. Maximum number of common normals of 2
y = 4ax and 2
x = 4by can be equal to
(A) 3 (B) 4
(C) 6 (D) 5
Ans. (D)
Solution. Normals to 2
y = 4ax and 2
x = 4by in terms of ’m’ are y = mx – 2am – a 3
m
and y = mx + 2b + 3
b
m
.
For a common normal
2b + 2
b
m
- 2am + a 3
m = 0
 a 5
m + 2a 3
m + 2b 2
m + b = 0
That means there can be atmost 5 common normals.
58. If two different tangents of 2
y = 4x are the normals to 2
x =4by, then (Ref.Parabola
P.K.Sharma Pg.C3.8 Q.18)
(A)
1
| b |
2 2
 (B)
1
| b |
2 2

(C)
1
| b |
2
 (D)
1
| b |
2

Ans. (B)
Solution. Tangent to 2
y = 4x in terms of m is
y = mx +
1
m
.
Normal to 2
x = 4by in terms of m is
y = mx + 2b + 2
b
m
.
If these are same lines, then
1
m
= 2b + 2
b
m
 2b 2
m – m + b = 0
For two different tangents, we must have
1 – 8 2
b > 0

31.  |b| <
1
8
59. Length of the latus rectum of the parabola whose parametric equation is x = 2
t + t +
1, y = 2
t - t + 1, where t R
 , is equal to (Ref.Parabola P.K.Sharma Pg.C3.9 Q.21)
(A) 8 (B) 4
(C) 2 (D) None of these
Ans. (D)
Solution. x = 2
t + t + 1, y = 2
t - t + 1
 x + y = 2( 2
t + 1) and x – y = 2t

(x y)
2

1 +
2
x y
2

 
 
 
 2
(x y)
 2 (x + y) – 4 = 2(x + y – 2)
Comparing it with 2
y = 4ax, we get
length of latus rectum = 2 units
60. Minimum distance between the curves 2
y = x – 1 and 2
x = y – 1 is equal to
(A)
3 2
4
(B)
5 2
4
(C)
7
2
4
(D)
1
2
4
Ans. (A)
Solution. Both curves are symmetrical about the line y = x. If line AB is the line of shortest
distance then at A and B slopes of curves should be equal to one.
For 2
y = x – 1,
dy 1
dx 2y
 = 1
 y =
1
2
, x =
5
4
 B
1 5
,
2 4
 
  
 
.
B
A
O
x
y
y=x
Hence minimum distance AB,
=
2 2
5 1 5 1 3 2
4 2 4 2 4
   
   
   
   
units.