1. Total No. of questions in Parabola are-
InchapterExamples.......................................38
SolvedExamples...........................................24
TotalNo. ofquestions............................. ....62
PARABOLA

2. 1. DEFINITION
A parabola is the locus of a point which moves in
such a way that its distance from a fixed point is
equal to its perpendicular distance from a fixed
straight line.
1.1 Focus : The fixed point is called the focus of the
Parabola.
1.2 Directrix : The fixed line is called the directrix of the
Parabola.
2. TERMS RELATED TO PARABOLA
2.1 Eccentricity : If P be a point on the parabola and
PM and PS are the distances from the directrix and
focus S respectively then the ratio PS/PM is called
the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1  Hyperbola, e = 0  circle, e < 1
 ellipse
2.2 Axis : A straight line passes through the focus and
perpendicular to the directrix is called the axis of
parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and
its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and
the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any
point P (x, y) on the parabola from the focus is called
the focal length. i.e.
The focal distance of P = the perpendicular distance
of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular
to the axis of Parabola or parallel to Directrix is called
double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing
through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through
the focus of parabola then it is called as latus rectum.
2.7.1Length of latus rectum :
The length of the latus rectum = 2 x perpendicular
distance of focus from the directrix.
3. STANDARD FORM OF EQUATION OF
PARABOLA
If we take vertex as the origin, axis as x- axis and
distance between vertex and focus as 'a' then equation
of the parabola in the simplest form will be-
y2 = 4ax
y
M L(a,2a)
L’
S (a,0)
x
A
Z
Directrix x = a
Focus
axis
Focal chord
Double ordinate
Latus Rectum
Vertex
F
o
c
a
l
d
is
ta
n
c
e
y’
x+
a
=
0
P
(a,-2a)
L’
L

3. 3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A  (0, 0)
(ii) Focus S  (a, 0)
(iii) Directrix  x + a = 0
(iv) Axis  y = 0 or x– axis
(v) Equation of Latus Rectum  x = a
(vi) Length of L.R.  4a
(vii) Ends of L.R.  (a, 2a), (a, – 2a)
(viii) The focal distance  sum of abscissa of
the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2  then coordinates of end points
of this Double ordinate are










,
4a
2
and 









,
4a
2
.
3.2 Other standard Parabola :
Examples
based on Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and
directrix is x + y – 2 = 0, then its vertex is
(A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Q1 P1
Y
S
X
M
Z S P1
Q1
X
M
A
Z
y2 = – 4 ax
x2 = 4ay x2 = – 4ay
Equationof Vertex Axis Focus Directrix Equation of Lengthof
Parabola Latus rectum Latus rectum
y2 = 4ax (0,0) y= 0 (a,0) x = –a x= a 4a
y2 = – 4ax (0,0) y= 0 (–a, 0) x = a x= –a 4a
x2 = 4ay (0,0) x=0 (0,a) y = a y = a 4a
x2 = – 4ay (0,0) x=0 (0, –a) y = a y = –a 4a

4. Sol. First we find the equation of axis of parabola,
which is perpendicular to directrix. So its equation
is x – y + k = 0. It passes through focus
S (3, –4)
 3 – (–4) + k = 0  k = –7
Let Z is the point of intersection of axis and
directrix.
Solving equation x + y –2 = 0 and x –y–7=0
gives Z (9/2, –5/2)
Vertex A is the mid point of Z and S
 A















2
2
5
4
,
2
9
3
2
= A
A 






4
13
,
4
15
Ans.[D]
Ex.2 The equation of the parabola whose vertex is
(–3, 0) and directrix is x + 5 = 0 is-
(A) y2 = 8(x + 3)
(B) y2 = – 8(x + 3)
(C) x2 = 8(y + 3)
(D) y2 = 8(x + 5)
Sol. A line passing through the vertex (–3, 0) and
perpendicular to directrix x + 5 = 0 is x-axis
which is the axis of the parabola by definition.
Let focus of the parabola is (a, 0). Since
vertex, is the middle point of Z(–5, 0) and
focus S, therefore
1
–
a
2
)
5
a
(
3 




 Focus = (–1,0)
Thus the equation to the parabola is
(x + 1)2 + y2 =(x + 5)2
 y2 = 8(x +3) Ans. [A]
Ex.3 The equation of the directrix of the parabola
y2 = 12x is-
(A) x + 3 = 0 (B) y + 3 = 0
(C) x – 3 = 0 (D) y – 3 = 0
Sol. Here a = 3, so the equation of the directrix
is given by x = – a  x= – 3
 x + 3 = 0
Ans. [A]
Ex.4 The equation of the latus rectum of the
parabola x2 = – 12y is-
(A) y = 3 (B) x = 3
(C) y = – 3 (D) x = –3
Sol. Here a = 3 so the equation of the L.R. is
given by y = – a  y = – 3 Ans. [C]
4. REDUCTION TO STANDARD EQUATION
If the equation of a parabola is not in standard form
and if it contains second degree term either in y or
in x (but not in both) then it can be reduced into
standard form. For this we change the given equation
into the following forms-
(y – k)2 = 4a (x – h) or (x – p)2 = 4b (y – q)
And then we compare from the following table for the
results related to parabola.
Examples
based on Reduction to Standard eqn. of a Parabola
Ex.5 The axis of the parabola
9y2 –16x –12y –57 = 0 is
(A) 3y = 2 (B) 2x = 3
(C) x +3y = 3 (D) y = 3
Equation of Vertex Axis Focus Directrix Equation of L.R. Length of L.R.
Parabola
(y– k)2 = 4a (x–h) (h,k) y = k (h + a, k) x + a – h = 0 x = a + h 4a
(x–p)2 = 4b ( y–q) (p, q) x = p (p,b+ q) y + b – q = 0 y = b + q 4b

5. Sol. 9y2 –12y = 16x + 57
 (3y – 2)2 = 16x + 57
 














16
61
x
9
16
3
2
y
2
Which shows that equation of the axis is
y –2/3 = 0 or 3y = 2 Ans. [A]
Ex.6 Vertex, focus, latus rectum, length of the
latus rectum and equation of directrix of the
parabola y2 = 4x +4y are
(A) (1, 2), (0, 2), y = 0, 4,x = – 2
(B) (–1, 2), (0, 2), x = 0, 4, x = –2
(C) (–1, 2), (1, 2), x = 0, 4, x = 2
(D) (–1, 2) (0, 2), y = 0, 2 , y = – 2
Sol. Given parabola y2 = 4x + 4y
or (y – 2)2 =4(x +1)
or Y2 = 4X
Here X = x +1, Y= y – 2
vertex = (X = 0, Y = 0)
or (x + 1 = 0, y – 2 = 0)  (–1, 2)
Focus (X = 1, Y = 0)
or (x + 1 = 1, y – 2 = 0)  (0, 2)
Latus rectum  X = 1 x = 0
Length of Latus rectum = 4
equation of the directrix is
X = –1  x +1 = –1  x = –2
Ans. [B]
Ex.7 The vertex of the parabola x2 – 8y – x + 19 = 0 is-
(A) 





2
1
,
32
75
(B) 





32
75
,
2
1
(C) 






32
75
,
2
1
(D) None of these
Sol. The given equation of Parabola can be written
as
2
2
1
x 





 – 8y + 19 –
4
1
= 0
2
2
1
x 





 = 8y –
4
1
76 

2
2
1
x 





 = 8 






32
75
y
 vertex = 





32
75
,
2
1
Ans.[B]
5. GENERAL EQUATION OF A PARABOLA
If (h,k) be the locus of a parabola and the equation
of directrix is ax + by + c = 0, then its equation is
given by
2
2
)
k
y
(
)
h
x
( 

 = 2
2
b
a
c
by
ax



which
gives (bx– ay)2 + 2gx + 2fy + d = 0 where g, f, d are the
constants.
Note:
(i) It is a second degree equation in x and y and the
terms of second degree forms a perfect square and it
contains at least one linear term.
(ii) The general equation of second degree
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a
parabola, if
(a) h2 = ab
(b)  = abc + 2fgh – af2 – bg2 – ch2  0
6. EQUATION OF PARABOLA WHEN ITS VERTEX
AND FOCUS ARE GIVEN
6.1 If both lie on either of the coordinate axis :
In this case first find distance 'a' between these
points and taking vertex as the origin suppose the
equation as y2 = 4ax or x2 = 4ay. Then shift the origin
to the vertex.
6.2 When both do not lie on any coordinate axes
In this case first find the coordinates of Z and
equation of the directrix, then write the equation of
the parabola by the definition.
Examples
based on
Equation of Parabola when its vertex and
focus are given
Ex.8 If (0,4) and (0,2) are vertex and focus of
parabola respectively, then its equation is-
(A) y2 + 8x = 32 (B) y2 – 8x = 32
(C) x2 + 8y = 32 (D) x2 – 8y = 32
Sol. Since vertex and focus are on y-axis, so
y-axis is the axis of the parabola.
Distance between vertex and focus = 'a' = 2.
So on taking vertex as origin, the equation of
the parabola is x2
= – 4ay
(negative because vertex lies above the
focus)
or x2
= – 8y.
Now shifting the origin to its original position,
the required equation will be
x2
= –8(y – 4)
 x2
+ 8y = 32
Ans. [C]

6. 7. PARAMETRIC EQUATION OF PARABOLA
The parametric equation of Parabola y2 = 4ax are
x = at2, y = 2at
Hence any point on this parabola is (at2, 2 at) which
is called as 't' point.
Note:
(i) Parametric equation of the Parabola x2 = 4ay is
x = 2at, y = at2
(ii) Any point on Parabola y2 = 4ax may also be written
as (a/t2, 2 a/t)
(iii) The ends of a double ordinate of a parabola can be
taken as (at2, 2 at) and (at2, – 2at)
(iv) Parametric equations of the parabola
(y – h)2 = 4a (x – k)2 is x – k = at2 and y – h= 2 at
Examples
based on
Parametric equation of Parabola
Ex.9 The parametric equation of the curve y2 = 8x are-
(A) x = 2t, y = 4t2 (B) x = 2t2, y = 4t
(C) x = t2, y = 2t (D) None of these
Sol. Here a = 2; y = 2at  y = 2.2.t = 4t
x = at2  x = 2t2 Ans. [B]
Ex.10 The parametric equation of the curve
(y– 2)2 = 12 (x – 4) are-
(A) 6t, 3t2 (B) 2 + 3t, 4 + t2
(C) 4 + 3t2,2 + 6t (D) None of these
Sol. Here a = 3
y – 2 = 2at  y = 2 + 2.3t = 2 + 6t
x – 4 = at2  x = 4 + 3.t2 = 4 + 3t2
Ans. [C]
Ex.11 Parameter t of a point (4,–6) of the parabola
y2 = 9x is-
(A) 4/3 (B) – 4/3 (C) – 3/4 (D) – 4/5
Sol. Parametric coordinates of any point on
parabola y2 = 4 ax are (at2, 2 at)
Here 4 a = 9  a = 9/4
 y coordinate 2 at = – 6
 2 (9/4) t= – 6 t = – 4/3 Ans. [B]
8. CHORD
8.1 Equation of chord joining any two points of a parabola
Let the points are (at1
2, 2 at1) and (at2
2, 2 at2) then
equation of chord is-
(y– 2at1) = 2
1
2
2
1
2
at
at
at
2
at
2


(x – at1
2)
 y – 2at1 =
2
1 t
t
2

(x – at1
2)
 (t1 + t2) y = 2x + 2at1 t2
Note :
(i) If 't1' and 't2' are the Parameters of the ends
of a focal chord of the Parabola y2 = 4ax, then
t1t2 = – 1
(ii) If one end of focal chord of parabola is
(at2, 2at) , then other end will be (a/t2, – 2a/t)
and length of focal chord = a (t + 1/t)2.
(iii) The length of the chord joining two points 't1
'
and 't2' on the parabola y2 =4ax is
a (t1 – t2) 4
)
t
t
( 2
2
1 

8.2 Length of intercept = )
mc
a
)(
m
1
(
a
m
4 2
2


Examples
based on
Chord
Ex.12 If one extremity of the focal chord of the
parabola y2 = 16x is (1,–4) then find other
extremity.
Sol. Here 2at = – 4 and a = 4
 t =
4
.
2
4

= –
2
1
other extremity is








2
/
1
4
.
2
,
4
/
1
4
= (16,16) Ans.
Ex.13 Find the length of a chord of a parabola
y2 = 4x which passes through vertex and
makes an angle of 45º with x- axis.
Sol. Let the equation of chord is y = mx + c
Here m = tan 45º = 1
as it passes through vertex (0,0) c = 0
Hence the equation is y = x and a = 1
 = 4 )
0
1
)(
1
1
(
1 
 = 4 2 Ans.
Ex.14 Length of intercept by the line 4y = 3x – 48
on the parabola y2 = 64x is-

7. (A)
1600
9
(B)
9
1600
(C)
9
160
(D) None of these
Sol. 4y = 3x – 48  m = 3/4, c = –12
y2 = 64x  a = 16
Length of intercept = )
mc
a
(
)
m
1
(
a
m
4 2
2


= 















4
3
12
16
16
9
1
16
16
9
4
=
9
1600
Ans. [B]
9. POSITION OF A POINT AND A LINE WITH
RESPECT TO A PARABOLA
9.1 Position of a point with respect to a parabola
A point (x1, y1) lies inside, on or outside of the region
of the parabola y2 = 4ax according as y1
2 – 4ax1 <
= or > 0
9.2 Line and Parabola :
The line y = mx + c will intersect a parabola
y2 = 4ax in two real and different, coincident or
imaginary point, according as a – mc >, = < 0
Ex.15 For the parabola y2 = 8x point (2, 5) is
(A) Inside the parabola
(B) Focus
(C) Outside the parabola
(D) On the parabola
Sol. (y2 – 8x)x = 2, y = 5 = (5)2 – 8 × 2 = 9 > 0
 Point (2, 5) is outside parabola y2 = 8x
Ans. [C]
10. TANGENT TO THE PARABOLA
10.1 Condition of Tangency : If the line y = mx + c touches
a parabola y2 = 4ax then c = a/m
Note:
(i) The line y = mx + c touches parabola x2 = 4ay
if c = – am2
(ii) The line x cos  + y sin  = p touches the parabola
y2 = 4ax if a sin2  + p cos  = 0
(iii) If the equation of parabola is not in standard form,
then for condition of tangency, first eliminate one
variable quantity (x or y) between equations of
straight line and parabola and then apply the condition
B2 = 4AC for the quadratic equation so obtained.
Examples
based on
Condition of tangency
Ex.16 The line y = mx + 1 is a tangent to the
parabola y2 = 4x if
(A) m = 1 (B) m = 4
(C) m = 2 (D) m = 3
Sol. Here a = 1, c = 1, m = m. Now applying
condition of tangency c = a/m, we have
1 = 1/m  m = 1
Ans. [A]
Ex.17 If the line 2x – 3y = k touches the parabola
y2 = 6x, then the value of k is
(A) 27/4 (B) –81/4
(C) – 7 (D) – 27/4
Sol. Given
2
k
y
3
x

 ...(1)
and y2 = 6x ...(2)
 




 

2
k
y
3
6
y2
 y2 = 3(3y + k)
 y2 – 9y – 3k = 0 ...(3)
If line (1) touches parabola (2) then roots of
quadratic equation (3) is equal
 (– 9)2 = 4 × 1 × (– 3k) 
4
27
k 

Ans. [D]
10.2Equation of Tangent
10.2.1 Point Form : The equation of tangent to the
parabola y2 = 4ax at the point (x1, y1) is
yy1 = 2a(x + x1) or T = 0
10.2.2 Parametric Form : The equation of the
tangent to the parabola at t i.e. (at2, 2at) is
ty = x + at2
10.2.3 Slope Form : The equation of the tangent of the
parabola y2 = 4ax is
y = mx +
m
a
(at ,2at)
2
Note :
(i) y = mx + a/m is a tangent to the parabola y2 = 4ax
for all value of m and its point of contact is (a/m2, 2a/m).
(ii) y = mx – am2 is a tangent to the parabola x2 = 4ay
for all value of m and its point of contact is
(2am, am2)

8. (iii) Point of intersection of tangents at points t1 and t2
of parabola is [at1t2, a(t1+t2)]
(iv) Two perpendicular tangents of a parabola meet on its
directrix. So the director circle of a parabola is its
directrix or tangents drawn from any point on the
directrix are always perpendicular.
(v) The tangents drawn at the end points of a focal
chord of a parabola are perpendicular and they meet
at the directrix.
Examples
based on
Equation of tangent
Ex.18 The equation of the tangent to the parabola
y2 = 4ax at the points (3, 2) is
(A) 3y + x +3 = 0 (B) 3x + y +3 = 0
(C) 3x = y +3 (D) 3y = x +3
Sol. Since (3, 2) lies on the parabola y2 = 4ax, so
4 = 12a  a = 1/3
Now using T = 0, the equation of the tangent
at (3, 2) is
y(2) = 2a(x + 3)
 y =
3
1
(x + 3)
 3y = x + 3
Ans. [D]
Ex.19 The equation of the tangent to the parabola
y2 = 12x drawn at the upper end of its latus
rectum is
(A) y = x + 3 (B) x + y = 3
(C) y = x – 3 (D) None of these
Sol. Here a = 3, so the upper end of LR = (3, 6).
Hence the equation of the tangent at this
point is y(6) = 6(x + 3)
 y = x + 3 Ans. [A]
Ex.20 The slope of tangent lines drawn from (3, 8)
to the parabola y2 = –12x are
(A) –3, –1/3 (B) 3, 1/3
(C) 3, –1/3 (D) –3, 1/3
Sol. Since 82 + 12 × 3  0, therefore the point
(3, 8) is not on the parabola
Now equation of any tangent to the parabola
y2 = –12x is written as
y = mx – 





m
3
Since this line passes through (3, 8)
 8 = 3m – 





m
3
 3m2 – 8m – 3 = 0
Solving this equation, we get m = 3, –
3
1
Ans. [C]
Ex.21 If a tangent to the parabola y2 = ax makes
an angle of 45º with x- axis, then its point of
contact is-
(A) 





2
a
,
4
a
(B) 





2
a
,
2
a
(C) 






4
a
,
2
a
(D) 






2
a
,
4
a
Sol. The slope of the tangent = tan 45º = 1
 m = 1 and a  a/4
 Point of contact = 





1
4
/
a
.
2
,
1
4
/
a
2 = 





2
a
,
4
a
Ans. [A]
Ex.22 Find the common tangent of the parabola
x2 =4ay and y2 = 4ax(m > 0).
Sol. Equation of tangent for x2 = 4ay is y = mx – am2
as this line also touches y2 = 4ax
 – am2 = a/m (c = a/m)
 m3 = – 1
 m = – 1
 equation of tangent  x + y + a = 0
Ans.
11. NORMAL TO THE PARABOLA
11.1 Equation of Normal :
11.1.1 Point Form : The equation to the normal at the
point (x1, y1) of the parabola y2 = 4ax is given by
y – y1 =
a
2
y1

(x – x1)
11.1.2 Parametric Form : The equation to the normal
at the point (at2,2at) is y + tx = 2at + at3
11.1.3 Slope Form : Equation of normal in terms of
slope m is y = mx – 2am – am3.
Note :
(i) the foot of the normal is (am2, –2am)
(ii) Condition for normal : The line y = mx + c
is a normal to the parabola y2 = 4ax if
c = –2am – am3 and x2 = 4ay
if c = 2a + 2
m
a
(iii) Normal Chord : If the normal at 't1' meets
the parabola y2 = 4ax again at the point ' t2'
then this is called as normal chord. Again for
normal chord t2 = – t1 –
1
t
2

9. (iv) If two normal drawn at point ' t1' and ' t2'
meet on the parabola then t1t2 =2
Examples
based on
Equation and properties of Normal
Ex.23 The equation of normal at the point (a/m2, 2a/m)
on the parabola y2 = 4ax is-
(A) m3y = m2x – 2am2 – a
(B) y = mx – 2am – am3
(C) m3y = 2am2 – m2x + a
(D) None of these
Sol. The equation of the normal at (at2, 2at) of the
parabola y2 = 4ax is written as
y + tx = 2at + at3
For the given point t = 1/m, therefore equation
of required normal is written as
3
m
1
a
m
1
a
2
x
m
1
y 




















 m3y = 2am2 – m2x + a
Ans. [C]
Ex.24 Find the equation of a normal at the parabola
y2 = 4x which is parallel to the line y = 3x + 4.
Sol. Let the equation is y = 3x + c
This is normal to the parabola y2 = 4x then
c = – 2 am – am3
= – 2.3 – 33 ( a = 1)
= – 33
Hence equation is y = 3x – 33
Ans.
Ex.25 If the line x + y = 1 is a normal to the parabola
y2 = kx, then the value of k is
Sol. As line y = –x + 1 is the normal to the
parabola y2 = kx then
1 = – 2(a)(– 1) – a(– 1)3
a = 1/3
 4a = k k = 4/3
Ans.
Ex.26 Find the equation of a normal at the parabola
y2 = 4x which passes through (3,0).
Sol. Equation of Normal y = mx – 2 am – am3
Here a = 1 and it passes through (3,0)
0 = 3m – 2m – m3
 m3 – m = 0
 m = 0, ± 1
for m = 0  y = 0
m = 1  y = x – 3
m = –1  y = – x + 3
Ans.
Ex.27 If the normal of the parabola y2 = 4ax drawn
at (a, 2a) meets the parabola again at the
point (at2, 2at) then t is equal to-
(A) 3 (B) 1 (C) –1 (D) –3
Sol. If t' be the parameter of the given point, then
2at' = 2a  t' = 1
Now t= – t' –
t
2

 t = –1 –
2
1
= – 3
Ans. [D]
Ex.28 Find the point where the normal at (4,4)
meets the parabola y2 = 4x
Sol. Here a = 1 and
at1
2 = 4  t1 = 2
t2 = – 2 –
2
2
= – 3
Point is (at2
2, 2 at2) = {(–3)2,2.1.(–3)}= (9,– 6)
Ans.
Ex.29 If two of the normal of the parabola y2
= 4x,
that pass through (15, 12) are 4x + y = 72,
and 3x – y = 33, then the third normal is
(A) y = x – 3 (B) x + y = 3
(C) y = x + 3 (D) None of these
Sol. Here, If m1
, m2
, m3
are slopes of normal, then
m1
+m2
+m3
=0 and m1
m2
m3
=
a
y1
a = 1 here m1
= –4, m2
= 3
 – 4 + 3 + m3
= 0  m3
= 1
Also (–4) (3) (1) = –
1
12
is satisfied.
But (15, 12) satisfies y = x – 3
Hence (A) is correct answer
Ans. [A]
Note: As seen value of m3
is not sufficient to locate
the correct answer. The answer should be
confirmed before it is reported.

10. Ex.30 If two normal drawn from any point to the
parabola y2
= 4ax makes angle  and  with the
axis such that tan .tan  = 2, then locus of this
point is-
(A) y2
= 4ax (B) x2
= 4ay
(C) y2
= – 4ax (D) x2
= –4ay
Sol. Let the point is (h, k). The equation of any
normal to the parabola y2
= 4ax is
y = mx – 2am – am3
passes through (h, k)
k = mk – 2am – am3
am3
+ m(2a – h) + k = 0 ...(i)
m1
, m2
, m3
are roots of the equation
then m1
.m2
.m3
= –
a
k
but m1
m2
= 2, m3
= –
a
2
k
m3
is root of (i)
a
3
a
2
k






 –
a
2
k
(2a – h) + k = 0  k2
= 4ah
Thus locus is y2
= 4ax
Ans. [A]
12. PAIR OF TANGENTS
If the point (x1, y1 ) is outside the parabola, then two
tangents can be drawn. The equation of pair of
tangents drawn to the parabola y2 = 4ax is given by
SS1 = T2
i.e. (y2 – 4ax) (y1
2 – 4ax1) = [yy1 – 2a(x + x1)]2
12.1Locus of point of Intersection of tangents
The locus of point of intersection of tangent to the
parabola y2 = 4ax which are having an angle  between
them is given by y2 – 4ax = (a+x)2 tan2 
Note :
(i) If  = 0º or  then locus is (y2 –4 ax) = 0
which is the given parabola.
(ii) If  = 90º, then locus is x + a = 0 which is
the directrix of the parabola.
Ex.31 The equation of pair of tangents drawn from
(1,4) to the parabola y2 =12x is-
(A) 3x2 + y2 – 10 x + 4y – 3 = 0
(B) 3x2 + y2 – 10 x + 4xy + 4y – 3 = 0
(C) 3x2 + y2 + 10 x – 4xy – 4y + 3 = 0
(D) x2 + 3y2 + 10 x + 4xy – 4y – 3 = 0
Sol. From formula SS1 = T2
(y2 – 12x). 4 = (4y – 6x – 6)2
 3x2 +y2 +10x– 4xy –4y+3 = 0
Ans. [C]
Ex.32 Find the locus of perpendicular tangents of
the parabola x2 – 4x + 2y + 3 = 0
Sol. The given equation of parabola can be
written as
x2 – 4x + 4– 4 + 2y + 3 = 0
 (x–2)2 +2y – 1 = 0
(x–2)2 = – 2 (y –1/2) 4a = 2
 a =1/2
Directrix  y – 1/2 = 1/2  y = 1
is the equation of locus.
Ans.
13. CHORD OF CONTACT
The equation of chord of contact of tangents drawn
from a point (x1, y1) to the parabola y2 = 4ax is
yy1 = 2a (x + x1)
This equation is same as equation of the tangents
at the point (x1, y1).
Note :
(i) The chord of contact joining the point of contact of
two perpendicular tangents always passes through
focus.
(ii) Lengths of the chord of contact is
a
1
)
a
4
y
)(
ax
4
y
( 2
2
1
1
2
1 

(iii) Area of triangle formed by tangents drawn from
(x1, y1) and their chord of contact is
a
2
1 2
/
3
1
2
1 )
ax
4
y
( 

11. Examples
based on
Chord of contact
Ex.33 The chord of contact drawn from (2,4) to the
parabola y2 = 4x is-
(A) 2y = x – 2 (B) y = 2x + 2
(C) y = 2x – 1 (D) 2y = x + 2
Sol. Here a = 1 equation of chord of contact is
yy1= 2a (x + x1)
4y = 2 (x + 2)  2y = x + 2 Ans.[D]
Ex.34 The area of triangle made by the chord of
contact and tangents drawn from point (4,6)
to the parabola y2 = 8x is-
(A)
2
1
(B) 2
(C)
2
1
3 (D) None of these
Sol. Area =
a
2
1
(y1
2 – 4ax1)3/2
Here a = 2, (x1, y1) = (4,6)
 Area of triangle =
2
.
2
1
(36– 32)3/2
=
( ) /
4
4
3 2
= 2
Ans. [B]
Ex.35 The length of chord of contact of tangents
drawn from point (–2, –1) to the parabola
y2 = 4x is-
(A) 2 2 (B) 3 5
(C) 8 (D) None of these
Sol. Here a = 1, (x1, y1) (–2, –1)
Hence length of chord of contact
=
1
a
(y 4ax )(y 4a )
1
2
1 1
2 2
 
= (1 4)(1 4.1.( 2))
   = 5 9
. = 3 5
Ans. [B]
14. EQUATION OFTHE CHORD WITH GIVEN
MIDPOINT
The equation of the chord at the parabola
y2 = 4ax bisected at the point (x1, y1) is given by
T = S1 i.e. yy1 – 2a(x + x1) = y1
2 – 4ax1
Ex.36 Find the equation of a chord of the parabola
y2 = 6x if mid point of the chord is (– 1,1).
Sol. Here a = 3/2
y = – 2
3
2
(x –1) = 1 – 6 (–1)
 y – 3x + 3 = 7  y – 3 x = 4
Ans.
15. DIAMETER OF THE PARABOLA
The locus of the mid points of a system of parallel
chords of a parabola is called a diameter of the
parabola.
The equation of a system of parallel chord
y = mx + c with respect to the parabola y2 = 4ax is
given by y
a
m

2
Ex.37 The equation of system of parallel chords of the
parabola y2 =
3
2
x is y + 2x + 1 = 0 then find its
diameter.
Sol. Here 4a =
3
2
 a =
6
1
and m = – 2
Diameter is y
a
m

2
 y =
2
6
1
.
2

 y = –
6
1
This is the required equation of diameter. Ans.
15.1 Properties of Diameter :
(i) Every diameter of a parabola is parallel to its
axis.
(ii) The tangents at the end point of a diameter
is parallel to corresponding system of parallel
chords.
(iii) The tangents at the ends of any chord meet
on the diameter which bisects the chord.
16.GEOMETRICAL PROPERTIES OF THE
PARABOLA
(i) The sub tangent at any point on the parabola
is twice the abscissa or proportional to
square of the ordinate of the point.
(ii) The sub normal is constant for all points on
the parabola and is equal to its semi latus
rectum 2a.

12. (iii) The semi latus rectum of a parabola is the
H.M. between the segments of any focal
chord of a parabola i.e. if PQR is a focal
chord, then
QR
PQ
QR
.
PQ
2
a
2


(iv) The tangents at the extremities of any focal
chord of a parabola intersect at right angles and
their point of intersection lies on directrix i.e. the
locus of the point of intersection of perpendicular
tangents is directrix.
(v) If the tangent and normal at any point P of
parabola meet the axes in T and G respectively,
then
(a) ST = SG = SP
(b) PSK is a right angle, where K is the point
where the tangent at P meets the directrix.
(c) The tangent at P is equally inclined to the
axis and the focal distance.
(vi) The locus of the point of intersection of the
tangent at P and perpendicular from the focus on
this tangent is the tangent at the vertex of the
parabola.
(vii) If a circle intersect a parabola in four points,
then the sum of their ordinates is zero.
(viii) The area of triangle formed inside the parabola
y2 = 4ax is
a
8
1
(y1 – y2)(y2 – y3) (y3 – y1) where y1,y2, y3
are ordinate of vertices of the triangle.
(ix) The abscissa of point of intersection R of tangents
at P(x1, y1) and Q(x2, y2) on the parabola is G.M. of
abscissa of P and Q and ordinate of R is A.M. of
ordinate of P and Q thus R 




 
2
x
x
,
x
x 2
1
2
1
(x) If vertex and focus of a parabola are on the
x-axis and at distance a and a' from origin respectively
then equation of parabola
y2 = 4(a' – a) (x – a)
(xi) The area of triangle formed by three points on a
parabola is twice the area of the triangle formed by
the tangents at these points.
Examples
based on
Geometrical properties of a parabola
Ex.38 The latus rectum of a parabola whose focal
chord is PSQ such that SP=3 and SQ= 2 is
given by-
(A) 24/5 (B) 6/5
(C) 12/5 (D) None of these
Sol. Since the semi-latus rectum of a parabola is
the HM of segments of a focal chord.
 Semi-latus rectum =
SQ
SP
SQ
.
SP
2

=
5
12
 Latus rectum =
5
24
Ans. [A]

13. Review Chart for Standard Parabolas
y2 = 4ax x2 = 4ay
Diagram
Vertex (A) (0, 0) (0, 0)
Focus (S) (a, 0) (0,a)
Axis y = 0 x = 0
Directrix x + a = 0 y + a = 0
Equation of LR x – a = 0 y – a = 0
Length of LR 4a 4a
Extremities of LR(L1, L2) (a, 2a); (a, –2a) (2a, a) ; (–2a, a)
Focal distance of (x, y) x + a y + a
Parametric equations x = at2, y = 2at x = 2at, y = at2
Parametric points (at2, 2at) (2at, at2)
Condition of tangency c = a/m c = –am2
(for y = mx + c)
Tangent at (x1, y1) yy1 = 2a (x + x1) xx1 = 2a (y + y1)
Tangent in slope form y = mx + a/m y = mx – am2
point of contact of above (a/m2, 2a/m) (2am, am2)
Tangent at ' t' point ty = x + at2 tx = y + at2
Slope of tangent at ' t ' 1/t t
Normal at (x1, y1) y – y1 =–
2a
y1
(x – x1) y – y1 =
1
x
a
2
 (x – x1)
Normal in slope form y = mx – 2am – am3 y = mx + 2a + a/m2
Foot of above normal (am2, –2am) (–2a/m, a/m2)
Normal at ' t ' point y + tx = 2at + at3 ty + x = 2at + at3
Slope of normal at 't' –t – 1/t
Condition of normal c = – 2am – am3 c = 2a + a/m2
(for y = mx + c)
Director circle x + a = 0 y + a = 0
Diameter w.r.t. (y = mx + ) y = 2a/m x = 2am

14. Ex.1 The vertex of the parabola y2
+6x – 2y +13 =0 is
(A) (1 , –1) (B) (–2, 1)
(C) (3/2, 1) (D) (– 7/2,1)
Sol. We have : y2
+ 6x – 2y + 13 = 0
 y2
– 2y = –6x – 13
 (y – 1)2
= – 6(x + 2)
Clearly, the vertex of this parabola is (–2, 1)
Ans. [B]
Ex.2 If vertex of parabola is (2, 0) and directrix is y-
axis, then its focus is-
(A) (2, 0) (B) (–2, 0)
(C) (–4, 0) (D) (4, 0)
Sol. Since the axis of the parabola is the line which
passes through vertex and perpendicular to the
directrix, therefore x-axis is the axis of the
parabola.
Obviously Z  (0, 0).
Let focus of the parabola is S (a, 0). Since vertex
(2,0) is mid point of ZS, therefore
2
0
a 
= 2  a = 4.
 Focus is (4, 0)
Ans. [D]
Ex.3 If the focus of a parabola is (1, 0) and its directrix
is x + y = 5, then its vertex is-
(A) (0, 1) (B) (0, –1)
(C) (2, 1) (D) (3, 2)
Sol. Since axis is a line perpendicular to directrix, so
it will be x – y = k. It also passes from focus,
therefore k = 1.
So equation of axis is x – y = 1.
Solving it with x + y = 5, we get
Z  (3, 2).
If vertex is (a, b), then a = 2, b = 1.
Hence vertex is (2, 1).
Ans. [C]
Ex.4 The directrix and axis of the parabola
4y2
– 6x – 4y = 5 are respectively.
(A) 8x + 11 = 0; y – 1 = 0
(B) 8x – 11 = 0, 2y – 1 = 0
(C) 8x + 11 = 0; 2y – 1 = 0
(D) None of these
Sol. Here 4y2
– 4y = 6x + 5
 4
2
2
1
y 





 = 6 (x + 1)
Put y –
2
1
= Y, x + 1 = X
The equation in standard form Y2
=
2
3
X
4a =
2
3
 a =
8
3
Directrix, X + a = 0
 x +1+
8
3
= 0 8x + 11 = 0
Axis is Y = 0  y –
2
1
= 0 2y – 1 = 0
Ans. [C]
Ex.5 The angle subtended by double ordinate of
length 8a at the vertex of the parabola
y2
= 4ax is-
(A) 45º (B) 90º (C) 60º (D) 30º
Sol. Let (x1
, y1
) be any point on the parabola
y2
= 4ax, then length of double ordinate
2y1
= 8a  y1
= 4a
y1
2
= 4ax1
 x1
= 4a
 vertices of double ordinate are
P(4a, 4a); Q(4a, –4a)
If A is the vertex(0, 0), then
Slope of AP = 1 = m1
Slope of AQ = –1 = m2
 m1
m2
= –1  PAQ = 90º Ans. [B]
Ex.6 The length of latus rectum of a parabola, whose
focus is (2, 3) and directrix is the line
x – 4y + 3 = 0 is-
(A)
17
7
(B)
21
14
(C)
21
7
(D)
17
14
SOLVED EXAMPLES
(3/8, 0) X
O
Y
(–3/8, 0)

15. Sol. The length of latus rectum
= 2 × perp. from focus to the directrix
= 2 ×
2
2
)
4
(
)
1
(
3
)
3
(
4
2



=
17
14

The numerical length =
17
14
Ans. [D]
Note:- The negative sign of the latus rectum may only
be ignored if its length is asked. For other
calculations it should be used.
Ex.7 The coordinates of an endpoint of the latus
rectum of the parabola (y – 1)2
= 4(x + 1) are
(A) (0,–3)(B) (0,–1)
(C) (0,1) (D) (1, 3)
Sol. Shifting the origin at (–1, 1) we have







1
Y
y
1
X
x
...(i)
Using (i), the given parabola becomes.
Y2
= 4X
The coordinates of the endpoints of latus rectum
are
(X = 1, Y = 2) and (X =1, Y= –2)
Using (i), the coordinates of the end point of the
latus rectum are (0,3) and (0, –1)
Ans. [B]
Ex.8 The locus of the middle points of all focal chords
of the parabola y2
= 4ax is-
(A) y2
= 2a(x + a) (B) y2
= 2ax ( C )
y2
= 2a (x – a) (D) None of these
Sol. Let the middle point be (h, k), then equation of
the chord is given by
k2
– 4ah = yk – 2a(x + h)
Since it passes through (a, 0),
 k2
– 4ah = – 2a(h + a)
 k2
= 2a(h – a)
Hence required locus is
y2
= 2a(x – a) which is parabola.
Ans. [C]
Ex.9 The length of the chord of parabola x2
= 4ay
passing through the vertex and having slope tan
 is-
(A) 4a cosec  cot  (B) 4a tan  sec  ( C )
4a cos  cot  (D) 4a sin  tan 
Sol. Let A be the vertex and AP be a chord of
x2
= 4ay such that slope of AP is tan . Let the
coordinates of P be(2at, at2
) Then,
Slope of AP =
at
2
at2
=
2
t
tan =
2
t
 
t 2tan
Now, AP =
2
2
2
)
0
at
(
)
0
at
2
( 

 = at 2
t
4 
= 2a tan  a
tan
4
4 2
 = 4a tan  sec 
Ans. [B]
Ex.10 The point on y2
= 4ax nearest to the focus has
its abscissa equal to
(A) –a (B) a (C) a/2 (D) 0
Sol. Let P(at2
, 2at) be a point on the parabola
y2
= 4ax and S be the focus of the parabola.
Then, SP = a + at2
[ focal distance = x + a]
Clearly, SP is least for t = 0.
Hence, the abscissa of P is at2
= a × 0 = 0
Ans. [D]
Ex.11 Tangents are drawn to the points of intersection
of the line 7y – 4x = 10 and parabola y2
= 4x,.then
the point of intersection of these tangents is
(A) 





2
5
,
5
7
(B) 





2
7
,
2
5
(C) 







2
7
,
2
5
(D) 




 

2
5
,
2
7
Sol. Here let the point be (x1
, y1
)
 chord of contact yy1
= 2 (x + x1
)
compare with the given line
10
x
2 1

=
7
y1

=
4
2

 x1
= 5/2, y1
= 7/2
Point reqd. 





2
7
,
2
5
Ans. [B]
Ex.12 If the line x + my + am2
= 0 touches the parabola
y2
= 4ax, then the point of contact is
(A) (am2
, – 2am) (B) 




 
m
a
2
,
m
a
2
(C) (–am2
, – 2am)
(D) The line does not touch.
Sol. Here x = –my – am2

16. Putting in y2
= 4ax, we get
y2
= 4a (–my – am2
)
or y2
+ 4amy + 4a2
m2
= 0
or (y + 2am)2
= 0
Which is a perfect square, hence the line is a
tangent  y = –2am, also x = am2
the point reqd. is (am2
, –2am) Ans. [A]
Note:- For condition of tangency, solve the equation of
line and curve and put B2
–4AC = 0 for the
quadratic equation obtained by the two,
(the line and curve equation)
Ex.13 The common tangent of the parabola
y2
= 8ax and the circle x2
+ y2
= 2a2
is
(A) y = x + a (B) y = x – a
(C) y = x –2a (D) y = x + 2a
Sol. Any tangent to parabola is
y = mx +
m
a
2
Solving with the circle
x2
+ (mx +
m
a
2
)2
= 2a2
B2
– 4AC = 0 gives m = ± 1
Otherwise
Perp. from (0, 0) = radius a 2

2
m
1
m
a
2







= a 2  m = ± 1
Tangent y = ± x ± 2a
 y = x + 2a is correct option.
Ans. [D]
Ex.14 If two parabolas y2
= 4x and x2
= 32y intersect in
the point (16, 8) at an angle , then  is equals-
(A) tan–1
(3/5) (B) 
(C) tan–1
(4/5) (D) /2
Sol. The slope of tangent at the point (16, 8) to the
parabola y2
= 4x is
2
1
a
y =
4
1
and slope of tangent
to the parabola x2
= 32y is 1.
 tan = )
4
/
1
(
1
)
4
/
1
(
1


=
3
5
Ans. [A]
Ex.15 The slope of tangents drawn from a point
(4, 10) to the parabola y2
= 9x are-
(A)
4
3
,
4
1
(B)
4
9
,
4
1
(C)
3
1
,
4
1
(D) None of these
Sol. The equation of a tangent of slope m to the
parabola y2
= 9x is
y = mx +
m
4
9
If it passes through (4,10), then
10 = 4m +
m
4
9
 16m2
– 40m + 9 = 0
(4m –1)(4m –9) = 0  m =
4
1
,
4
9
Ans. [B]
Ex.16 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) 2 (C) 1/3 (D) 1/2
Sol. Any tangent to y2
= 4x is
y = mx + 1/m
If it is drawn from (–2, –1), then
–1 = –2m + 1/m
 2m2
– m – 1 = 0
If m = m1
, m2
then m1
+ m2
=1/2,
m1
m2
=–1/2
 tan a =
2
1
2
1
m
m
1
m
m


=
2
1
2
1
2
2
1
m
m
1
m
m
4
)
m
m
(



2
/
1
1
2
4
/
1


= 3 Ans. [A]
Ex.17 If the straight line x + y = 1 is a normal to the
parabola x2
= ay, then the value of a is
(A) 4/3 (B) 1/2 (D) 3/4 (D) 1/4
Sol. We know that equation of normal to the parabola
x2
= ay is
y =mx +
4
a
2
+ 2
m
4
a
Given that x + y = 1 or y = –x + 1 is normal to
the parabola therefore
m = –1 and
2
a
+ 2
m
4
a
= 1

2
a
+
4
a
= 1
4
a
3
= 1 a =
3
4
Ans. [A]

17. Ex.18 If line y = 2x + k is normal to the parabola y2
=
4x at the point (t2
, 2t), then-
(A) k = –12, t = –2
(B) k = 12, t = –2
(C) k = 12, t = 2
(D) None of these
Sol. Since normal to the parabola y2
= 4x at (t2
, 2t) is y +
tx = 2t + t3
.
Comparing it with y = 2x + k, we get
t = –2, k = 2t + t3
= –12
Ans. [A]
Ex.19 Which of the following lines, is a normal to the
parabola y2
= 16x
(A) y = x – 11 cos – 3 cos 3
(B) y = x – 11 cos – cos 3
(C) y = (x – 11) cos + cos 3
(D) y = (x – 11) cos – cos 3
Sol. Here a = 4
condition of normality c = –2am – am3
(1) and (2) are not clearly the answer as
m = 1 for (3), (4) m = cos
c = –2(4) cos – 4 cos3

= – 8 cos – (3 cos + cos 3)
= –11 cos – cos 3
Hence (D) is correct
Ans. [D]
Ex.20 A normal is drawn to the parabola y2
= 4ax at the
point (2a, –2 2 a) then the length of the normal
chord, is
(A) 4 2 a (B) 6 2 a
(C) 4 3 a (D) 6 3 a
Sol. Here comparing (2a,–2 2 a) with
(am2
, –2am) we get m = 2
Now length of normal chord
= 2
m
a
4
(1 +m2
)3/2
=
2
a
4
(1 + 2)3/2
= 2a 3 3
= 6 3 a
Ans. [D]
Ex.21 If a normal chord of the parabola y2
=4ax subtend
a right angle at the vertex, its slope is-
(A) ± 1 (B) ± 2
(C) ± 3 (D) None of these
Sol. If P(at1
2
, 2at1
) be one end of the normal, the other
say Q(at2
2
, 2at2
)
then t2
= – t1
–
1
t
2
...(1)
Again slope of
OP = 2
1
1
at
at
2
=
1
t
2
Slope of OQ =
2
t
2

1
t
2
×
2
t
2
= –1
 t1
t2
= –4 ...(2)
From (1) and (2)
–
1
t
4
= –t1
–
1
t
2
 +
1
t
2
= t1
 t1
2
= 2  t1
= ± 2 Ans. [B]
Note:- Slope of the normal at P is (–t1
)
Ex.22 If the tangents at P and Q on a parabola (whose
focus is S) meet in the point T, then SP, ST and
SQ are in-
(A) H.P. (B) G.P.
(C) A.P. (D) None of these
Sol. Let P (at1
2
, 2at1
) and Q (at2
2
, 2at2
) be any two
points on the parabola y2
= 4ax, then point of
intersection of tangents at P and Q will be
T  [at1
t2
,a(t1
+ t2
)]
Now SP= a(t1
2
+ 1)
SQ = a(t2
2
+ 1)
ST =a )
1
t
)(
1
t
( 2
2
2
1 

 ST2
= SP.SQ
 SP, ST and SQ are in G.P.
Ans. [B]
Ex.23 If the distance of 2 points P and Q from the
focus of a parabola y2
= 4ax are 4and 9
respectively, then the distance of the point of
intersection of tangents at P and Q from the
focus is
(A) 8 (B) 6 (C) 5 (D) 13
O
90°
Q
P
X
Y

18. Sol. If S is the focus of the parabola and T is the
point of intersection of tangents at P and Q, then
ST2
= SP × SQ  ST2
= 4 × 9  ST = 6
Ans. [B]
Ex.24 From the point (–1, 2) tangents are drawn to
parabola y2
=4x, then the area of the triangle
formed by the chord of contact and the tangent
is-
(A) 2 2 (B) 3 2
(C) 4 2 (D) 8 2
Sol. Here chord of contact y(2) = 2(1) (x –1)
or y = x – 1
Length of the chord = 2
1
4
)
1
1
)(
1
1
(
1 
 = 8
Area required =
2
1
× 8 ×
(perpendicular from (–1, 2) to chord)
= 4 ×
1
1
1
1
2



= 8 2
Ans. [D]
Note:- Direct formula may be used.