1. SOLVED SUBJECTIVE EXAMPLES
Example 1 :
Show that the line x cos + ysin = p touches the parabola y2
= 4ax if p cos + asin2
 = 0 and
that the point ofcontact is (a tan2
, – 2a tan)
Solution :
The givenline is
x cos + y sin = p
or y = – x cot  + p coses 
 m = – cot  and c = p cosec 
since the givenline touches the parabola
 c =
m
a
or cm = a
or (p cosec ) (– cot  = a
and point ofcontact is 





m
a
2
,
m
a
2
i.e. 







 cot
a
2
,
cot
a
2
or (atan2
, – 2atan)
Example 2 :
Prove that the line

x
+
m
y
= 1 touches the parabola y2
= 4a (x + b) if m2
( + b) + a2
= 0.
Solution :
The givenparabola is
y2
= 4a (x + b) .......(1)
Vertex ofthis parabola is (– b, 0)
Now shifting (0, 0) at (– b, 0) then
then x = X + (– b) and y = Y + 0 ........(2)
or x + b = X and y = Y ........(3)
From (1), Y2
= 4aX
and the line

x
+
m
y
= 1
reduces to

b
X 
+
m
Y
= 1
or Y = m 




 


b
X
1

2.  Y = 







m
X + m 







b
1 .......(4)
The line (4) willtouch the parabola (3), if
m 







b
1 = 







m
a
or

2
m








b
1 = – a
or m2
( + b) + a2
= 0
Example 3 :
Find the equations of the straight lines touching both x2
+ y2
= 2a2
and y2
= 8ax.
Solution :
The givencurves are
x2
+ y2
= 2a2
and y2
= 8ax
The parabola (2) is
y2
= 8ax
or y2
= 4(2a)x
 Equation oftangent of(2) is
y = mx +
m
a
2
or m2
x – my + 2a = 0 ......(3)
It is also tangent of(1), then the lengthofperpendicular fromcentre of(1) i.e. (0, 0)to (3)
must be equalto the radius of(1) i..e., 2
a .

 2
2
2
m
)
m
(
a
2
0
0




= a 2
or
2
2
4
2
a
2
m
m
a
4


or m4
+ m2
– 2 = 0
or (m2
+ 2) (m2
– 1) = 0
 m2
+ 2  0 (gives theimaginaryvalues)
 m2
– 1 = 0
 m = ± 1
Hence from(3) the required tangents are
x ± y + 2a = 0.

3. Example 4 :
Tangents are drawnfromthe point (x1
, y1
) to the parabola y2
= 4ax, show that the length oftheir
chord ofcontact is
|
a
|
1
)
a
4
y
)(
ax
4
y
( 2
2
1
1
2
1 
 .
Solution :
Given parabola is
y2
= 4ax ......(1)
Let P  (x1
, y1
)
Let the tangent fromP touch the parabola at Q (at1
2
, 2at1
) and R(at2
2
, 2at2
) then P is the point of
intersectionoftangents
 x1
= at1
t2
& y1
= a(t1
+ t2
) or t1
t2
=
a
x1
and t1
+ t2
=
a
y1
Now QR = 2
2
1
2
2
2
2
1 )
at
2
at
2
(
)
at
at
( 


= ]
4
)
t
t
[(
)
t
t
(
a 2
2
1
2
2
1
2


 = |a| |t1
– t2
| }
4
)
t
t
{( 2
2
1 

= | a | 








a
x
4
a
y 1
2
2
1








 4
a
y
2
2
1
{fromequation2}
= |a|
|
a
|
)
ax
4
y
( 1
2
1 
.
|
a
|
)
a
4
y
( 2
2
1  P (x , y )
1
1
R
Q
Chord of
Contact
(at , 2at )
1 1
2
(at , 2at )
2
2
2
=
|
a
|
1
)
a
4
y
)(
ax
4
y
( 2
2
1
1
2
1 
 .
Example 5 :
A ray of light is coming along the line y = b from the positive direction of x-axis and strikes a
concave mirror whose intersectionwith the x-yplane is a parabola y2
= 4ax. Find the equation of
the reflected ray and show that it passes through the focus of the parabola. Both a and b are
positive.
Solution :
A rayoflight along y= b intersects the parabola at 







b
,
a
4
b
P
2
Equation to the parabola is y2
= 4ax ... (1)
Differentiating (1)
dx
dy
=
y
a
2

4. Slope of normalat P is
a
2
b

 Ifthe angle between rayoflight and the normalis  then tan =
a
2
b
Let PQ be reflected ray
Angle between PQ and the normalat P is also 
Let the slope of PQ be m
tan =
m
a
2
b
1
a
2
b
m
a
2
b
m
a
2
b
1
m
a
2
b








a
2
b
m
m
a
4
b
a
2
b
2
2




X
Normal
Y
S
A
y=b
P








b
,
a
4
b2

a
2
b
2
a
4
b
a
4
m 2
2
2









 
 m = – 2
2
b
a
4
ab
4

 Equation to the reflected ray is y– b = – 








 a
4
b
x
b
a
4
ab
4 2
2
2







 




a
4
b
ax
4
b
a
4
ab
4
b
y
2
2
2
4abx + (5a2
– b2
) y – 4a2
b = 0
Substitute S(a, 0)
 Reflected raypasses through the focus.
Example 6 :
Find theequationofaxis, vertex, directrix, extremities oflatus rectumand lengthof latus rectumof
the parabola x2
– 3y + 2x + 3 = 0.
Solution :
x2
–3y + 2x + 3 = 0
 x2
+ 2x + 1 = 3y – 2  (x + 1)2
= 3 






3
2
y . Its axis is the line x + 1 = 0. It’s vertex is







3
2
,
1 . It’s focus is x + 1 = 0, y–
3
2
=
4
3
i.e., 






12
17
,
1 . It’s directrix is the line y–
3
2
= –
4
3
,
i.e., y+
12
1
= 0. It’s latus rectumis the line y–
3
2
=
4
3
. i.e., y–
12
17
= 0. Extremities ofit’s latus

5. rectum are x+ 1 =
2
3
, y–
3
2
, i.e., 





12
17
,
2
1
. Lengthof it’s latus rectumis 3 units.
Example 7 :
Ifthe line ax+ by+ c = 0 is a tangent to the parabola y2
= 4ax thenfind thecorresponding point of
contact. If the line ax + by + c = 0 is a normal to the parabola y2
= 4ax then find the foot of its
normal.
Solution :
Let the point of contact of tangent be P1
(at1
2
, 2at1
). Lines ax + by + c = 0 and yt1
– x – at1
2
= 0
must be identical,

b
t1
=
a
1

=
c
at
2
1

 t1
= –
a
b
.
Thus P1
 







 b
2
,
a
b2
. If foot of normal be P2
(at2
2
, 2at2
) then y + t2
x – 2at2
– at2
3
= 0 and
ax + by + c = 0 must be identical,

b
1
=
a
t2
=
c
)
at
at
2
(
3
2
2 

 t2
=
b
a
Thus P2
 







b
a
2
,
b
a 2
2
3
.
Example 8 :
Prove that the area of the triangle formed by three points on a parabola is twice the area of the
triangle formedbythe tangents at these points.
Solution :
Let the three points on the parabola be at ( 1
2
1 at
2
,
at ),
2
2
at
( , 2at2
), (at3
2
, 2at3
). The area of the  formed bythese points
1
at
2
at
1
at
2
at
1
at
2
at
|
2
1
3
2
3
2
2
2
1
2
1

| = |a2
(t1
– t2
) (t2
– t3
) (t3
– t1
)|
The intersectionofthe tangents, at these points, are the points
{at1
t2
, a(t1
+ t2
)}, {at2
t3
, a(t2
+ t3
)}, {at3
t1
, a(t3
+ t1
)}
The area of the  formed bythese points =
2
1
|a2
(t1
– t2
) (t2
– t3
) (t3
– t1
)|.

6. Example 9 :
Find thelocus ofthe point ofintersectionoftwo normals to a parabolawhichare at right angles to
one another.
Solution :
The equationofthe normal to the parabola y2
= 4ax is
y = mx – 2am – am3
It passesthrough the point (h, k) if
k = mh – 2am – am3
 am3
+ m(2a - h) + k = 0 ... (1)
Let the roots ofthe above equation be m1
, m2
and m3
.
Let the perpendicular normals correspond to the values ofm1
and m2
so that m1
m2
= –1.
From the equation (1), m1
m2
m3
=
a
k

Since m1
m2
= –1, m3
=
a
k
Since m3
is a root of (1), we have a
a
k
a
k
3







(2a - h) + k = 0
 k2
+ a(2a - h) + a2
= 0  k2
= a(h – 3a)
Hence the locus of(h, k) is y2
= a(x– 3a).
Example 10 :
Three normals froma point to the parabola y2
= 4ax meet the axisofthe parabola inpoints whose
abscissa are inA.P. Find the locus of the point.
Solution :
The equationofanynormalto the parabola is
y = mx – 2am – am3
It passesthrough the point (h, k) if
am3
+ m (2a - h) + k = 0 ... (1)
The normalcuts the axis ofthe parabola viz., y= 0 at point where x = 2a + am2
Hence the abscissaofthe points inwhich the normals through(h, k) meet the axisofthe parabola
are
x1
= 2a + am1
2
, x2
= 2a + am2
2
, x3
= 2a + am3
2
Since x1
, x2
, x3
are inA.P.
(2a + 2
1
am ) + (2a + 2
3
am ) = 2 (2a + 2
2
am )
 2
2
2
2
2
1 m
2
m
m 
 ... (2)
Also, from (1), m1
+ m2
+ m3
= 0, ... (3)
m2
m3
+ m3
m1
+ m1
m2
=
a
h
a
2 
... (4)

7. and m1
m2
m3
= –
a
k
... (5)
From(3),
(m1
+ m3
)2
= 2
2
m  2
3
2
1 m
m  + 2m1
m3
= 2
2
m

2
2
2
2
2
2
2
2
am
k
2
m
m
am
k
.
2
m
2 



 k
2
am3
2 
Since m2
is a root of (1), am2
3
+ m2
(2a – h) + k = 0
 2k + m2
(2a – h) + k = 0
 {m2
(h–2a)}3
= 27k3

a
k
2
(h – 2a)3
= 27k3
 27 ak2
= 2 (h – 2a)3
.
Hence the locus of (h, k) is 27 ay2
= 2(x – 2a)3
.
Example 11 :
P, Q are the points t1
, t2
on the parabola y2
= 4ax. The normals at P, Q meet on the parabola.
Show that the middle point of PQ lies on the parabola y2
= 2a(x + 2a).
Solution :
The points P and Q are (at1
2
, 2at1
) and (at2
2
, 2at2
)
As the normals at t1
& t2
meet onthe parabola, t1
t2
= 2 ... (i)
Also if (x1
, y1
) be the midpoint ofPQ, then
x1
=
2
1
(at1
2
+ at2
2
) and y1
=
2
1
(2at1
+ 2at2
) ... (ii)
From (iii) we get (t1
+ t2
)2
= (y1
/a)2
 (y1
/a)2
= t1
2
+ t2
2
+ 2t1
t2
= (2x1
/a) + 4, using (i) and (ii)
 y1
2
= 2a (x1
+ 2a)
Hence the locus of (x1
, y1
) is y2
= 2a(x + 2a).
Example 12 :
Ifa circle intersects the parabola y2
= 4axin four points thenshow that the algebraic sumofthe
ordinatesiszero.Also show that theline joiningonepair ofthese fourpoints andthe linejoiningthe
other pair have slopes equalinmagnitude.
Solution :
Solving y2
= 4ax and x2
+ y2
+ 2gx + 2fy+ c = 0 we have y1
+ y2
+ y3
+ y4
= 0 (y1
, y2
, y3
, y4
are
ordinates of the points ofintersection)  y1
+ y2
= – (y3
+ y4
)
Slope ofline joining (x1
, y1
) and (x2
, y2
) is m(say)
 2
1
2
1
2
2
1
2
1
2
1
2
y
y
a
4
a
4
y
a
4
y
y
y
x
x
y
y
m








Slope ofline joining (x3
y3
) and (x4
, y4
) is 
m (say)

8.  4
3
2
3
2
4
3
4
3
4
3
4
y
y
a
4
a
4
y
a
4
y
y
y
x
x
y
y
m









Since y1
+ y2
= –(y3
+ y4
),
We have m = – 
m  Slopes are equalin magnitude.
Example 13 :
Tangents aredrawn to the parabolaat three distinct points. Prove that these tangent lines always
make a triangle and that the locusofthe orthocentre ofthe triangle is the directrixofthe parabola.
Solution :
Let the three points beA(at1
2
, 2at1
), B(at2
2
, 2at2
), C(at3
2
, 2at3
).
Tangents at these point are t1
y = x + at1
2
, t2
y = x + at2
2
, t3
y = x + at3
2
.
Since t1
, t2
, t3
are distinct, no two tangents are parallel or coincident. Hence these tangents will
form a triangle. The vertices of the triangle are [at1
t2
, a(t1
+ t2
)], [at2
t3
, a(t2
+ t3
)] and
[at3
t1
, a(t3
+ t1
)]. Equation ofthe two altitudes are
[y – a(t2
+ t3
)] = –t1
[x - at2
t3
] ... (1)
and [y – a (t3
+ t1
)] = –t2
[x – at3
t1
] ... (2)
Subtracting (2) from(1), we get x = –a
Hence the locus ofthe orthocentere is x + a = 0 which is the directrixofthe parabola.
Example 14 :
Tangents are drawn to a parabola from a point T. IfP, Q are the points of contact, prove that the
perpendicular distances fromP, T and Q upon anytangent to the parabola are in G.P.
Solution :
Let the equation ofthe parabola be y2
= 4ax. Let P(at1
2
, 2at1
) and Q(at2
2
, 2at2
) be the two points
ofcontact. Then the coordinatesofT willbe (at1
t2
, a(t1
+ t2
)). Theequation ofthe tangent to the
parabola at anypoint (at2
, 2at) is ty= x+ at2
. Let p1
, p2
, p3
be the length ofthe perpendicular on
this tangent fromthe points P, T and Q respectively. Then
|
t
1
)
t
t
(
a
|
|
t
1
att
2
at
at
|
p
2
2
1
2
1
2
2
1
1







|
t
1
)
t
t
(
a
|
|
t
1
att
2
at
at
|
p
2
2
2
2
2
2
2
2
3






 so that
|
)
t
1
(
)
t
t
(
)
t
t
(
a
|
p
p 2
2
2
2
1
2
3
1



 =
2
2
2
1
2
2
1
t
1
)
t
t
(
at
at
t
at










= p2
2
 p1
, p2
, p3
are in G.P.
Example 15 :
Three normals aredrawnfromthe point (c, 0)to the curve y2
=x. Show that c must begreater than
1/2. One normalis always the x-axis. Find c for whichthe other two normalsare perpendicular to
eachother.

9. Solution :
Equation to normalto the parabola
y2
= x is y = mx –
2
m
–
4
m3
is passing through (c, 0)
0 = cm –
4
m
2
1
c
m
4
m
2
m 3
3










m = 0 or
4
m
2
1
c
2


0
2
1
c 
 







 0
4
m
ce
sin
2
c 
2
1
. If c =
2
1
, then m = 0
Then onlyone normalwillbethere i.e., x-axis.
Since three normals are there,  c >
2
1
Then 


















2
1
c
2
2
1
c
2 = –1
 c –
4
1
2
1
  c =
4
3

10. SOLVED OBJECTIVE EXAMPLES
Example 1 :
Equation ofparabola having it’s focus at S(2, 0) and one extremityofit’slatus rectumas (2, 3) is
(A) y2
= 4(3 – x) (B) y2
= 4(1 – x)
(C) y2
= 8(3 – x) (D) y2
= 8(1 – x)
Solution :
Clearlythe other extremityoflatus rectumis (2, – 2). It’s axis is x-axis. Corresponding value of
a =
2
0
2
=1. Hence it’s vertexis (1, 0) or(3, 0). Thus it’sequationis y2
=4(x– 1)
or y2
= – 4(x – 3).
Example 2 :
Equation ofparabola having the extremities ofit’s latus rectumas (3, 4) and (4, 3) is
(A)
2
3
7
x 





 +
2
3
7
y 





 =
2
2
6
y
x





 

(B)
2
2
7
x 





 +
2
2
7
y 





 =
2
2
8
y
x





 

(C)
2
2
7
x 





 +
2
2
7
y 





 =
2
2
4
y
x





 

(D) None ofthese
Solution :
Focus is 





2
7
,
2
7
and it’s axis is the line y= x. Corresponding value of‘a’is
4
1
)
1
1
(  =
4
2
.
Let the equation of it’s directrix be y + x + = 0. 
2
|
4
x
3
| 


= 2.
4
2
 = – 6, – 8
Thus equationofparabola is
2
2
7
x 





 +
2
2
7
y 





 =
2
)
6
y
x
( 2


or
2
2
7
x 





 +
2
2
7
y 





 =
2
)
8
y
x
( 2


.
Example 3 :
An equilateral triangle is inscribed in the parabola y2
= 4ax, such that one vertex ofthis triangle
coincides with the vertexofthe parabola. Side length ofthistriangle is -
(A) 4a 3 (B) 6a 3
(C) 2a 3 (D) 8a 3

11. Solution :
If triangle OAB is equilateral then OA = OB = AB. Thus AB will be a double ordinate of the
parabola.
Thus AOX = XOB =
6

.
Let A = (at1
2
, 2at1
) then B  (at1
2
, 2at1
)
mOA
=
1
t
2
=
3
1
 t1
= 3
2  AB = 4at1
= 8a 3 units.
Example 4 :
If the normal drawn to parabola y2
= 4ax at the point A(at1
2
, 2at1
) meets the curve again at
B(at2
2
, 2at2
) then
(A) |t2
|  2 2 (B) |t2
|  2 2
(C) |t1
|  2 2 (D) |t1
|  2 2
Solution :
We have t2
= – t1
–
1
t
2
 t1
2
+ t1
t2
+ 2 = 0.
Since ‘t1
’ is real, thus t2
2
– 8  0  |t2
|  2 2 .
Example 5 :
Locus oftrisectionpoint ofanydouble ordinate ofy2
= 4ax is
(A) 3y2
= 4ax (B) y2
= 6ax
(C) 9y2
= 4ax (D) None ofthese
Solution :
LetAB be a double ordinate, whereA  (at2
, 2at), B  (at2
, –at). IfP(h, k) be it’s trisection point
then 3h = 2at2
+ at2
, 3k =4at – 2at  t2
=
a
h
, t =
a
2
k
3
Thus locus is 2
2
a
4
k
9
=
a
h
, i.e., 9y2
= 4ax.
Example 6 :
Tangents drawn to parabola y2
= 4ax at the pointAand B intersect at C. If‘S’be the focus ofthe
parabola then SA, SC and SB are in
(A)A.P. (B) G.P.
(C) H.P. (D) noneofthese

12. Solution :
IfA  (at1
2
, 2at1
), B  (at2
2
, 2at2
),  C  (at1
at2
, a(t1
+ t2
)).
Now SA = a + at1
2
, SB = a + at2
2
,
SC = 2
2
1
2
2
2
1 )
t
t
(
a
)
a
t
at
( 


 a 2
2
2
1
2
2
2
1 t
t
1
t
t 

 = a )
t
1
)(
t
1
( 2
2
2
1 

Clearly SC2
= SA. SB
Example 7 :
Double ordinatesAB of the parabola y2
= 4axsubtends an angle /2 at the focus ofthe parabola
then tangents drawnto parabola atAand B willintersect at
(A) (– 4a, 0) (B) ( –2a, 0)
(C) (–3a, 0) (D) None ofthese
Solution :
Let A  (at2
, 2at), B  (at2
, – 2at). mOA
=
t
2
, mOB
= –
t
2
. Thus
t
2
. –
t
2
= – 1  t2
= 4.
Thus, tangentswillintersect at (–4a, 0).
Example 8 :
Ifa normal chord ofy2
= 4ax subtends anangle /2 at the vertex ofthe parabola thenit’s slope is
equalto
(A) ± 1 (B) ± 2
(C) ± 2 (D) noneofthese
Solution :
Let AB be a normal chord, whereA (at1
2
, 2at1
) and B  (at2
2
, 2at2
).
We have t2
= – t1
–
1
t
2
and t1
t2
= 4  t1
t2
= – t1
2
– 2 = – 4  t1
2
= 2.
Now slope of chord AB =
2
1 t
t
2
 = – t1
= ± 2
Example 9 :
Length ofthe shortest normalchord ofthe parabola y2
= 4ax is
(A) a 27 (B) 3a 3
(C) 2a 27 (D) noneofthese
Solution :
Let AB be a normal chord where A (at1
2
, 2at1
), B  (at2
2
, 2at2
). We have t2
= – t1
–
1
t
2
.
AB2
= [a2
(t1
– t2
)2
(t1
+ t2
)2
+ 4] = a2
2
1
1
1
t
2
t
t 







 







 4
t
4
2
1
= 4
1
3
2
1
2
t
)
t
1
(
a
16 

13. 
1
2
dt
)
AB
(
d
= 16a2







 


8
1
3
1
3
2
1
1
2
2
1
4
1
t
t
4
.
)
t
1
(
]
t
2
.
)
t
1
(
3
[
t
= 5
1
2
2
1
2
t
)
t
1
(
32
.
a 
(t1
2
– 2)
t1
= 2 is indeed the point ofminima ofAB2
.
Thus ABmini
=
2
a
4
(1 + 2)3/2
= 2a 27 units.
Example 10 :
Ifthree distinct and real normals can be drawn to y2
= 8x fromthe point (a, 0) then
(A) a > 2 (B) a > 4
(C) a (2, 4) (D) noneofthese
Solution :
Equation of normal in terms of m is y = mx – 4m – 2m3
. If it passes through (a, 0) then
am – 4m – 2m3
= 0
 m(a – 4 – 2m2
) = 0  m = 0, m2
=
2
4
a 
.
For three distinct normal, a – 4 > 0  a > 4
Example 11 :
An equilateraltriangle SAB is inscribed inthe parabola y2
= 4axhaving it’s focus at ‘S’. Ifchord
AB lies towards the left ofS, thenside length ofthis triangle is
(A) 3a(2 – 3 ) (B) 4a(2 – 3 )
(C) 2(2 – 3 ) (D) 8a(2 – 3 )
Solution :
Let A (at1
2
, 2at1
), B  (at2
2
, – 2at1
). We have mAS
= tan 




 
6
 a
at
at
2
2
1
1
 = –
3
1
 t1
2
+ 2 3 t1
= – 1 = 0  t1
= – 3 ± 2
Clearly t1
= – 3 – 2 is rejected. Thus t1
= (2 – 3 ). HenceAB = 4at1
= 4a(2 – 3 ).
Example 12 :
Locus of the midpoint of anyfocalchord ofy2
= 4ax is
(A) y2
= a(x – 2a) (B) y2
= 2a(x – 2a)
(C) y2
= 2a(x – a) (D) noneofthese

14. Solution :
Let the midpoint be P(h, k). Equationofthis chord isT = S1
. i.e., yk – 2a(x +h) = k2
–4ah. It must
pass through(a, 0)
 2a(a + h) = k2
– 4ah. Thus required locus is y2
= 2ax – 2a2
.
Example 13 :
Slope of the normal chord ofy2
= 8xthat gets bisected at (8, 2) is
(A) 1 (B) – 1
(C) 2 (D) – 2
Solution :
Let AB be the normal chord where A (2t1
2
, 4t1
), B  (2t2
2
, 4t2
). It’s slope =
2
1 t
t
2
 We also
have t2
= – t1
–
1
t
2
and 16 = 2(t1
2
+ t2
2
), 4 = 4(t1
+ t2
)  t1
+ t2
= 1. Thus slope is 2.
Example 14 :
The lengthoflatus rectumofhe parabola, whosefocus is (a sin2, a cos2) anddirectrixisthe line
y = a, is
(A) |ra cos2
| (B) |4a sin2
|
(C) |4a cos2| (D) |a cos2|
Solution :
Distance offocus fromdirectrix is |a cos 2 – a|. Thus lengthoflatus rectumis|4a sin2
|.
Example 15 :
In the adjacent figure a parabola is drawn to pass through the vertices B, C and D ofthe square
ABCD. IfA(2, 1), C(2, 3) then focus ofthis parabola is -
(A) 





4
11
,
1 (B) 





4
11
,
2
(C) 





4
13
,
3 (D) 





4
13
,
2
Solution :
ClearlyAC is parallelto y-axis. It’s midpoint is (2, 2). Thus B  (1, 2).
Parabola will be in the form of (x – 2)2
= (y– 3).
It passes through (3, 2)
 l = – . Thus parabola is (x – 2)2
= – 1(y – 3).
It focus is x – 2 = 0. y – 3 = –
4
1
, i.e., 





4
11
,
2 .