Mathematics

SECTION-1 : (Only One Options Correct Type)

This section contains 30 multiple choice type questions. Each question has four choices (A), (B), (C) and
(D) out of which ONLY ONE is correct.

1. If the normals drawn at the end points of a variable chord PQ of the parabola
2
y = 4 ax intersect at parabola, then the locus of the point of intersection of the tangent
drawn at the points P and Q is
(A) x +a = 0 (B) x – 2a = 0
(C) y2 − 4x + 6 = 0 (D) none of these .

2. If tangents at A and B on the parabola y2 = 4ax intersect at the point C, then ordinates of A, C
and B are
(A) always in A.P. (B) always in G.P.
(C) always in H.P. (D) none of these

3. Equation of the circle of minimum radius which touches both the parabolas y = x2 + 2x + 4 and
x = y2 + 2y + 4 is
(A) 2x2 + 2y2 – 11x – 11y – 13 = 0 (B) 4x2 + 4y2 – 11x – 11y – 13 = 0
2 2
(C) 3x + 3y – 11x – 11y – 13 = 0 (D) x2 + y2 – 11x – 11y – 13 = 0

4. The locus of the centre of a circle which cuts orthogonally the parabola y2 = 4x at (1, 2) will pass
through points
(A) (3, 4) (B) (4, 3)
(C) (5, 3) (D) (2, 4)

5. A parabola y = ax2 + bx + c crosses the x-axis at (,0)and(,0) both to the right of the origin. A circle
also passes through these two points. The length of a tangent from the origin to the circle is :
bc
(A) (B) ac2
a
b c
(C) (D)
a a

6. Centre of locus of point of intersection of tangents to the parabola y2 = 4ax, the angle between them
being always 45 is
(A) (3a, 0) (B) (–3a, 0)
(C) (3a, a) (D) (a, 3a)

2  1 1
7. All the tangents to the parabola y = – 4(x – 1) whose slope lie in the interval R –   ,  are
 2 2
chords bisected by the line x = 1 to a circle. Equation of the circle is
(A) x2 + y2 = 5 (B) x2 + y2 = 4
2 2
(C) x + y – 2x = 0 (D) x2 + y2 – x – 4 = 0
2
8. In a parabola y = 4ax, two points P and Q are taken such that the tangents drawn to parabola at
these points meet at directrix in R. If a circle is drawn circumscribing the PQR then the focus of the
conic described by centre of circumscribing circle lies on the line
3 3
(A) y = x – a (B) y = –x – a
2 2
(C) y = x – a (D) y = x + a

9. If 2p2 – 3q2 + 4pq – p = 0 and a variable line px + qy = 1 always touches a parabola whose axis is
parallel to x-axis. Then the equation of the parabola is
2 2
(A) (y – 4) = 24(x – 2) (B) (y – 3) = 12(x – 1)
2 2 2
(C) (y – 4) = 12(x – 2) (D) (y – 2) = 24(x – 4)

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10. From a point P on x2 + 3y2 = 1 tangents PA and PB are drawn to y2 = 4x such that angle between
the tangent is 30°, then the co-ordinate of point where circumcircle of  PAB intersect the x-axis is :
7   5 
(A)  ,0  (B)  ,0 
 3  3 3 
(C)  3,0  (D) none of these

11. Two equal circles of largest radii having following property

(i) They intersect each other orthogonally.
(ii) They touch both the curves 4(y+2) = x2 and 4(2y) = x2 in the region x[2 2 , 2 2 ].
Then radius of this circle is
(A) 2 (B) 3
1 3
(C) (D)
3 2

12. If 2x + 3y = 7 and x – y = 1 are two normals of a parabola y2 = 4ax from a point then third normal
may be
(A) x – 3y + 1 = 0 (B) x + 3y = 5
(C) x + 3y + 1 = 0 (D) x – 3y – 7 = 0

13. The length of sub tangent, ordinate and subnormal to the parabola y2 = 4ax at a point (different from
origin) are in
(A) AP (B) GP
(C) HP (D) AGP

14. Consider a point P(at2, 2at) on the parabola y2 = 4ax, A focal chord PS (S focus) is drawn to meet
parabola again at Q. From Q, a normal is drawn to meet parabola again at R. From R, a tangent is
drawn to the parabola to meet focal chord PSQ (extended) at T. The area of QRT is
3 4
 1  1
(A) 8a2  t 2  2  (B) a2  t  
 t   t
3 4
8a2  1  1
(C)  t  t  (D) 4a2  t  
3  t

15. P is a point ‘t’ on the parabola y2 = 4ax and PQ is a focal chord. PT is a tangent at P and QN is a
normal at Q. The minimum distance between PT and QN is equal to
 1 
(A) 0 (B) a  1  t 2  
 1  t2 
3/2
a  t 2  1 a 1  t2
(C) (D)
t2 t2

16. Equation of a normal to the curve y  x 2  6x  6 which is perpendicular to the straight line joining
the origin to the vertex of the parabola is
(A) 4x – 4y – 11 = 0 (B) 4x – 4y + 1 = 0
(C) 4x – 4y – 21 = 0 (D) 4x – 4y + 21 = 0
2
17. If normals to parabola y = 4ax at three points, P, Q, R meet at A and S be the focus, then
SP  SQ  SR
is (where SP, SQ, SR, SA are distances)
 SA 2
(A) 4a (B) 2a
(C) a (D) none of these

18. Let y2 = 4x be a parabola and P(h, k) be a point on it. The values of h for which three different
normals can be drawn from P to parabola is
(A) h > 2 (B) h > 8
(C) h > 6 (D) none of these

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2
19. Consider a parabola x = 4y, then the number of positive integral points on the parabola through
which exactly one normal can pass is/are
(A) 1 (B) 2
(C) 3 (D) none of these

a  b 
20. The line 2y  ax  b is a tangent to parabola y 2  4ax (a  N), then for max value of   the
 a 
value of ‘a’ is ([.] denotes the greatest integer function)
(A) 2 (B) 3
(C) 1 (D) 4
2 2 2
21. If (h, k) is a point on the axis of the parabola 2(x 1) + 2(y  1) = (x + y + 2) from where three
distinct normals may be drawn then
(A) h > 2 (B) h < 4
(C) h > 8 (D) h < 8

22. A circle has centre on the x-axis, and touches the parabola y 2  8  x  2  internally at one point. A
tangent is drawn from (–4, –6) to the circle. The minimum slope of the tangent is
93 5 9  2 14
(A) (B)
5 5
9  2 11 9  23
(C) (D)
5 5

23. Normal at P to the parabola (12x + 5y + 3)2 = 52(5x12y+1) meets the line 12x + 5y + 3 = 0 at G,
perpendicular GN is drawn to SP, S being focus, then NP is equal to
(A) 1 (B) 2
(C) 3 (D) 26

24. Tangents at points P and Q to the parabola y2 = 4ax intersect at T, where ST  2 10 , S being
focus, if product of length of perpendiculars from focus to these tangents is 8 10 , then length of
latus rectum of parabola is
(A) 4 (B) 2 3
(C) 2 5 (D) 16

25. A tangent is drawn to the parabola y2 = 4(x + 2) and cut the coordinate axis at A and B. If rectangle
OAPB is completed then locus of P is
3 2
(A) x + y + 2 = 0 (B) y + 3x = 0
2 2
(C) x + y (x + 2) = 0 (D) none of these

26. From the point P (2, ), pair of tangents PA and PB are drawn to the extremities of the chord x + y
– 2 = 0 of the parabola y2 + 2y – 8x + 33 = 0 If  APB = /2, then exhaustive set of values of  and 
are
(A)   (0, ),  R (B)   (, 0),   (0, )
(C)   R,   R (D)  R,  (0, )
2 2
27. Let A be the area between co-ordinate axis, y = x – 1, x = y – 1 and the line which makes the
shortest distance between two parabolas and A be the area between x = 0, x2 = y – 1, x = y and the
2 2
shortest distance between y = x – 1 and x = y – 1, then
1
(A) A=A (B) A  (A ) 2
(C) A  2A  (D) can’t say anything

28. The circum circle of ABC is x2 + y2 – 5x – 4y + 6 = 0, if a parabola (k + 1)y2 = x have sides AB, BC,
CA as tangents then the value of k is/are:
7 11
(A) – (B) –
8 13
2 4
(C) (D)
3 7
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2
29. The point P on the parabola y = 4ax for which |PR – PQ| is maximum, where R  (– a, 0), Q  (0,
a), is
(A) (a, 2a) (B) ( a, -2a)
(C) (4a, 4a) (D) (4a, -4a)

30. The locus of the centre of the circle described on any focal chord of a parabola
y2 = 4ax as diameter is
2 2
(A) x = 2a (y – a) (B) x = –2a(y – a)
2 2
(C) y = 2a (x – a) (D) y = –2a(x –a)

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HINTS & SOLUTIONS

1. B
Let the points P and Q be ( at12 , 2at1) and ( at22, 2at2) respectively. Since the normals at P
and Q intersects at parabola , t1t2 = 2
Let ( h, k) be the point of intersection of the tangents at P and Q .
Then at1t2 = h and a( t1 + t2) = k
 h = 2a
 x – 2a = 0

2. A
2
A  (at1 , 2at1)
B  (at22, 2at2)
Tangents at A and B will intersect at the point C, whose coordinate is given by
(at1t2, a(t1+t2)) clearly ordinate of A, C and B are always in A.P.

3. B
Given parabolas are symmetric about the line y = x so they have a common normal with slope–1 it
 1 13   13 1  2 2 11 11 13
meets the parabolas at  ,  , ,  hence the required circles is x + y  x  y  0
 2 4 4 2 4 4 4

4. A
Tangent to parabola y2 = 4x at (1, 2) will be the locus
i.e. y . 2 = 2(x + 1)
y = x + 1.

5. D
Use power of a point;
c c
OT2 = OA. OB =    OT  .
a a , 0)
, 0)
O A B

6. B
a
Let tangent for parabola y2 = 4ax is y = mx +
m
2 y a
 m x – my + a = 0  m1 + m2 = , m1m2 =
x x
2 2
m1  m2   a    y  4a 
Then tan 45 =  1          
1  m1m2   x   x  x 

Locus is x2 – y2 + 6ax + a2 = 0
 (x + 3a)2 – y2 = 8a2
 Rectangular Hyperbola.
 Centre is (–3a, 0).

7. A
y2 = – 4(x – 1)
1
y = m(x – 1) –
m
y 1
 x  1 2
m m
1
Let – =t
m
x + yt = 1 + t2 …(1)
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Maths-6

let the equation of circle be x2 + y2 + 2gx + 2fy + c > 0

let (1, ) be the bisecting point
x + y + g(x + 1) + f(y + ) + c = 1 + 2 + 2g + 2f + c
2
x(1 + g) + y( + f) = 1 +  + g + f …(2)
comparing (1) and (2)
f
t
1 g
1   2  g   f
 1  t2 .
1 g
Note that  and t are parameters
2    f 2
 + (g + f) + 1 = 1 + g +
1 g
2 2 2 2
 (1 + g) + (1 + g)(g + f) = g + g + f +  + 2f
 g2 + [(1 + g)(g + f) – 2f] – g – g2 – f 2 = 0 is true for more than two value of .
 g = 0, g + g2 + f 2 = 0, (1 + g)(g + f) – 2f = 0
 g = 0, f = 0
 equation of the circle is x2 + y2 = a2
x + yt = 1 + t2
1
t=–
m
 1 1
 – 2 < t < 2 as m  R –   ,  .
 2 2
(1, t) should be an interior point
 1 + t2 – a2 < 0
 a2 > 1 + t2  a2 = 5.
8. A
Let P(t1) and Q(t2) be required
points.
Now at1t2 = – a P(at12, 2at1)
 t1t2 = – 1 …(1)
Also, PRQ = 90 since R lies on S(2a+a(t12+t22+t1t2, –at1t2(t1+t2))
the directrix x = – a (at1t2, a(t1+t2))R
 PQ is the diameter, coordinates
of centre are (using (1))
a 2 1  1 
  t1  2  , a  t1    2
Q(at2 , 2at2)
2
  t1   t1  

x=–a
Now, locus described by centre (h, k) is
a 1
h =  t12  2 
2 t1 
 1
k = a  t1  
 t1
2
2h  k 
   2
a a
 y2 = 2a(x – a)
3 
so, focus is  a, 0  .
2 

9. C
2
The parabola be (y – a) = 4b(x – c)
p bq
Equation of tangent is y – a = –  x  c  
q p
Comparing it with px + qy = 1, we get
cp2 – bq2 + apq – p = 0
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 Maths-7

c b a
   1
2 3 4
 The equation is (y – 4)2 = 12(x – 2).

10. A
 1  1 2 
Co-ordinates of P  0,   , A(0 , 0) and B  , 
 3 3 3 

11. A
r Y
From Geometry, P(r 2, ).
2 r
P(r 2, )
Now this point lies on the curve 4(2 y) = x2 2
r
 r  2
 42 

 r 2
2
  X
L

 r 2

12. B
Clearly point of intersection of normals is (2, 1) and as sum of slopes is zero
 m1 + m2 + m3 = 0
2
   1  m3  0
3
1
 m3  
3
1
 equation is y  1    x  2 
3
 x + 3y = 5
13. B
dx
Clearly length of sub tangent = y A
dy
Length of ordinate = |y| = B
dy
Length of subnormal = y C
dx
 B2  AC
 A, B, C are in GP
14. C
 1  a 2a 
PQ is a focal chord Q      2 ,  
 t  t t 
 2 
Normal drawn at P(t) meets the curve at Q   t,
 t 
 1  1
 Normal drawn at Q    meets the parabola again at R  2t  t 
 t
  2
1  1
R a  2t   , 2a  2t   
  t  t 
2
 1  1
Equation of tangent at R is y  2t    x  a 2t   ….. (1)
 t  t
 2  
Equation of focal chord PQ is t  1 y  2t x  a  ….. (2)
T is the point of intersection of line (1) and (2)
a  1  2a  1 
 T   4t 2  2  ,  4t  
3
  t  3  t 
3
8a2  1
 Area of  TQR = t
3  t 

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Maths-8

15. C
a 2a 
P  at 2 , 2at  , Q  2 , 
t t 
Equation of tangent PT is ty = x + at2
Equation of normal at Q is xt2 – yt3 – a – 2at2 = 0
 PT || QN
3/2
at 4  a  2at 2 a  t 2  1
d 
t4  t6 t2

16. C
2
y  3   x  3   vertex (3, –3)
Slope of the line joining vertex and origin = –1
 Slope of normal = 1
dy dx 5 11
Now  2x  6 and   1  x  and y  
dx dy 2 4
11  5
 Equation of normal is y   1 x  
4  2

17. C
Normal  y + tx = at3 + 2at
Let A = (h, k)
 at3 + 2at – ht – k = 0
SP = a 1  t12  , SQ = a 1  t 22  , SR = a 1  t 23 
2 2
 t   t 
2
1 1 2 t t 1 2    2a  h 
a
2  2a  h 2
 t t   t t 
2 2
1 2 1 2 2 t t 12   t 2t 3  
a2
 SP.SQ.SR = a   h  a   k 2   a  SA 
2 2

18. B
y = mx – 2m – m3
y2 h
x=h
3
m + 2m – mh + 2 h = 0 has three real roots for h > 8.

19. B
Let, P(t) be a point on the given parabola and Q(t1) is point on which normal is drawn which passes
2
through P(t) then t1   t 
t
2
 t + t1t + 2 = 0 ..... (1)
Now, for exactly one normal there will be no real solution for ‘t’ in equation (1)
 t12  8  0  t12  8
2 2  t1  2 2

20. C
ax b
Equation of tangent is y  
2 2
b a2
  b4
2 a
 a=1

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 Maths-9

21. A
1 1 1 2
a = 2
2 2
2
h–1>
2
h>1+1
S(1,1)
h>2

22. B
Shift the origin to (2, 0). The parabola becomes y 2  8x and the point is  2, 6  .
The circle touches the parabola at one point. Therefore only one normal can be drawn. The
maximum distance of the centre from origin is 2a
2
Equation of the biggest circle is  x  4   y 2  4
Again shift the origin to  4,0  then circle and point are x 2  y 2  42 ,  6, 6 
Equation of tangent to the circle is y  mx  4 m2  1, which passes through  6, 6  .

 6  6m  4 m2  1
18  224
Solving, we get m 
10
18  4 14
 Minimum slope = .
10

23. B
NP is equal to semilatus rectum.

24. D
Product of length of perpendiculars from focus to tangents = a(ST)

25. C
For y2 = 4(x + 2) parametric coordinates are (t2 – 2, 2t) so equation of tangent is
2
y(2t) = 2(x + t – 2) + 8
2
ty = x + t + 2
2
A  (–t – 2, 0)
 t2  2 
B   0, 
 t 
t2  2
 h = – t2 – 2, k =.
t
h h
k=– t=–
t k
 h2 
so locus is h = – 2 –  2 
k 
 h2 
h+2+  2 =0
k 
 xy2 + x2 + 2y2 = 0
 x2 + y2(x + 2) = 0.

26. C
2
Parabola is (y + 1) = 8 (x – 4)
Focus = (2, 0)
Directrix : x = 2.

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Maths-10

Clearly P (2, ) lies on the directrix of the parabola and x + y  2 = 0 is the equation of any focal
chord.

27. C
Line of shortest distance will be x2 =y–1 x=y
perpendicular to line x=y. Also both
parabolas are symmetric about x=y. C
A=area OABCDO D
E
A=area OECDO B
(0,1) x+y=k
 A =A/2.
O
(1,0) A

y2 =x–1

28. A
 1 
Circumcircle of ABC passes through focus  ,0  of the given parabola .
 4k  4 
1 1
Putting y = 0, in equation of circumcircle, we get either = 2 or = 3.
4k  4 4k  4

29. A
We know any side of the triangle is more than the difference of the remaining two sides so that |PR
– PQ|  RQ
 The required point P will be the point of intersection of the line RQ with parabola which is (a,
2a) as PQ is a tangent to the parabola.

30. C
If A(at12 ,2 at1) , B(at22 ,2 at2) be the extremities of a focal chord for the parabola y2 = 4ax, then
t1t2 = –1 . . . . . (1)
y A 2
y = 4ax

C
O S x

B
We want to find locus of point C(, ) where C is the centre of the circle having AB as the
diameter.
a 2
=
2
 2

t 1  t 2 ;  = a( t1+ t2)
To eliminate t1, t2
 2 
2 = a2(t12 +t22 + 2t1t2) = a2  a  2 
 
 2 = 2a( - a)
2
i.e. y = 2a(x –a) .

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