ConicSectionsExercise-3.pdf - MT Educare

Q. No. 1 The angle subtended by double ordinate of length 8a at the vertex of the parabola

y2 = 4ax is

Option 1 450

Option 2 900

Option 3 600

Option 4 300

Correct Answer 2

Explanation

AB = 8a

A 4 ,4 B 4 ,4a a a a� ��

AOB=90�� 0

Q. No. 2 The length of the chord of parabola x2

= 4ay passing through the vertex and having

slope than is ��Option 1 4a cosec �� cot ��Option 2 4a tan �� sec ��Option 3 4a cos �� cot ��Option 4 4a sin �� tan ��Correct Answer 2

Explanation

OA: = tany x��

S: = 4x ay2

A 4 tan ,40tana� � �� 2

OA = 4 tan .seca� � ��

Q. No. 3 If the line x + my + am2

= 0 touches the parabola y2

= 4ax, then the point of contact is

Option 1 (am2, 2am)

Option 2 2,

a a

mm

� �� 2

Option 3 (am2, 2am)

Option 4 The line does not touch

Correct Answer 1

Explanation =

xy am

m is tangent to y

2 = 4ax. Point of touch �� (am

2, 2am)

Q. No. 4 A circle described on any focal chord of the parabola, y2

= 4ax as its diameter will

touch :

Option 1 The axis of the parabola

Option 2 The directrix of the parabola

Option 3 the tangent drawn at the vertex of the parabola

Option 4 Latus rectum

Correct Answer 2

Explanation Directrix property of parabola

Q. No. 5 The locus of the middle points of the focal chords of the parabola, y2

= 4x is :

Option 1 y2

= x 1

Option 2 y2

= 2(x 1)

Option 3 y2

= 2(1 x)

Option 4 None of theses

Correct Answer 2

Explanation A ,2at at�� 2

1 1

–2B ,

a a

tt

� ��

211

1+

1h ; =

2

a tt

k a tt

� ��

� ��

2

1 2

1

1

1

12h = + 2a t

t

� ��

2

1

1

12h = + 2a a t a

t

� ��

2

2 2

1

1

�� 2x = y2 + 2 [compare with y

2 = 4x]

Q. No. 6 The length of the side of an equilateral triangle inscribed in the parabola, y2

= 4x so

that one of its angular point is at the vertex is :

Option 1 8 3

Option 2 6 3

Option 3 4 3

Option 4 2 3

Correct Answer 1

Explanation

2OA : tan30 =

t

t

0 1

2

1

= 2 3t�� 1

AB = 8 3��B is image of s in axis of parabola.

Q. No. 7 The condition that the line, x. cos �� + y. sin �� = p touches the parabola, y2

= 4a (x + a) is

:

Option 1 cos = 0a p ��

Option 2 + cos = 0a p ��

Option 3 cos = 0a p��

Option 4 cos + = 0a p��

Correct Answer 2

Explanation S = y2

= 4a(x + a)

Any tangent can be taken as :

ty = x + a + at2

cos + cos =x y p� �� is a tangent.

1 +1= =

sin cos� ��

2a tt

p

.sec– sec =

a

p

��

2

+ cos 0a p� � ��

Q. No. 8 If the normal to the parabola y2

= 4ax at the point (at2, 2at) cuts the parabola again at

(aT2, 2aT), then

Option 1 2 �� T �� 2

Option 2 Τ ( ,8) (8, )� � � ��

Option 3 Τ 8��2

Option 4 Τ 8��2

Correct Answer 4

Explanation S : y2

= 4ax

Normal at (at2, 2at) cuts the parabola again at

2 2 ,2 – –a t a t

t t

� ��

2

2= T t

t

+ T + 2 = 0 , 1 0t t t R� �� 2

0� ��

T – 8 0��2

Q. No. 9 If the straight line x + y = 1 is a normal to the parabola x2

= ay, then the value of a is

Option 1 4/3

Option 2 1/2

Option 3 3/4

Option 4 1/4

Correct Answer 1

Explanation S : x2

= ay

x + y = 1 is normal to S.

Any normal to S is

2 9+ = +

4 4

ax ty t t

3

at 2 9

+4 4

at t

� ��

2

2 9= 1and + = 1

4 4

at t t�� 3

4=

3a��

Q. No. 10 A normal is drawn to the parabola y2 = 4ax at the point 2 ,2 2a a then the length of

the normal chord, is

Option 1 4 2 a

Option 2 6 2 a

Option 3 4 3a

Option 4 6 3a

Correct Answer 4

Explanation S = y2

= 4ax

2 ,2 2 = 2P a a t� �� 1 1

2= + 1

2t1

8 ,4 2P a a��2

= 6 3P P a1 2

Q. No. 11 Which of the following lines, is a normal to the parabola y2

= 16 x

Option 1 y = x 11 cos �� 3cos 3 ��Option 2 y = x 11 cos �� cos 3 ��Option 3 y = (x 11) cos �� + cos 3 ��Option 4 y = (x 11) cos �� cos 3 ��Correct Answer 4

Explanation S : y2

= 16x �� a = 4

Normal with stop = 1

y = x 12

None of the option can be compared

Normal with slope cos ��

= cos 8cos 4cosy x � � �� 3

= cos 11 +cos3x� ��

Q. No. 12 If the distance of 2 points P and Q on parabola y2

= 4ax from the focus are 4 and 9

respectively, then the distance of the point of intersection of tangents at P Q from the

focus is

Option 1 8

Option 2 6

Option 3 5

Option 4 13

Correct Answer 2

Explanation ,2 ,2P at at Q at at�� 2 2 2

1 1 2 2

Point of intersection of tangent at P and Q

R , +at t a t t�� 1 2 1 2

PF = 4 = a(1 + t1

2)

QF = 9 = a(1 + t2

2)

RF = 1 + +a t t a t t2 22

1 2 1 2

= 1 1 + = 6a t a t2 2

1 2

Q. No. 13 If the normal at two points P, Q of the parabola, y2

= 4x intersect at a third point R on

the parabola, then the product of the ordinates of P and Q is :

Option 1 4

Option 2 6

Option 3 16

Option 4 8

Correct Answer 4

Explanation Any normal with slope m :

y = mx 2am am3, passing through (h, k)

k = hm 2am am3 intersect the parabola at (am

2, –2am)

�� m1 + m2 + m3 = 0

m1 m2 + m3 (m1 + m2) = 2 a h

a

m1 m2 m3 = k

a

Let (h,k) �� (am1

2, –2am1)

�� m2m3 = 2

�� Product of ordinates = 4a2m2m3

= 8[as a = 1]

Q. No. 14 The normal chord at a point ‘t’ on the parabola y2

= 4ax subtends a right angle at the

vertex, then t2 is equal to :

Option 1 3

Option 2 1

Option 3 4

Option 4 2

Correct Answer 4

Explanation

A �� (at2, 2at)

2 2B = + ,2 +a t a t

t t

� ��

2

1 11 1

AOB = 90�� 0

22 +

2= 1

2+

a ttat

ata t

t

� ��

� ��

1

1

2 2

1

1

2 + = 4t t

t

� ��

1 1

1

= 2t�� 2

1

Q. No. 15 The locus of the foot of the perpendiculars drawn from the vertex on a variable

tangent to the parabola y2 = 4ax is :

Option 1 x (x2

+ y2) + ay

2= 0

Option 2 y (x2

+ y2) + ax

2= 0

Option 3 x (x2

y2) + ay

2= 0

Option 4 None of these

Correct Answer 1

Explanation Let ty = x + at2

be the tangent

1+ =

1 + 1 + 1 +

ty atx

t t t

��2

2 2 2 [x cos �� + y sin �� = p]

Let (h, k) be foot of perpendicular

; =

1 + 1 +

at ath k

t t��

2 3

2 2

�� x(x2 + y

2) + ay

2 = 0 required locus.

Q. No. 16 Length of the focal chord of the parabola y2

= 4ax at a distance p from the vertex is

Option 1 2a

p

2

Option 2 a

p

3

2

Option 3 4a

p

3

2

Option 4 p

a

3

Correct Answer 3

Explanation

AB : 2x + (t1 + t2)y = 2 at1t2

+2 2+ =

2 + + 2 + + 2 + +

t tx a

t t t t t t

�� 1 2

2 2 2 2 2 21 2 1 2 1 2

2=

2 + +

aP

t t

��2 2

1 2

… (i)

AB = 2a t t a t t� � �� 2 22 2 2

1 2 1 2

= 4 2 + 2a t t a t t a t t� � � �� 22 2 2 2 2 2 2 21 2 1 2 1 2

4=

a

P

3

2[offer substituting values]

Q. No. 17 If P, Q, R are three conormal points on the parabola y2

= 4ax then the centroid of the

triangle PQR always lies on :

Option 1 The xaxis

Option 2 The yaxis

Option 3 The line y = x

Option 4 The direct of the padrblola

Correct Answer 1

Explanation Sum of ordinates of three conormal points = 0

Centroid lies on x axis in case of y2 = 4ax

Q. No. 18 Two tangents to the parabola y2

= 4ax make angle �� 1 and �� 2 with the xaxis. The

locus of their point of intersection if cot

= 2cot

��

��1

2

is :

Option 1 2y2

= 9 ax

Option 2 4y2

= 9 ax

Option 3 y2

= 9 ax


Correct Answer 1

Explanation Let the tangents be

t1y = x + at22

t2y = x + at22

= tan ; = tant t� �� 1 1 2 2

Let (h, k) be point of intersection

= . = tan .tan .h a t t a � �� 1 2 1 2

= tan .tan .k a � �� 1 2

4 = tan .tan

k b

aa� ��

22

1 22

tan .tan=

tan tan4

k

a

k b

aa

� ��

2

22

1 2

21 2

2

… (i)

cot cot + cot= 2 = 9

cot cot cot

� ��

2

1 1 2

2 1 2

… (ii)

= 9 2 = 9 4

ky ax

k ah��

22

2 required locus

Q. No. 19 From an external point P, pair of tangent lines are drawn to the parabola, y2

= 4x. if ��1

and ��2 are the inclination’s of these tangents with the axis of x such that,

+ =4

�� 1 2 then the locus of P is :

Option 1 x y + 1 = 0

Option 2 x + y 1 = 0

Option 3 x y 1 = 0

Option 4 x + y + 1 = 0

Correct Answer 3

Explanation Let two tangents be

t1y = x + at12

1= tan

t��1

1

t2y = x + at22

1= tan

t��2

2

External point could at1t2, a(t1 + t2)

=tan .tan

ah

� �� 1 2

… (i)

1 1= +

tan tan

k

a � �� 1 2

… (i)

Also cot .cot 1

+ = . 1 =4 cot cot

� ��

� � �� 1 2

1 21 2

… (ii)

= = 1k h a y x� �� required locus.

Q. No. 20 If (t2, 2t) is one end of a focal chord of the parabola, y

2= 4x then the length of the focal

chord will be :

Option 1 1+t

t

� ��

2

Option 2 1 1+ +t t

t t

� ��

2

2

Option 3 1 1 +t t

t t

� ��

2

2


Correct Answer 1

Explanation (t2, 2t) one end of focal chord

1 2,

tt

� �� 2

other end.

Length of focal chord

= 1 1

4 +t ttt

� ��

2 22

2

= 1 1

+ 4t tt t

� ��

2

1 11 2

= 1

+tt

� ��

2

1 1

Q. No. 21 The equation of common tangent to the parabola, y2

2x and x2

= 16y is :

Option 1 Ax + By + C = 0, where A, B, C �� I then A = _______, B = _______, C = _______:

Option 2 1,2,2

Option 3 2,2,1

Option 4 2,1,2

Correct Answer 1

Explanation S1

: y2

= 2x

S2 : x

2 = 16y.

Let y = mx + c be common tangent for S1,c =

1

2m.

1= +

2y mx

m�� … (i) is a tangent to

x2 = 16y.

Any tangent to x2 16y is

4=

' '2

xy

m m … (ii)

Comparing (i) and (ii)

1

2= = 11 4

' '

m m

m m

��2

1=

2m��

Required tangent eg : x + 2y + 2 = 0

A = 1 ; B = 2, C = 2

Q. No. 22 A tangent to the parabola x2

+ 4ay = 0 cuts the parabola x2

= 4by at A and B the locus of

the mid point of AB is :

Option 1 (a + 2b) x2

= 4 b2y

Option 2 (b + 2a) x2

= 4 b2y

Option 3 (a + 2b) y2

= 4 b2x

Option 4 (b + 2x) x2

= 4 a2y

Correct Answer 1

Explanation S1

: x2

= 4ay

S2 : x

2 = 4by

Let the midpoint be (h,k). equation to chord with midpoint to S2

hx 2b(y + k) = h2 4bk

2 + 42= +

bk h bkbx y

h h��

2

is tangent to S1

2 +2= , =

bk hb am

h m h

��

2

2 2 . =

bh bk a

h

� ��

2

2

�� x2(2b + a) = 4b

2y required locus

Q. No. 23 The equation of the other normal to the parabola y2

= 4ax which passes through the

intersection of those at (4a, 4a) and (9a, 6a) is :

Option 1 5x y + 115a = 0

Option 2 5x + y 135a = 0

Option 3 5x y 115a = 0

Option 4 5x + y + 115 = 0

Correct Answer 2

Explanation S ; y2

= 4ax.

Two conormal points are (4a, 4a) and (9a, 6a) ordinate of third point = 10 a

Let the third point be (am2, 2am)

Then, m = 5

Equation to normal :

y = sx + 10a + 124a

�� 5x + y 135a = 0.

Q. No. 24The distance of a point on the ellipse + = 1

6 2

x y2 2

from the centre is 2. The eccentric

angle of the point is

Option 1

2

��

Option 2 � ��

Option 3 3,

2 4

� ��

Option 4

4

��

Correct Answer 4

Explanation: + =1

6 2

x yS

2 2

Let the point be 6 cos , 2 sin� �� .

6 cos + 2 sin = 2� �� 2 2

2

34cos = 2 = ,

4 4

� �� 2

Q. No. 25 The equation of tangents to the ellipse 9x2

+ 16y2

= 144 which pass through the point

(2,3)

Option 1 y = 3

Option 2 x + y = 2

Option 3 x y = 3

Option 4 y = 3; x + y = 5

Correct Answer 4

ExplanationS : + =1

16 9

x y2 2

Let the tangent be

= + 16 + 9y mx m2

passes throught

2,3 3 = 2 + 16 + 9m m�� 2

= 0,1m��Required tangent equation

y = 0, y + x = S

Q. No. 26If the points of intersection of the ellipse + = 1,

x y

a b

2 2

2 2and + = 1,

x y

p q

2 2

2 2form

quadrilateral ABCD such that AC and BD be the conjugate diameter of first ellipse, then

Option 1+ = 2

x y

p q

2 2

2 2

Option 2+ = 1

a b

p q

2 2

2 2

Option 3+ = 1

a b

p q

Option 4+ = 2

a b

p q

2 2

2 2

Correct Answer 4

Explanation: + = 1x y

Sa b

2 21

2 2

: + = 1x y

Sp q

2 22

2 2

1 1 1 1 : = 0S S x y

a p q b

� � � ��

1 2 2 2

2 2 2 2 is a pair of straight lines through origin.

They are conjugate diameters of S1 so

1 1

=1 1

ba p

a

q b

�� 22 2

2

2 2

Product of stapes.

1 = +1a b

p q

� ��

2 2

2 2

+ = 2a b

p q��

2 2

2 2

Q. No. 27If tan ��1

tan ��2 = a

b

2

2 , then the chord joining two point ��1

and ��2on the ellipse

+ = 1,x y

a b

2 2

2 2 will subtend a right angle at

Option 1 Focus

Option 2 Centre

Option 3 End of the major axes

Option 4 End of minor axes

Correct Answer 2

Explanationtan .tan =

a

b� ��

2

1 2 2

cos cos; = 1

sin sin

a a

b b

� ��

� �� 1 2

1 2

Clearly, they subtend right angle at centre.

Q. No. 28The line x = at

2 meets the ellipse + = 1,

x y

a b

2 2

2 2 in the real points if

Option 1 2t ��

Option 2 1t ��

Option 3 1t


Correct Answer 2

Explanation Solving = b2t

4+ y

2 b

2= 0

For really : 0b b t� �� 2 2 4

1 1t t� � � �� 4 2

Q. No. 29The eccentric angles of the extremities of latus rectum of the ellipse + = 1,

x y

a b

2 2

2 2 is

Option 1tan

ae

b

��

1

Option 2tan

be

b

��

1

Option 3tan

b

ae

��

1

Option 4tan

a

be

��

1

Correct Answer 3

ExplanationEnds of latus rectum are , 1 ae e b

� ��

2

= cos ,sin = 1 e e� � � �� 2

1 tan = =

e b

e ae� � � ��

2

Assumed that a b � ��

Q. No. 30 Product of the perpendiculars from the foci upon any tangent to the ellipse

+ = 1,x y

a b

2 2

2 2 is

Option 1 b

Option 2 a

Option 3 a2

Option 4 b2

Correct Answer 4

ExplanationTangent at ' '�� point to + = 1

x y

a b

2 2

2 2is cos + sin = 1

x y

a b� �� . Let the foci be ,0ae�� .

Product of foci = cos 1 cos +1

=

cos sin+

e eb

a b

��

� ��

2

2 2

2 2

Q. No. 31 The equation of the common tangents to the parabola y2

= 8x and the hyperbola 3x2

y2 = 3 is

Option 1 2x �� y + 1 = 0

Option 2 x �� y + 1 = 0

Option 3 x �� 2y + 1 = 0

Option 4 x �� y + 2 = 0

Correct Answer 1

Explanation S1

: y2

= 8x

S2 : 3x

2 – y

2 = 3

Let the tangent be

2= +y mx

m which is always tangent to S

1.

2= +y mx

m is a tangent to S

2.

23 + = 3x mx

m

� ��

2

2

43 4 + 3 = 0m x x

m

� ��

2 2

2

4= 0 16 4 + 3 3 = 0 = 2m m

m

� ��

2

2

equation to tangents :

2 1 = 0y x� �� = 2 +1y x��

Q. No. 32 The area of a triangle formed by the lines x y = 0, x + y = 0 and any tangent to the

hyperbola x2 y

2 = a

2 is

Option 1 a2

Option 2 2a2

Option 3 3a2

Option 4 4a2

Correct Answer 1

Explanation S : x2

y2

= a2

Equation of tangent at (x1, y1)

xx1 yy1 = a2

Point of intersection with x = y line ,

a a

x y x y

� ��

2 2

1 1 1 1

Point of intersection with y = x line

,+ +

a a

x y x y

� ��

2 2

1 1 1 1

1

1 = 1 = 2

2 + +

0 0 1

a a

x y x y

a aa

x y x y��

2 2

1 1 1 1

2 22

1 1 1 1

Q. No. 33 The locus of the mid point of the chords of the circle x2

+ y2

= a2

, which are tangent to

the hyperbola = 1x y

a b

2 2

2 2 is

Option 1 x2

+ y2

= a2

b2

Option 2 (x2

+ y2)

2= a

2 b

2

Option 3 (x2

+ y2)

2= a

2x

2 b

2y

2

Option 4 (x2

+ y2)

2= a

2+ b

2

Correct Answer 3

Explanation S1

: x2

+ y2

= 16

S2 : 9x

2 16y

2 = 144

Let the midpoint be (h, k) of chords of S1 equation to chords

hx + ky = h2 + k

2 … is

If hx + ky = h2 + k

2 is tangent at (x1, y1) cos

2

+= =

9 16 144

h k h k

x y

2 2

1 1

�� 16x2

9y2 = (x

2 + y

2)

2 . required locus.

Q. No. 34 The circles on focal radii of parabola as diameter touch

Option 1 The tangent at the vertex

Option 2 The axis

Option 3 The directrix


Correct Answer 1

Explanation A ,2 ; O 0,0 ;S ,0at at a� � �� 2

As is focal radii

equation to circle as AS is diameter :

(x a)(x at2) + y(y.t.2a) = 0

The circle touches x = 0 line, i.e. tangent at vertex

Q. No. 35 The ends of a line segment are P (1, 3) and Q (1, 1). R is a point on the line segment PQ

such that PR : QR = 1 : !!!! if R is in interior point of a parabola y2 = 4x, then

Option 1 (0, 1)! �! �! �! �

Option 2 3 ,1

5

� �� !�!�!�!��

Option 3 1 3

,2 5

� �� !�!�!�!��


Correct Answer 2

Explanation P 1,3 ; Q 1,1��

3 +1R 1,

+1

!!!!� �� !!!! lies in side y

2 4x = 0

3 + 1 3 4 0 ,1

+ 1 5

!!!!� � � �� !�� !�� !�� !�� !!!!

2

Q. No. 36 If y1, y2 are the ordinates of two points P and Q on the parabola and y3 is the ordinate

of the point of intersection of tangents at P and Q, then

Option 1 y1, y2, y3 are in A.P

Option 2 y1, y3, y2 are in A.P

Option 3 y1, y2, y3 are in G.P

Option 4 y1, y3, y2 are in G.P

Correct Answer 2

Explanation Let y1 = 2at1 y2 = 2at2 . then

y3 = a(t1 +t2)

�� 2y3 = 2at1 + 2at2 = y1 + y2

�� y1 , y3, y2 + AP

Q. No. 37 The two parabola y2

= 4ax and y2

= 4c (x b) cannot have a common normal, other

than the axis unless, if

Option 1 2

a b

a

Option 22

b

a c

Option 32

+

b

a b


Correct Answer 2

Explanation Any normal to y2

= 4c(x b) can be taken as

y = m(x b) 2cm cm3 … (i) is also normal to

y2 = 4ax

y = mx 2am am3

… (ii)

(i) and (ii) are equivalent

2am + am3 = bm + 2am + cm

30m """"

= 2 2

b bm

a c a c� � � � � � � � 2

Q. No. 38 If the normal drawn from any point to the parabola y2

= 4ax cuts the line x = 2a in

points whose ordinates are in arithmetic progression, then tangents of the angles

which the normal makes with the axis, are

Option 1 In A. P.☒

Option 2 In G.P.☒

Option 3 In H.P.☒

Option 4 None of these ☐

Explanation Let the normal be y = mx 2am am3

; let it pass through (x1, y1).

y1 = mx1 2am am3 i) intersects with x = 2a

�� y = am3. Let the ordinates of point of intersection be , , AP# # ## # ## # ## # #1 2 3

, , APam am am�� 3 3 3

1 2 3

Also, m1 + m2 + m3 = 0

2 = +m m m�� 3 3 3

2 1 3

2 = + ,m m m m m�� 2 2 2

2 1 3 1 2

=m m m�� 2

2 1 3

, , G.Pm m m� $$%� $$%� $$%� $$%1 2 3

Q. No. 39 If the normals from any point to the parabola x2

= 4y cuts the line y = 2 in points whose

abscissae are in A.P., then the slopes of the tangents at the three conormal points are

in

Option 1 A.P.☒

Option 2 G.P.☒

Option 3 H.P.☒

Option 4 none of these☐





, , APam am am�� 3 3 3

1 2 3

Also, m1 + m2 + m3 = 0

2 = +m m m�� 3 3 3

2 1 3

2 = + ,m m m m m�� 2 2 2

2 1 3 1 2

=m m m�� 2

2 1 3

, , G.Pm m m� $$%� $$%� $$%� $$%1 2 3

Q. No. 40 The points on the ellipse such that the tangent at each of them makes equal angles

with the axes is

Option 1

,

+ +

a b

a b a b

� ��

2 2

2 2 2 2

Option 2

,

+ +

b a

a b a b

� ��

2 2

2 2 2 2

Option 31 1

,

+ +a b a b

� ��

2 2 2 2


Correct Answer 1

Explanation x y

a bS : + =1

2 2

2 2

Let the point be a bcos , sin� ��

x y

a b

cos sin+ = 1

� �� tangent

Slope b

a

– cos= = 1

sin

��

��

bb a

a

cos=1 cos = 1 – cos

sin

2 22 2 2 2

2 2

��

��a

a

a b

cos =

2

2 2� � ��

��

Points are a b

a b a b

,+ +

2 2

2 2 2 2

� ��

Q. No. 41 Tangent are drawn from the points on the line x y 5 = 0 to x2

+ 4y2

= 4, then all the

chords of contact pass through a fixed point, whose coordinate are

Option 1 4 1,

5 5

� ��

Option 2 4 1,

5 5

� ��

Option 3 4 1 ,

5 5

� ��


Correct Answer 1

Explanation Let the point be (h, k) on x – y – 5 = 0

�� h – k – 5 = 0

chords of contact

hx + 4k – 4 = 0 will pass through 4 1

,5 5

� ��

Q. No. 42 Tangents are drawn to y2

= 4 a x from a variable point P moving on x + a = 0, then the

locus of foot of perpendicular drawn from P on the chord of contract of P is

Option 1 y = 0

Option 2 (x a)2

+ y2

= a2

Option 3 (x a)2

+ y2

= 0

Option 4 y(x a) = 0

Correct Answer 3

Explanation Let (h, k) be the point.

Chord of contact

hx ky

a b+ = 1

2 2

b h by x

ka k

= +

2 2

2�� is a tangent with radius ‘r’ (ray). x

2 + y

2 = r

2

b b h x yr

k a k a b r

1= 1 + =

2 22 2 2 2

2

2 4 4 2

� ��

None of these

Q. No. 43 The eccentricity of the conic 4(2y x 3)2

9 (2x + y 1)2

= 80 is

Option 1 3

13

Option 2 13

3

Option 3 13

Option 4 3

Correct Answer 2

Explanation 4(2y x 3)2

9 (2x + y 1)2

= 80

y x x y2 3 2 + 14 5 9 5 = 80

5 5

2 2� � � ��

� � ��

y x x y2 3 2 + 1

5 5 = 1

164

9

2 2� � � ��

e

16

139= 1+ =4 3

��

Q. No. 44 The points (s) on the parabola y2

= 4x which are closest to the circle : x2

+ y2

24y + 128

= 0 is/are

Option 1 (0, 0)

Option 2 2, 2 2

Option 3 (4, 4)

Option 4 None of thses

Correct Answer 3

Explanation Let (t2, 2t) be the point on parabola.

Let f(t) be the square of the length of tangent.

f(t) = t4 + 4t

2 – 48t + 128

Longer the tangent, longer the distance

For f(t)min, f ’(t) = 4t3 + 8t – 48 = 0

�� t = 2

�� Required point �� (4, 4)

Q. No. 45 If M is the foot of the perpendicular from a point P of a parabola y2

= 4ax to its

directrix and SPM is an equilateral triangle, where S is the focus, then SP is equal to :

Option 1 a

Option 2 2a

Option 3 3a

Option 4 4a

Correct Answer 4

Explanation Let at atP ,22��

a atM ,2��

aS ,0��

PM = SP = SM [equilateral �� ]

a at a at a at at+ = 2 + 2 = + 222 2 22 2��

t a= 3 SP = 4� ��

Q. No. 46 From the focus of the parabola, y2

= 8x as centre, a circle is described so that a

common chord of the curves is equidistant from the vertex & focus of the parabola.

The equation of the circle is :

Option 1 (x 2)2

+ y2

= 9

Option 2 (x 2)2

+ y2

= 6

Option 3 (x 2)2

+ y2

= 4

Option 4 (x 2)2

+ y2

= 18

Correct Answer 1

Explanation S : y2

– 8x = 0

F 2,0 V 0,0� ��

Equation to desired chord : x = 1 end points of chord

Chord 1, 2 2 end points of diameter� � ��

�� equation to circle : (x 2)2 + y

2 = 9

Q. No. 47 IF the chord of contact of tangents from a point P to the parabola y2

= 4ax touches the

parabola x2 = 4by, the locus of P is :

Option 1 Circle

Option 2 Parabola

Option 3 Ellipse

Option 4 Hyperbola

Correct Answer 1

Explanation Let h kP ,�� .

Chord of contact to y2 – 4ax = 0

ky – 2ax – 2ah = 0 is a tangent to x2 = 4by of form

bx my

m= +

k bm h

a m= , =

2��

bhk

a+ = 0

2��

bxy

a+ = 0

2�� hyperbola

Q. No. 48 AB, AC are tangents to a parabola y2

= 4ax. p1, p2 and p3 are the lengths of the

perpendiculars from A, B and C respectively on any tangent to the curve, then p2, p1, p3

are in :

Option 1 A.P.

Option 2 G.P.

Option 3 H.P.


Correct Answer 4

Explanation Let at at C at A at t a t tB ,2 ; , ; , +21 1 2 1 2 1 2� � ��

Let a

y mxm

= + be any tangent

aat mat

mtam

mm m

2 1

P = =1 + 1 +

221 1

1

22 2

aat mat

mtam

mm m

2 1

P = =1 + 1 +

222 2

2

32 2

aamt t a t t

mt mtam

mm m

+ +1 1

P = =1 + +1

1 2 1 21 2

12 2

Clearly, P2, P1, P3 $$%$$%$$%$$% G.P

Q. No. 49 If two normal to a parabola y2

= 4ax intersect at right angles then the chord joining

their feet passes through a fixed point whose coordinates are :

Option 1 (2a, 0)

Option 2 (a, 0)

Option 3 (2a, 0)


Correct Answer 2

Explanation Let at at at at,2 and ,22 31 1 2 2 be two feet of normal.

Since, normal are at right angles

�� t1t2 = 1

�� Chord joining feet is focal chord

Q. No. 50 The triangle PQR of area ‘A’ is inscribed in the parabola y2

= 4ax such that the vertex P

lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the

difference of the ordinates of the points Q and R is :

Option 1 A

2a

Option 2 A

a

Option 3 2A

a

Option 4 4A

a

Correct Answer 3

Explanation a aat at

tt

2P 0,0 Q ,2 R ,

21 1 2

11

� ��

at at

a a

tt

0 0 11

A = mol 2 12

21

21 1

211

a aat

a t

2 22 1

1

� ��

Q. No. 51 Three normal, none of which is the axis of parabola, drawn from any point to the

parabola y2 = 4ax, are cut to the line x = 2a in points whose ordinates are in

arithmetical progression. Then the tangents of the angles which the normal make the

axis of the parabola are in

Option 1 A.P.

Option 2 G.P.

Option 3 H.P.


Correct Answer 2





, , APam am am�� 3 3 31 2 3

Also, m1 + m2 + m3 = 0

2 = +m m m�� 3 3 32 1 3

2 = + ,m m m m m�� 2 2 22 1 3 1 2

=m m m�� 22 1 3

, , G.Pm m m� $$%� $$%� $$%� $$%1 2 3

Q. No. 52 T is a point on the tangent to a parabola y2

= 4ax at its point P. TL and TN are the

perpendiculars on the focal radius SP and the directrix of the parabola respectively.

Then :

Option 1 SL = 2 (TN)

Option 2 3 (SL) = 2 (TN)

Option 3 SL = TN

Option 4 2 (SL) = 3 (TN)

Correct Answer 3

Explanation Let at atP ,22��

Equation to tangent PQ :

ty = x + at2.

Let at t a t tT , , +1��

Equation to SP :

(t2 1)y – 2tx + 2at = 0

t a t t t at t at

t t

1 . + 2 . 2TL=

1 2 + 4

21 1

2 2

��

a t t= 1

a at t a t t a t tSL = + + 22 22

1 1 1

a t tTN = 1 + = SL1

Q. No. 53 A circle is described whose centre is the vertex and whose parabola y2

= 4ax If PQ is

the common chord of the circle and the circle and the parabola and L1 L2 is the latus

rectum, then the area of the trapezium PL1 L2 Q is

Option 1 3 2 a2

Option 2 2 2 a2

Option 3 4 a2

Option 4 2 + 2

2a

� ��

2

Correct Answer 4

Explanation

The equation to circle

ax y

3+ =

2

2

2 2 � ��

Point of intersection is

y ay

a

3+ =

4 2

2 222

� �� y

4 + 16a

2y

2 36a

4 = 0

�� (y2 – 2a

2) (y

2 + 18a

2) = 0

y a= 2� �� a

x =2

aa a

1Area = 4 2 2

2 2� � � ��

a 2 + 2=

2

2

Q. No. 54 From the point (15, 12) three normal are drawn to the parabola y2

= 4x, then centroid

of triangle formed by three conormal points is

Option 1 16,0

3

� ��

Option 2 (4, 0)

Option 3 26,0

3

� ��

Option 4 (6, 0)

Correct Answer 3

Explanation Let the normal be

y = mx – 2m – m3 passes through (15, 12)

�� 12 = 15m – 2m – m3

�� m3

+ 12 – 13m = 0

�� m = 1, 3, 4

m m m+ +centroid ,0

3

2 2 21 2 3

� ��

26,0

3

� ��

Q. No. 55 A ray of light travels along a line y = 4 and strikes the surface of a curve y2

= 4(x + y)

then equation of the line along reflected ray travel is

Option 1 x = 0

Option 2 x = 2

Option 3 x + y = 4

Option 4 2x + y = 4

Correct Answer 1

Explanation The reflected ray would pass through the focus of parabola

y2 – 4y + 4 = 4(x + 1)

�� (y 2)2 = 4(x + 1)

Point of incident �� (0, 4)

Focus �� (0, 2)

Reflected ray : x = 0 line

Q. No. 56 Through the vertex O of the parabola y2

= 4ax two chords OP and OQ are drawn and

the circles on OP and OQ as diameter intersect in R. if ��1 , ��2 and &&&& are the angles

made with the axis by the tangents at P and Q on the parabola and by OR then cot

+� �� 1 2 is equal to

Option 1 2 tan &&&&

Option 2 2 tan ( )� &� &� &� &

Option 3 0

Option 4 2 cot &&&&

Correct Answer 1

Explanation Let at atP ,221 1��

at atQ ,222 2��

OP as diameter

x at x y at y + 2 = 02 2 2

1 1

OQ as diameter

x at x y at y + 2 = 02 2 2

2 2

t t x yOR + +2 = 01 2��

t t cot + cot

cot = ;cot = tan =2

1 21 1 2 2

� ��

cot +cot =2tan1 2� � � ��

Q. No. 57 Normals at three points P, Q , R at the parabola y2

= 4ax meet in a point A and S be its

focus, SP . SQ . SR = (SA) , !!!! 2if then= !!!! is equal to

Option 1 a3

Option 2 a2

Option 3 a

Option 4 1

Correct Answer 3

Explanation Let the point A be (h, k), Normal y = mx 2am am3

pass through (h, k)

�� k = ml1 2am am3

am am Q am am R am amP ,2 ; ,2 ; ,22 2 21 1 2 2 3 3� � ��

a am am a m a mSP = + 2 = 1+ = 1+2 222 2 2

1 1 1 1

a m m mSP.SQ SR = 1+ + 1+ 1+2 2 2

3 2 2 21 2 3��

a m m m m m m= 1 + + +3 2 2 2 2 2 2

1 1 2 1 2 3' '' '' '' '

Which on simplification reduces to a(b a)2 + k

2

i.e SP. SQ. SR = a.SA2

Q. No. 58 A hyperbola has centre ‘C’ and one focus at P (6, 8) if its two directories are

3x + 4y + 10 = and 3x + 4y 10 = 0 then CP =

Option 1 14

Option 2 8

Option 3 10

Option 4 6

Correct Answer 3

Explanation x yS: + =1

25 16

2 2

Let P 5cos +4sin� � ��

Tangent : x ycos sin

+ = 15 4

� �� meets line x = 0

at ecQ 0,4cos� ��

So, ecR 4cos ,0� ��

Equation to circle :

x x ec y y ec 4cos + 4cos = 0� �� which passes through (0,0) inrespective of ' '�� .

Q. No. 59Any ordinate MP of on ellipse + = 1

25 9

2x y

meets the auxiliary circle in Q, then locus of

point of intersection of normal at P and Q to the respective curves, is

Option 1 x2

+ y2

= 8

Option 2 x2

+ y2

= 34

Option 3 x2

+ y2

= 64

Option 4 x2

+ y2

= 15

Correct Answer 3

Explanation Let P 5cos ,3sin� � ��

Q 5cos ,5sin� � ��

Normal at ax by

a b x y ecP 5 sec 3 cos = 16cos sin

2 2� � � � ��

Normal at Q : y x= tan .��Required locus

y xx y

x y5 1+ 3 1+ =16

2 2

2 2

x y+ = 642 2��

Q. No. 60 If P is a moving point in the xy plane in such a way that perimeter of triangle PQR is 1

whereQ 3, 5 , R 7,3 5� �� then maximum area of triangle PQR is

Option 1 6 sq. unit

Option 2 12 sq. unit

Option 3 18 sq. unit

Option 4 9 sq. unit

Correct Answer 2

Explanation In PQR��

PQ + QR + RP = 16 ; QR = 6

�� PQ + RP = 10

�� P lies on ellipse whose foci are Q, R with major axis equal to so

Ar PQR�� is maximum if P lies on minor axis

PQ = PR = 5

CP = 5 3 = 42 2��

PQR1

Ar = 4 6 = 122

� � � ��

Q. No. 61If + =3� # �� # �� # �� # � then the chord joining the points �� and #### for the hyperbola = 1

2 2

2 2

x y

a b

passes through

Option 1 Focus

Option 2 Centre

Option 3 One of the end points of the transverse axis

Option 4 On of the end points of the conjugates axis

Correct Answer 2

Explanation Equation to chord of hyperbola joining ,� #� #� #� # points :

x y

a b

+ +cos sin = cos

2 2 2

# � # � ## � # � ## � # � ## � # � #� � � � � ��

+ = 3� # �� # �� # �� # �

x y

a b

+cos + = 0

2

� #� #� #� #� ��

passes through origin irrespective of +� #� #� #� # .

Origin is centre

Q. No. 62 The combined equation of the asymptotes of the hyperbola

2x2 + 5xy + 2y

2 + 4x + 5y = 0 is

Option 1 2x2

+ 5xy + 2y2

+ 4x + 5y + 2 = 0

Option 2 2x2

+ 5xy + 2y2

+ 4x + 5y 2 = 0

Option 3 2x2

+ 5xy + 2y2

= 0


Correct Answer 1

Explanation S : 2x2

+ 5xy + 2y2

+ 4x + 5y = 0

Let the asymptotes be y = mx + c.

2x2 + 5x(mx + c) + 2(mx + c)

2 + 4x + 5(mx + c) = 0

= x2(2 + 5m + 2m

2) + x(5c + 4mc + 4 + 5m) + 2c

2 + 5c = 0

For y = mx + c to be asymptotes

2 + 5m + 2m2 = 0 and 5c + 4mc + 4 + 5m = 0

m1

= ,22

��

m c m c1 1

= = , = 2 ; = 22 2��

The asymptotes are

2y + x + 1 = 0 and y + 2x + 2 = 0

Equation to any hyperbola whose asymptotes are above lines :

y x y x2 + +1 +2 + 2 = , 1R 02! !�! !�! !�! !�

Q. No. 63 Consider a circle with its centre lying on the focus of the parabola, y2

= 2 px such that it

touches the directrix of the parabola. Then a point of intersection of the circle and the

parabola is :

Option 1,

2

pp

� ��

☒

Option 2,

2

pp

� ��

☒

Option 3 ,

2

pp

� ��

☐

Option 4 ,

2

pp

� ��

☐

Explanation S : y2

= 2px.

If the circle is centred at focus of parabola and touches the directrix then focal chord

will be one of its diameters

Q. No. 64 Let V be the vertex and L be the latus rectum of the parabola x2

= 2y + 4x 4. Then the

equation of the parabola whose vertex is at V, latus rectum is L

2 and axis is

perpendicular to the axis is perpendicular to the axis of the given parabola.

Option 1 y2

= x – 2☒

Option 2 y2

= x – 4☒

Option 3 y2

= 2 – x☐

Option 4 y2

= 4 – x☐

Explanation x2

= 2y + 4x – 4

�� (x 2)2 = 2y

v = (2, 0) L = 2

For the required parabola

v = (2, 0), L = 1 axis 11 to old one is

y x= 22 �� required parabola

Q. No. 65 If equation of tangent at P, Q and vertex A of a parabola are 3x + 4y 7 = 0,

2x + 3y 10 = 0 and x y = 0 respectively, then

Option 1 focus is (4, 5)☒

Option 2 length of latus rectum is 2 2 ☒

Option 3 axis is x + y – 9☒

Option 4vertex is

9 9,

2 2

� ��

☒

Explanation Let the parabola equation be

x y c x yk

+ + =

2 2

2� � � �� 3x + 4y – 7 = 0 is a tangent

y yy c y

k

7 4 7 4+ +

3 3=2 2

2� � � ��

is a quadratic

k c c k= 0 21 2 14 6 = 4 7 +3 21 22 2� ��

� �� … (i)

Similarly, for tangent 2x + 3y – 10 = 0.

k c c k= 0 10 2 4 20 = 4 10 + 2 20 22 2� ��

� �� … (ii)

From (i) and (ii)

k = 2 2 , c = 9

Hence, all options are correct.

Q. No. 66 If A and B are points on the parabola y2

= 4ax with vertex O such that OA perpendicular

to OB and having lengths r1 and r2 respectively, then the value of

+

r r

r r

4 4

3 31 2

2 2

3 31 2

is

Option 1 16a2

Option 2 a2

Option 3 4a


Correct Answer 1

Explanation t tOA OB = 41 2( �( �( �( � [t1, t2 are A and B points]

r a t a t r a t a t= +4 = + + 42 2 2 2 2 2 2 2

1 1 1 2 2 2

Let r r

yr r

.=

+

4/3 4/31 22/3 2/3

1 2

r ry

r r r r r r

.=

+ +3 +

4 43 1 2

2 2 2/3 2/3 2/3 2/31 2 1 1 2 2

��

After simplifying :

t y a t a y y a y a 8 + 3 4 2.4 +16 4 = 04 3 4 6 2 2 2 2 3 6 6 3 6 61 1��

y a=16 2��

Q. No. 67 Let P, Q and R are three conormal points on the parabola y2

= 4ax. Then the correct

statement (s) is/are

Option 1 Algebraic sum of the slopes of the normal at P, Q and R vanishes.☒

Option 2 Algebraic sum of the ordinates of the points P, Q and R vanishes.☒

Option 3 Centroid circumscribing the triangle PQRE lies on the axis of the parabola.☒

Option 4 Circle circumscribing the triangle PQR passes through the vertex of the

parabola.

☒

Explanation Let the normal be

y = mx 2am am3 pass through (h, k)

�� k = hm 2am am3

Clearly, m1 + m2 + m3 = 0 sum of scopes

a hm m

a

2 =1 2

km m m

a

=1 2 3

The conormal points are am am i,2 , = 1,2,321 i

sum of ordinates = 2a(m1 + m2 + m3) = 0

a m m m+ +Centroid ,0

3

2 2 21 2 3

� ��

Equation to area of circle:

a m a m yam am

am am x y x a m a m y

am am a m a m y

x a m a m x y a m a m

y x a m a m x y a m a m

x a m a m x y a m a m

+ 4 12 1

2 1 + + + 4 1

2 1 + 4 1

+ 4 1 + 4

+ + 4 1 + + 4 = 0

+ 4 1 + 4

2 4 2 22

1 1 11 1

2 2 2 2 4 2 22 2 2 2 2

22 4 2 2

3 33 3 3

2 4 2 2 2 4 2 21 1 1 1 1 1 1 1

2 4 2 2 2 4 2 22 2 2 2 2 2 2

2 4 2 2 2 4 2 23 3 3 3 3 3 3

To pass through vertex :

x y a m a m

x y a m a m

x y a m a m

+ 4

+ 4 = 0

+ 4

2 4 2 21 1 1 1 1

2 4 2 22 2 2 2

2 4 2 23 3 3 3

There, x am= 2i i , yi = 2ami

am am a m am am a m

am am a m am am a m

am am a m am am a m

2 4 2

2 4 + 2 = 0

2 4 2

2 2 2 42 21 1 1 1 2 1

2 2 2 42 22 2 2 2 2 2

2 2 2 42 23 3 3 3 3 3

m m

m m m m m

m m

1

1 = 0

1

31 1

31 2 3 2 2

33 3

��

�� m1m2m3(m1m2)(m2m3)(m3m4) (m1 + m2 + m3) = 0

Which is always true as m1 + m2 + m3 = 0

Hence , a, b, c, d

Q. No. 68 The locus of the mid point of the focal radii of a variable point moving on the

parabola, y2 = 4ax is a parabola whose

Option 1 Latus rectum is half the latus rectum of the original parabola☒

Option 2Vertex is , 0

2

a� ��

☒

Option 3 Directrix is yaxis☒

Option 4 Focus has the coordinates (a, 0)☒

Explanation S : y2

= 4ax

at atP ,22��

Midpoint �� (h, k)

at ah k at

+= ; =

2

2

a kh

a= +1

2

2

2

� ��

� ��

ha a ak k a h

a

2= = 2

2

2 32 2��

is parabola

Latus rectum = 2a; a

vertex ,02

� ��

Directrix : $$%$$%$$%$$% x = 0 ; focus �� (a,0)

Q. No. 69If P is a point of the ellipse + = 1

x y

a b

2 2

2 2, whose foci are S and S’. Let PSS' =� �� and

PS'S =� #� #� #� # , then

Option 1 PS + PS' = 2 ,ifa a b ☒

Option 2 PS + PS' = 2 ,ifb a b��☒

Option 3 1tan tan =

2 2 1+

e

e

� #� #� #� # ☒

Option 4

tan = whena >2 2

a ba a b b

b

� #� #� #� # � ��

2 22 2

2

☐

Explanation x yS

a b: + =1

2 2

2 2

In PSS'��

PS + SS'PS'cos =

2PSSS'

2 2

��

+ PS +PS' SS' PS'PS + SS'tan =

2 PS +PS'+ SS' PS'PS + SS'

2 ��

Similarly,

PS +PS' SS' PS PS'+ SS'tan =

2 PS +PS'+ SS' PS'PS + SS'

2 ####��

e

e

PS +PS' SS' 1 tan tan = =

2 2 1+PS+PS'+ SS'

2 22 2

2

� #� #� #� # � ��

AS : PS + PS’ = major axis

SS’ = e major axis

Q. No. 70 If the chord through the points whose eccentric angles are �� and &&&&on the ellipse ,

+ = 1

2 2

2 2

x y

a bpasses through a focus, then the value of tan tan

2 2

� &� &� &� & is :

Option 1 + 1

1

e

e

☒

Option 2 1

+ 1

e

e

☒

Option 3 1 +

1

e

e

☐

Option 4 1

1 +

e

e

☐

Explanation Closed joining points with eccentric angles �� and &&&& is

x y

a b

+ + +cos + sin = cos

2 2 2

� & � & � &� & � & � &� & � & � &� & � & � &� � � � � ��

Pass through a focus (a, 0) or (ae, 0)

e+ +

cos = cos2 2

� & � &� & � &� & � &� & � &� � � ��

ee

e

cos

12= tan .tan

+ 1 2 2cos

2

1� &� &� &� &� ��

� �� & && && && &� �� &� &� &� & ��

e

e

1 tan .tan =

2 2 1+

� ��

Q. No. 71 The equation, 3x2

+ 4y2

18x + 16y + 43 = c.

Option 1 cannot represent a real pair of straight line for any value of c☒

Option 2 represents an ellipse, if 0 c☒

Option 3 represent empty set, if 0��c☒

Option 4 a point , if c = 0☒

Explanation S : 3x2

+ 4y2

– 18x + 16y + 43 = c

Pair of straight lines : h ab No volue of 'c'2 � ��

Ellipse : 3(x 3)2 + 4(y + 2)

2 = c. The equation represents ellipse if c > o

Empty set : 3(x 3)2 + 4(y + 2)

2 = c represents empty set if c < 0 as.

LHS 0 andRHS 0 � � � �Point if c = 0 point is (3, 2)

Q. No. 72If foci of + = 1

2 2

2 2

x y

a bcoincide with the foci + = 1

25 9

2 2x y

and eccentricity of the

hyperbola is 2, hen

Option 1 a2

+ b2

= 16☒

Option 2 there is no director circle to the hyperbola☒

Option 3 centre of the director circle is (0, 0)☐

Option 4 length of latus rectum of the hyperbola = 12☒

Explanation x ye

a bS = = 1 = 2

2 21

2 2

x ye

b

4S = + =1 =

25 5

2 22

2

b

a4 =1 +

2

2 …(i)

ae = 4 �� a = 2 [equating coordinates of focus]

�� a2 + b

2 = 16

Is b2 = 12

Director circle of hyperbola :

x2 + y

2 = 4

Latus rectum = b e2 1 = 122

Q. No. 73 If (5, 12 and (24, 7) are the foci of a conic passing through the origin then the

eccentricity of conic

Option 1 386

12

☒

Option 2 386

13

☐

Option 3 386

25

☐

Option 4 386

38

☒

Explanation Let the conic be

ax2 + 2hxy + by

2 + 2gx + 2fy = 0. Which passes through origin. Since there are two

focus, it must be either ellipse or hyperbola.

Ellipse : SS’ = 2ae = 386

OS + OS’ = 2a = 13 + 25 = 38

Hyperbola : OS’ – OS = 12 = 2a,

SS’ = 386 = 2ae

e386

=12

��

Q. No. 74 Assertion : If straight line x = 8 meets the parabola y2 = 8x at P and Q then PQ

subtends a right angle at the origin.

Reason : Double ordinate equal to twice of latus rectum of a parabola subtends a right

angle at the vertex.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion.

Option 3 If ASSERTION is true but REASON is false.

Option 4 If ASSERTION is false but REASON is true.

Correct Answer 1

Explanation S : y2

= 8x �� a = 2

at2 = 8 �� t

2 = 4 $$%$$%$$%$$% t1, t2 roots

t1, t2 = tl

Assertion is farce and reason is true and correct explanation

Q. No. 75 Assertion : Circum circle of a triangle formed by the lines x = 0 , x + y + 1 = 0 and

x y + 1 = 0 also passes through the point (1, 0)

Reason : Circum circle of a triangle formed by three tangents of a parabola passes

through its focus


assertion.


the assertion.



Correct Answer 1

Explanation x = 0;

x + y + 1 = 0

x y + 1 = 0

The above the are tangent to y2 = 4x

Let the three tangents be

ty x at= +21

t y x at= +2

2 2

t y x at= +2

2 3

Three vertices of �� are

(at1t2 , a(t1 + t2)), (at2t3 , a(t1 + t3)) and (at3t1 , a(t3 + t1))

Equation to circum circle :

a t t a t t a t tat t a t t

x y at t a t t x a t t a t t a t t

at t a t ta t t a t t a t t

+ + + 1+ 1

+ + 1 + + + + 1

+ 1 + + + 1

22 2 2 21 2 1 2 1 2

1 2 1 222 2 2 2 2 2

2 3 2 3 2 3 3 2 2 3

3 1 3 1 22 2 2 23 1 3 1 3 1

at t a t t a t t at t a t t a t t t t

y at t a t t a t t at t a t t a t t t t

at t a t t a t t at t a t t a t t t t

+ + 1 + + +

+ + + 1 + + + + = 0

+ + 1 + + +

2 22 2 2 2 2 2 21 2 1 2 1 2 1 2 1 2 1 2 1 2

2 22 2 2 2 2 2 22 3 2 3 3 2 2 3 2 3 2 3 3 2

2 22 2 2 2 2 2 23 1 3 2 3 1 3 1 3 1 3 2 3 1

Check for (2, 0)

t t t t t tt t t t

a t t t t a t t t t t t

t t t tt t t t t t

+ + + 1+ 1

+ 1 + + + + 1

+ 1 + + + 1

22 21 2 1 2 1 2

1 2 1 224 4 2 2

2 3 2 3 2 3 3 2 2 3

3 1 3 1 22 23 1 3 1 3 1

t t t t t t t t

a t t t t t t t t

t t t t t t t t

+ + +

+ + + + = 0

+ + +

22 2

1 2 1 2 1 2 1 2

24 2 2

2 3 2 3 2 3 3 2

22 23 1 3 1 3 1 3 1

Q. No. 76 Assertion : The perpendicular bisector of the line segment joining the point (a, 2 at)

and (a, 0) is tangent to the parabola y2 = 4ax, where R��t

Reason : Number of parabolas with a given point as vertex and length of latus rectum

equal to 4, is 2


assertion.


the assertion.



Correct Answer 3

Explanation Equation to bisector

(y at) = x

t

�� ty = x + at2 which is tangent to y

2 = 4ax

If vertex is given and length of takes rectum is given 4 parabola can be obtained

Q. No. 77 Assertion : Feet of perpendiculars drawn from foci of an ellipse 4x2

+ y2

= 16 on the

line 2 3 + = 8x y lie on director circle x2 + y

2 = 16

Reason : If perpendicular are drawn from foci of an ellipse to its any tangent then feet

of these perpendiculars lie on director circle of the ellipse.


assertion.


the assertion.




assertion.

Correct Answer 3

ExplanationLet the ellipse be

x y

a b+ = 1

2 2

2 2

Director circle : x2 + y

2 = a

2 + b

2 … (i)

Tangent : x y

a b

cos cos+ = 1

� ��

Feet of perpendiculars :

ee aae

a b a b

coscos 1

cos 1 sin ,+

cos sin cos sin+ +

2 2 2 2

� ��

e eb aae

a b a b

sin coscos +1 cos +1

,cos sin cos sin

+ +

2 2 2 2

� ��

Satisfying the above points in LHS,

LHS = x2 + y

2 LHS = a

2

So, they lie on axillary circle

Attenuate way :

The above tangent can be written :

x y Pcos + cos =� �� …(i)

a bP : cos + sin2 2 2 2� ��

Equation to perpendicular from (ae, 0)

x y aecos + cos = sin� � �� …(ii)

The feet lies on (i) and (ii) both

x y a b a e+ = cos + sin + sin2 2 2 2 2 2 2 2 2� � ��

a b a b= cos + sin + sin2 2 2 2 2 2 2� � ��

= a2

Hence, lies on axillary circle

Assertion is true, reason in false

Q. No. 78 Assertion : In a triangle ABC, if based BC is fixed and perimeter of the triangle is

constant, then vertex A moves on an ellipse.

Reason : If sum of distances of a point ‘P’ from two fixed points is constant then locus

of ‘P’ is a real ellipse.


assertion.


the assertion.



Correct Answer 3

Explanation AB + BC + CA = P

Let BC = C

AB + CA = P – C = K.

A lies on ellipse with foci at B and C

Reason is true and correct explanation

Q. No. 79 Match the following

No. Column A Column B Column C Id of Additional

Answer

1 (A) Area of triangle

formed by the tangents

drawn from a point

(2, 2) to the parabola

y2 = 4(x + y)

(p) 8

2 (B)Length of the

latusrectum of the conic

25{(x + 2)2 + (y 3)

2}

= (3x + 4y 6)2 is

(q) 4 3

3 (C) If focal distance of a

point on the parabola

y = x2 4 is

25

4 and

points are of the form

+ ,a b the value of a +

b is

(r) 4

4 (D) Length of a side of an

equilateral triangle

inscribed in a parabola y2

– 2x – 2y 3 = 0 whose

one angular point is

vertex of the parabola is

(s) 25

4

Explanation A) S : y2

4y = 4x

�� (y 2)2 = 4(x + 1)

Focus : (0, 2) and directrix : x + 2 = 0

(2, Z) lies on directrix :

Tangents are perpendicular

Let the points of contact be

t ttt

1 21+ ,2+2 and L ,2

21 1 2

11

� ��

Intersection point

tt

11+1,2+ 1

1

� ��

t tt

12 + = 2 11 1

1

� ��

Points on parabola are :

(0, 4) and (0, 0)

Area of 1

= 2 + 2 2 + 22

2 2 2 2� �� = 4

B) x y

x y3 + 4 6

2 + 3 2 =5

22 � ��

� ��

�� 16x2+ 9y

2 – 24xy – 64x – 102y + 289 = 0

The above equation is parabola with directrix 3x + 4y 6 = 0 and focus (2,3)

distance of focus to directrix

12=

25

24Latus rectum =

25��

C) x2 = y + 4 Let the point (h,k)

k k k17 25

+ = = 24 4

��

h= 6� �� a b+ = 8��

D) (y 1)2 = 2(x + 2)

Latus rectum = 2

Length of side = 2 3 latus rectum = 4 3

2OA : tan30 =

t

t

0 1

21

= 2 3t�� 1

AB = 8 3��B is image of s in axis of parabola.

Q. No. 80 Match the following

No. Column A Column B Column C Id of Additional

Answer

1 (A) If the midpoint

of a chord of the

ellipse + = 116 25

x y2 2

is (0,3), then length

of the chord is 4

5

k,

then k is

(p) 6

2 (B) If the line

= +y x !!!! touches

the ellipse

9x2 + 16y

2 = 144,

then the sum of

values of !!!! is

(q) 8

3 (C) If the distance

between a focus

and corresponding

directrix of an

ellipse be 8 and the

eccentricity be 1

2,

then length of the

mirror axis is 3

k,

the k is

(r) 0

4 (D) Sum of

distances of a point

on the ellipse

+ = 19 16

x y2 2

from

the foci

(s) 16

Explanation A) Equation to chord with (0, 3) as mid point :

yy

3 9 = 0 3

25 25� �� intersects the curve at

16,3

5

� ��

.

�� Length of chord k

k32 4

= = = 85 5

��

B) x y

S : + =116 9

2 2

m m=1 16. +9 52 2! � �! � �! � �! � �! � �! � �! � �! � �

Sum of values of !!!! = 0

C) Let x y

a ba b

+ =1,2 2

2 2

2 2 , be the ellipse

aae e

e

1 + = 8, = given

2

a16

=3

��

kb a e k

8= 1 = = = 8

3 3

2� ��

Hence, 2k = 16

D) Sum of distances from foci = major axis

= 2.9 = 8

Passage Text If the locus of the circumcentre of a variable triangle having sides y axis, y = 2 and

lx + my = 1, where (l, m) lies on the parabola y2 = 4ax is a curve C, then

Q. No. 81 Coordinates of the vertex of this curve C is

Option 1 32 ,

2a

� ��

Option 2 32 ,

2a

� ��

Option 3 32 ,

2a

� ��

Option 4 32 ,

2a

� ��

Correct Answer 3

Explanation

circumcentre is midpoint of AC

m m h kl

12 +

1 2= , ,

2 2

� ��

��

…(i)

Also, (l, m) lies on y2 = 4a

�� m2 = 4al …(ii)

Locus of (h, k)

y x aa

3 1 = + 2

2 8

2� ��

a3

vertex 2 ,2

� ��

Q. No. 82 The length of smallest focal chord of this curve C is :

Option 1 1

12a

Option 2 1

4 a

Option 3 1

16a

Option 4 1

8 a

Correct Answer 4

ExplanationSmallest focal chord is equal to latus rectum =

1

8 a

Q. No. 83 The curve C is symmetric about the line :

Option 1 3=

2y

Option 2 3=

2y

Option 3 3=

2x

Option 4 3=

2x

Correct Answer 2

ExplanationThe curve is symmetric about the line y

3 = 0

2

Passage Text If P is a variable point and F1 and F2 are two fixed points such that PF PF = 2a1 2 .

Then the locus of the point P is a hyperbola, with points F1 and F2 as the foci

F F 2a 1 2 . If = 1x y

a b

2 2

2 2is a hyperbola. Then its conjugate hyperbola is = 1

x y

a b

2 2

2 2.

Let P(x,y) is a variable point such that 1 + 2 – 5 + 5 = 3.x y x y2 2 2 2

Q. No. 84 If the locus of the point P represents a hyperbola of eccentricity e, then the

eccentricity e’ of the corresponding conjugate hyperbola is :

Option 1 5

3

Option 2 4

3

Option 3 5

4

Option 4 3

7

Correct Answer 3

ExplanationFor x y x y1 + 2 5 + 5 = 3

2 2 2 2��

Transverse axis = 2a = 3

Distance between foci = 15 252 2��

�� 2ae = 5

e5

=3

��

Now, e e

1 1+ = 1

2 2

c

[Property of hyperbola]

e5

=4

c��

Q. No. 85 Locus of intersection of two perpendicular tangents to the given hyperbola is

Option 1 7 55 3 + =

2 4x y

� ��

22

Option 2 7 25 3 + =

2 4x y

� ��

22

Option 3 7 7 3 + =

2 4x y

� ��

22

Option 4 none of these

Correct Answer 4

ExplanationConjugate axis = 2b = a e2 1 = 4

2

Transverse axis = 3.

Since, conjugate axis > Transverse axis, no real director circle exists

Q. No. 86If origin is shifted to point

73,

2

� ��

and the axes are rotated through and angle �� is

clockwise sense so that equation of given of given hyperbola changes to the standard

form = 1,then isx y

a b��

2 2

2 2 :

Option 1 4tan

3

� ��

1

Option 2 3tan

4

� ��

1

Option 3 5tan

3

� ��

1

Option 4 3tan

5

� ��

1

Correct Answer 2

Explanation Centre of hyperbola �� midpoint of foci

73,

2

� ��

Equation to transverse : 3x 4y + 5 = 0

Equation to conjugate axes : 4x + 3y 45

2 = 0

�� Equation to hyperbola :

x y

x y

454 +3

23 4 + 5 5

5 = 1

9 4

4

2

2

� ��

After shifting origin to7

3,2

� ��

x y x y3 4 4 +3

5 5 =19 4

4

2� � � ��

…(i)

Clearly, (i) change to x y

a b = 1

2 2

2 2 if

3= tan

4

1 � ��

Q. No. 87 The equation to the parabola whose axis parallel to the yaxis and which passes

through the points (0,4), (1, 9) and (4, 5). If latus rectum of parabola is !!!! , then the

value of 361!!!! must be

Explanation Let the parabola be

(x a)2 = y b!!!!

For point (0, 4)

a b= 4 2 !!!! … (i)

Similarly, for (1, 9) and (4, 5)

(1 a)2 = b9!!!! …(ii)

and (4 a)2 = b5!!!! …(iii)

From (i), (ii) and (iii).

24 12 12= = = = latusrectum

38 19 19

� �� ! � !! � !! � !! � !

12361 = 361 = 288

19� ! �� ! �� ! �� ! �

Q. No. 88 The locus of a point that divides a chord of slope 2 of the parabola = 4y x2

internally in

the ratio 1 : 2 is a parabola. If the vertex of parabola is ,! )! )! )! ) then the value of 729

+! )! )! )! ) 2 must be

Explanation Let ends of the chord be

t t t t,2 and ,22 2

1 1 2 2

t t t th k

2 + 4 +2= ; =

3 3

2 2

1 2 1 2

Slope = t t

2= 2

+1 2

�� t1 + t2 = 1

Eliminating t1 and t2 :

12h = 27k2 – 48k + 24

�� Locus : x y2 8

4 = 9 9 9

2� � � ��

2 8= ; li =

9 9!!!!

li729 + = 9002� !� !� !� !

Q. No. 89

Tangents are drawn to the ellipse + =19 5

x y2 2

at ends of latusrectum. If the area of an

quadrilateral by !!!! sq unit, then the value of !!!! must be

Explanation x yS e

2: + =1, =

9 5 3

2 2

5P 2,

3

� ��

Tangent at P :

x y2+ = 1

9 3

OB = 3, OA = 9

2

1 9Ar AOB = 3

2 2� � ��

By symmetry,

Ar(quad. ABCD) 1 9

= 4 3 = 272 2

� ��

Q. No. 90 If a circle cuts a rectangular hyperbola =xy c2

in A, B, C and D and the parameters of

these four points be , ,t t t1 2 3 and t4 respectively, then the value of 16 t t t t1 2 3 4 must be

Explanation Let x2

+ y2

= a2

be the circle.

Point of intersection with ay = c2

cx a

x+ =

42 2

2��

x a x c+ + =02 2 2 4��

x x x x c=4

1 2 3 4��

c t t t t c=4 4

1 2 3 4��

t t t t16 =161 2 3 4��

ConicSectionsExercise-3.pdf - MT Educare

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