Topic Question Set


Q 1 :

Let P be a point on the parabola y2=4ax, where a>0. The normal to the parabola at P meets the x-axis at a point Q. The area of the triangle PFQ, where F is the focus of the parabola, is 120. If the slope m of the normal and a are both positive integers, then the pair (a,m) is                 [2023]

  • (2, 3)

     

  • (1, 3)

     

  • (2, 4)

     

  • (3, 4)

     

(1)

Let point P(am2,-2am)

Equation of normal at P(am2,-2am) is

y=mx-2am-am3

Area of PFQ

=12(a+am2)×2am=120

a2m(1+m2)=120

Pair (a,m)=(2,3) satisfies the above equation.



Q 2 :

Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0, 0) to (x,y) in the ratio 1 : 3. Then the locus of P is            [2011]

  • x2=y

     

  • y2=2x

     

  • y2=x

     

  • x2=2y

     

(3)

Let A(x,y)=(t2,2t) be any point on parabola y2=4x.

Let P(h,k) divide OA in the ratio 1:3

 (h,k)=(t24,2t4)h=t24,  k=t2h=k2

 Locus of P(h,k) is x=y2



Q 3 :

The axis of a parabola is along the line y=x and the distances of its vertex and focus from origin are 2 and 22 respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is                     [2006]

  • (x+y)2=(x-y-2)

     

  • (x-y)2=(x+y-2)

     

  • (x-y)2=4(x+y-2)

     

  • (x-y)2=8(x+y-2)

     

(4)

Since distance of vertex and focus of the parabola from origin is 2 and 22,

 Vertex is (1, 1) and focus is (2, 2), directrix x+y=0

Equation of parabola is

      (x-2)2+(y-2)2=(x+y2)2

2(x2-4x+4)+2(y2-4y+4)=x2+y2+2xy

x2+y2-2xy=8(x+y-2)(x-y)2=8(x+y-2)



Q 4 :

Tangent to the curve y=x2+6 at a point (1, 7) touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are                [2005]

  • (-6,-11)

     

  • (-9,-13)

     

  • (-10,-15)

     

  • (-6,-7)

     

(4)

The given curve is y=x2+6

Equation of tangent at (1, 7) is 12(y+7)=x·1+62x-y+5=0            ...(i)

Since tangent (i) touches the circle x2+y2+16x+12y+c=0 with centre C(-8,-6) at Q.

Equation of CQ which is perpendicular to (i) is

y+6=-12(x+8)x+2y+20=0                   ...(ii)

On solving equations (i) and (ii), we get the coordinates of Q as x=-6,  y=-7

Coordinates of Q is (-6,-7)



Q 5 :

The angle between the tangents drawn from the point (1, 4) to the parabola y2=4x is                      [2004]

  • π6

     

  • π4

     

  • π3

     

  • π2

     

(3)

If m be the slope of the tangent to the parabola, then its equation is y=mx+1m

Since the tangent passes through (1, 4)

  4=m+1mm2-4m+1=0

If the angle between two tangents to the parabola be θ, then

tanθ=|m1-m21+m1m2|=(m1+m2)2-4m1m21+m1m2

=16-41+1=3θ=π3



Q 6 :

The focal chord to y2=16x is tangent to (x-6)2+y2=2, then the possible values of the slope of this chord are           [2003]

  • {-1,1}

     

  • {-2,2}  

     

  • {-2,-12}  

     

  • {2,-12}

     

(1)

Given parabola y2=16x, its focus = (4, 0). Let m be the slope of focal chord then its equation is

             y=m(x-4)               ...(i)

But it is given that equation (i) is a tangent to the circle

(x-6)2+y2=2 with centre, C(6, 0) and radius r=2

 Length of perpendicular from (6, 0) to (i) =2

6m-4mm2+1=22m=2(m2+1)

2m2=m2+1m=±1



Q 7 :

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2=4ax is another parabola with directrix        [2002]

  • x=-a  

     

  • x=-a2  

     

  • x=0  

     

  • x=a2  

     

(3)

If (h,k) is the mid point of line joining focus (a,0) and Q(at2,2at) on parabola then h=a+at22,  k=at

Eliminating t, we get 2h=a+a(k2a2)

k2=a(2h-a)      k2=2a(h-a2)

Locus of (h,k) is        y2=2a(x-a2), which is equation of a parabola

whose directrix is (x-a2)=-a2x=0



Q 8 :

The equation of the directrix of the parabola y2+4y+4x+2=0 is                      [2001]

  • x=-1  

     

  • x=1  

     

  • x=-32  

     

  • x=32  

     

(4)

Given equation of the parabola is y2+4y+4x+2=0

y2+4y+4=-4x+2(y+2)2=-4(x-12)

It is of the form Y2=-4AX

Equation of whose directrix is given by X=A

 Equation of required directrix is x-12=1x=32



Q 9 :

The equation of the common tangent touching the circle (x-3)2+y2=9 and the parabola y2=4x above the x-axis is             [2001]

  • 3y=3x+1  

     

  • 3y=-(x+3)

     

  • 3y=x+3  

     

  • 3y=-(3x+1)  

     

(3)

Let the equation of tangent to the parabola y2=4x be

y=mx+1m, where m is the slope of the tangent.

If y=mx+1m is also tangent to the circle (x-3)2+y2=9,

then length of perpendicular to the tangent from centre (3, 0) should be equal to radius 3.

i.e.,   3m+1mm2+1=39m2+1m2+6=9m2+9m=±13

 Tangents are x-3y+3=0 and x+3y+3=0

out of which x-3y+3=0 meets the parabola at (3,23), i.e., above x-axis.



Q 10 :

If the line x-1=0 is the directrix of the parabola y2-kx+8=0, then one of the values of k is                   [2000]

  • 1/8

     

  • 8

     

  • 4

     

  • 1/4

     

(3)

The directrix of the parabola y2=4a(x-x1) is given by x=x1-a.

Now given parabola is

y2=kx-8y2=k(x-8k)

 Directrix of parabola is x=8k-k4;

Now, x=1 also coincides with x=8k-k4

On comparing, 8k-k4=1k2-4k-32=0

  k=4



Q 11 :

If x+y=k is normal to y2=12x, then k is                    [2000]

  • 3

     

  • 9

     

  • -9

     

  • -3

     

(2)

y=mx+c is normal to the parabola y2=4ax if c=-2am-am3

Now given line x+y=k is normal to y2=12x

 m=-1, c=k and a=3

c=k=-2(3)(-1)-3(-1)3=9



Q 12 :

A normal with slope 16 is drawn from the point (0,-α) to the parabola x2=-4ay, where a>0. Let L be the line passing through (0,-α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r:s=1:16, then the value of 24a is ________ .                            [2024]



(12)

x2=-4ay

Equation of normal

y=mx-2a-am2

-α=-2a-a16=-8a

α=8a

Equation of required line passing through (0,-α) and parallel to directrix is

y=-αy=-8a, Solving with x2=-4ay

x2=32a2x=±42a=±α2

A(α2,-α),  B(-α2,-α)AB=2α

rs=4a2α2=1164a2×64a2=116

a=1224a=12



Q 13 :

Let the curve C be the mirror image of the parabola y2=4x with respect to the line x+y+4=0. If A and B are the points of intersection of C with the line y=-5, then the distance between A and B is                [2015]



(4)

Let (t2,2t) be any point on y2=4x and (h,k) be the image of (t2,2t) in the line x+y+4=0.

 h-t21=k-2t1=-2(t2+2t+4)2

h=-(2t+4) and k=-(t2+4)

For its intersection with y=-5, we get

   -(t2+4)=-5t=±1

 A(-6,-5) and B(-2,-5),    AB=4.



Q 14 :

If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle (x-3)2+(y+2)2=r2, then the value of r2 is           [2015]



(2)

End points of latus rectum of y2=4x are (1,±2)

Equation of normal to y2=4x at (1, 2) is

y-2=-1(x-1)x+y-3=0

As it is tangent to circle (x-3)2+(y+2)2=r2

 |3+(-2)-32|=rr2=2



Q 15 :

Consider the parabola y2=8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point P(12,2) on the parabola and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then Δ1Δ2 is                      [2011]



(2)

Δ1=area(PLL')=12×8×32=6

Equation of AB is y=2x+1; equation of AC is y=x+2 and equation of BC is -y=x+2.

On solving the above equations pairwise, we get A(1,3),  B(-1,-1) and C(-2,0)

 Δ2=12|131-1-11-201|=3

  Δ1Δ2=2



Q 16 :

Let A1,B1,C1 be three points in the xy-plane. Suppose that the lines A1C1 and B1C1 are tangents to the curve y2=8x at A1 and B1, respectively. If O=(0,0) and C1=(-4,0), then which of the following statements is (are) TRUE?                      [2024]

  • The length of the line segment OA1 is 43

     

  • The length of the line segment A1B1 is 16

     

  • The orthocenter of the triangle A1B1C1 is (0,0)

     

  • The orthocenter of the triangle A1B1C1 is (1,0)

     

Select one or more options

(3, 4)

Let parametric coordinates of A1 and B1 are

A1=(2t12,4t1),  B1=(2t22,4t2)

C(-4,0)(2t1t2,2(t1+t2))

t2=-t1 and t1(-t1)=-2

t1=2,  t2=-2

A1(4,42),  B1(4,-42)

 OA1=42+(42)2=43

Length of line segment A1B1=82

Altitude C1M: y=0              ...(i)

Altitude B1N: 2x+y=0                            ...(ii)

 Orthocentre (0,0)



Q 17 :

Consider the parabola y2=4x. Let S be the focus of the parabola. A pair of tangents drawn to the parabola from the point P=(-2,1) meet the parabola at P1 and P2. Let Q1 and Q2 be points on the lines SP1 and SP2 respectively such that  PQ1SP1 and PQ2SP2. Then, which of the following is/are TRUE?       [2022]

  • SQ1=2  

     

  • Q1Q2=3105

     

  • PQ1=3  

     

  • SQ2=1

     

Select one or more options

(1, 2, 3, 4)

Let a rough graph to refer the question

Let parametric point at P1(t2,2t), then tangent at P1

ty=x+t2

Since it passes through (-2,1)

 t2-t-2=0

 t=2,-1

 P1(4,4) and P2(1,-2)

 SP1:4x-3y-4=0  and  SP2:x-1=0

and for Q1:x1+24=y1-1-3=-(-8-3-4)25=35

 x1=25,  y1=-45 and Q2=(1,1)

So, SQ1=(1-25)2+(45)2=1

Q1Q2=925+8125=9025=3105

PQ1=14425+8125=155=3,  SQ2=1



Q 18 :

Let E denote the parabola y2=8x. Let P=(-2,4) and let Q and Q' be two distinct points on E such that the lines PQ and PQ' are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE?                            [2021]

  • The triangle PFQ is a right-angled triangle

     

  • The triangle QPQ' is a right-angled triangle

     

  • The distance between P and F is 52

     

  • F lies on the line joining Q and Q'

     

Select one or more options

(1, 2, 4)

Given that

E:y2=8x,  

P=(-2,4)

Point P(-2,4) lies on directrix of parabola.

So, QPQ'=π2 and chord QQ' is a focal chord and segment PQ subtends right angle at the focus. So, PFQ=π2

Slope of QQ'=2t1+t2=1

Slope of PF=-1   QQ'PF

PF=42



Q 19 :

If a chord, which is not a tangent, of the parabola y2=16x has the equation 2x+y=p and midpoint (h,k), then which of the following is (are) possible value(s) of  p,h and k?                       [2017]

  • p=-2, h=2, k=-4

     

  • p=-1, h=1, k=-3

     

  • p=2, h=3, k=-4

     

  • p=5, h=4, k=-3  

     

(3)

If (h,k) is the mid point of chord of parabola y2=16x, then equation of chord will be given by

T=S1yk-8(x+h)=k2-16h

8x-ky=8h-k2                  ...(i)

But given, the equation of chord is

2x+y=p                         ...(ii)

 (i) and (ii) are identical lines

82=-k1=8h-k2p

k=-4 and p=2h-4

which are satisfied by option (3)



Q 20 :

Let P be the point on the parabola y2=4x which is at the shortest distance from the center S of the circle x2+y2-4x-16y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then                       [2016]

  • SP=25

     

  • SQ:QP=(5+1):2

     

  • the x-intercept of the normal to the parabola at P is 6

     

  • the slope of the tangent to the circle at Q is 12

     

Select one or more options

(1, 3, 4)

Let point P on parabola y2=4x be (t2,2t)

 PS is shortest distance, therefore PS should be the normal to parabola.

Equation of normal to y2=4x at P(t2,2t) is

y-2t=-t(x-t2)

It passes through S(2,8)

  8-2t=-t(2-t2)

t3=8t=2,     P(4,4)

Also slope of tangent to circle at Q=-1slope of PS=12

Equation of normal at t=2 is 2x+y=12

Clearly x-intercept = 6, Now SP=25 and SQ=r=2

 Q divides SP in the ratio SP:PQ

=2:2(5-1) or (5+1):4



Q 21 :

The circle C1:x2+y2=3, with centre at O, intersects the parabola x2=2y at the point P in the first quadrant. Let the tangent to the circle C1, at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 23 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y-axis, then                    [2016]

  • Q2Q3=12

     

  • R2R3=46

     

  • area of the triangle OR2R3 is 62

     

  • area of the triangle PQ2Q3 is 42

     

Select one or more options

(1, 2, 3)

Given circle, C1:x2+y2=3                ...(i)

and parabola: x2=2y                 ...(ii)

Intersection point of (i) and (ii) in first quadrant

y2+2y-3=0y=1         (y-3)

 x=2P(2,1)

Equation of tangent to circle C1 at P is 2x+y-3=0

Let centre of circle C2 be (0,k) r=23

 |k-33|=23k=9 or -3

 Q2(0,9), Q3(0,-3)

(1) Q2Q3=12

(2) R2R3=length of transverse common tangent =(Q2Q3)2-(r1+r2)2=(12)2-(43)2=46

(3) Area (OR2R3)=12×R2R3×length of  from origin to tangent
      =12×46×3=62

(4) ar(PQ2Q3)=12×Q2Q3distance of P from y-axis
     =12×12×2=62



Q 22 :

Let P and Q be distinct points on the parabola y2=2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle OPQ is 32, then which of the following is (are) the coordinates of P?                   [2015]

  • (4,22)

     

  • (9,32)

     

  • (14,12)

     

  • (1,2)

     

Select one or more options

(1, 4)

Let point P be in first quadrant and lying on parabola y2=2x be (a22,a). Let Q be the point (b22,b). Clearly a>0.

 PQ is the diameter of circle through P,O and Q

 POQ=90°aa2/2×bb2/2=-1ab=-4

b is negative           (a>0)  Given area (POQ)=32

12|001a22a1b22b1|=32

14ab(a-b)=±32a-b=±32

As a is positive and b is negative, we have a-b=32

a+4a=32         (ab=-4)

a2-32a+4=0(a-22)(a-2)=0

  a=22, 2

 Point P can be ((22)22,22) or ((2)22,2)

i.e. (4,22) or (1,2)



Q 23 :

Let L be a normal to the parabola y2=4x. If L passes through the point (9, 6), then L is given by                [2011]

  • y-x+3=0  

     

  • y+3x-33=0

     

  • y+x-15=0  

     

  • y-2x+12=0

     

Select one or more options

(1, 2, 4)

The equation of normal to y2=4x is y=mx-2m-m3

Since the normal passes through (9, 6),  6=9m-2m-m3

m3-7m+6=0(m-1)(m2+m-6)=0

(m-1)(m+3)(m-2)=0m=1,2,-3

 Normals are y-x+3=0 or y-2x+12=0 or y+3x-33=0



Q 24 :

Let A and B be two distinct points on the parabola y2=4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be                  [2010]

  • -1r  

     

  • 1r  

     

  • 2r  

     

  • -2r

     

Select one or more options

(3, 4)

Given equation of parabola is y2=4x

Its axis is x-axis.

Let A(t12,2t1) and B(t22,2t2)

Then centre of circle drawn with AB as diameter is

(t12+t222, t1+t2)

As circle touches axis of parabola i.e., x-axis

  r=|t1+t2|t1+t2=±r

  Slope of AB=2(t2-t1)t22-t12=2t2+t1=±2r



Q 25 :

The tangent PT and the normal PN to the parabola y2=4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose              [2009]

  • vertex is (2a3,0)  

     

  • directrix is x=0

     

  • latus rectum is 2a3  

     

  • focus is (a,0)

     

Select one or more options

(1, 4)

Let P(at2,2at) be any point on the parabola y2=4ax.

  Tangent to the parabola at P is y=xt+at,

which meets the axis of parabola i.e. x-axis at T(-at2,0).

Also normal to parabola at P is tx+y=2at+at3

which meets the axis of parabola at N(2a+at2,0).

Let G(x,y) be the centroid of PTN, then

x=at2-at2+2a+at23   and   y=2at3

x=2a+at23  ...(i)   and   y=2at3  ...(ii)

Eliminating t from (i) and (ii), we get the locus of centroid G as

3x=2a+a(3y2a)2  y2=4a3(x-2a3),

which is a parabola with vertex (2a3,0), directrix as x-2a3=-a3x=a3, latus rectum as 4a3 and focus as (a,0).



Q 26 :

The equations of the common tangents to the parabola y=x2 and y=-(x-2)2 is/are                   [2006]

  • y=4(x-1)

     

  • y=0

     

  • y=-4(x-1)

     

  • y=-30x-50

     

Select one or more options

(1, 2)

If y=mx+c is tangent to y=x2 then x2-mx-c=0 has equal roots

m2+4c=0c=-m24

 y=mx-m24 is tangent to y=x2

 This is also tangent to y=-(x-2)2

mx-m24=-x2+4x-4

x2+(m-4)x+(4-m24)=0 has equal roots

 m2-8m+16=-m2+16m=0,4

 y=0 or y=4x-4 are the tangents.



Q 27 :

Match the following: (3, 0) is the point from which three normals are drawn to the parabola y2=4x which meet the parabola in the points P, Q and R. Then            [2006]

  Column I   Column II
(A) Area of PQR (p) 2
(B) Radius of circumcircle of PQR (q) 52
(C) Centroid of PQR (r) (52,0)
(D) Circumcentre of PQR (s) (23,0)

 

  • (A)(q); (B)(p); (C)(r); (D)(s)

     

  • (A)(r); (B)(s); (C)(q); (D)(p)

     

  • (A)(r); (B)(q); (C)(s); (D)(p)

     

  • (A)(p); (B)(q); (C)(s); (D)(r)

     

(4)

Let y=mx-2m-m3 be the equation of normal to y2=4x.

Since it passes through (3,0),    m=0,1,-1

Hence three points on parabola are given by (m2,-2m) for m=0,1,-1

  P(0,0), Q(1,2) and R(1,-2)

  Area(PQR)=12|0011211-21|=2

Radius of circumcircle, R=abc4Δ=5×5×44×2=52

                                         (where, a,b,c are the sides of PQR)

Centroid of PQR=(23,0); Circumcentre=(52,0)



Q 28 :

Let a,r,s,t be nonzero real numbers. Let P(at2,2at), Q, R(ar2,2ar) and S(as2,2as) be distinct points on the parabola y2=4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a,0).                        [2014]

Q.   The value of r is

  • -1t

     

  • t2+1t

     

  • 1t

     

  • t2-1t

     

(4)

  PQ is a focal chord, Q(at2,-2at)

Also QRPK,    mQR=mPK

-2at-2arat2-ar2=0-2at2a-at2

-2a(1t+r)a(1t+r)(1t-r)=-2ata(2-t2)

2-t2=t(1t-r)       [r-1t otherwise Q will coincide with R]

2-t2=1-trr=t2-1t



Q 29 :

Let a,r,s,t be nonzero real numbers. Let P(at2,2at), Q, R(ar2,2ar) and S(as2,2as) be distinct points on the parabola y2=4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a,0).                              [2014]

Q.  If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

  • (t2+1)22t3

     

  • a(t2+1)22t3

     

  • a(t2+1)2t3

     

  • a(t2+2)2t3

     

(2)

Tangent at P is ty=x+at2             ...(i)

Normal at S is sx+y=2as+as3              ...(ii)

Given st=1s=1t

 xt+y=2at+at3xt2+yt3=2at2+a

Now putting the value of x from equation (i) in above equation, we get

t2(ty-at2)+yt3=2at2+a

2t3y=a(t4+2t2+1)

 y=a(t4+2t2+1)2t3=a(t2+1)22t3



Q 30 :

Let PQ be a focal chord of the parabola y2=4ax. The tangents to the parabola at P and Q meet at a point lying on the line y=2x+a, a>0.              [2013]

Q.  Length of chord PQ is

  • 7a

     

  • 5a

     

  • 2a

     

  • 3a

     

(2)

PQ=(at2-at2)2+(2at+2at)2

=a(t+1t)2(t-1t)2+4(t+1t)2

=a(t+1t)(t-1t)2+4

=a(t+1t)2=5a