Q 11 :

Let a tangent to the curve y2=24x meet the curve xy=2 at the points A and B. Then the mid-points of such line segments AB lie on a parabola with the        [2023]

  • length of latus rectum 3/2

     

  • length of latus rectum 2

     

  • directrix 4x = 3

     

  • directrix 4x = - 3

     

(3)

Let the equation of tangent to y2=24x  is ty=x+6t2

Now, ty=x+6t2 meets the curve xy=2  at points A and B.

Let mid-point of AB be (h,k)

ty=2y+6t2               [x=2y]

ty2-6t2y-2=0

y1+y2=6t

t·2x=x+6t2          [y=2x]

x2+6t2x-2t=0

x1+x2=-6t2

Midpoint P is (-3t2,3t)

h=-3t2, k=3t (h-3)=(k3)2y2=-3x



Q 12 :

The equations of the sides AB and AC of a triangle ABC are (λ+1)x+λy=4 and λx+(1-λ)y+λ=0 respectively. Its vertex A is on the y-axis and its orthocentre is (1, 2). The length of the tangent from the point C to the be part of the parabola y2=6x in the first quadrant is               [2023]

  • 22

     

  • 2

     

  • 6

     

  • 4

     

(1)

AB:(λ+1)x+λy=4

AC:λx+(1-λ)y+λ=0

Since vertex A is on y-axis x=0

So, y=4λ and y=λλ-1
4λ=λλ-1λ=2

AB:3x+2y=4

AC:2x-y+2=0A(0,2)

Let coordinates of C be (α,2α+2).

Now, [slope of altitude through C]×(-32)=-1

(2αα-1)(-32)=-1α=-12

So, coordinates of C are (-12,1).

Equation of tangent be y=mx+32m

  m=1,-3

So, tangent which touches in first quadrant at T is  

T(am2,2am)(32,3)

  CT=(32+12)2+(3-1)2=4+4=22



Q 13 :

The equations of two sides of a variable triangle are x=0 and y=3, and its third side is a tangent to the parabola y2=6x. The locus of its circumcentre is     [2023]

  • 4y2-18y+3x+18=0

     

  • 4y2-18y-3x+18=0

     

  • 4y2+18y+3x+18=0

     

  • 4y2-18y-3x-18=0

     

(1)

Equation of the parabola is y2=6x.

Here 4a=6

         a=32

Equation of tangent is y=mx+am



y=mx+32m                              ...(i)

Putting x=0 in (i), we get A(0,32m)

Putting y=3 in (i), we get B(6m-32m2,3)

The centre of the circle will lie on the line AB as midpoint.

   h=6m-34m2, k=3+6m4mm=34k-6

On substituting h=6m-34m2 to eliminate m, we get 3h=2(-2k2+9k-9)

4k2-18k+3h+18=0

So, locus is 4y2-18y+3x+18=0



Q 14 :

If the tangent at a point P on the parabola y2=3x is parallel to the line x+2y=1and the tangents at the points Q and R on the ellipse x24+y21=1 are perpendicular to the line x-y=2, then the area of the triangle PQR is               [2023]

  • 35

     

  • 95

     

  • 325

     

  • 53

     

(1)

If tangent at a point P on y2=3x is parallel to the line x+2y=1 and tangent at point Q and R on ellipse x24+y21=1 are perpendicular to the line x-y=2.

Firstly we have equation of parabola y2=3x  (i)

Tangent at P(x1,y1) is parallel to x+2y=1

        2y=1-x

        y=-x2+12

{On comparing it with y=mx+C}

Then slope, m at P=-12  (ii)

On differentiating equation (i) with respect to 'x'

           2y·dydx=3  dydx=32y=-12  (from (ii))

y1=-3

Co-ordinates of P are (3,-3).

Similarly, Q(45,15), R(-45,-15)

So, we have three points P, Q and R by which area of PQR

=12|3-3145151-45-151|                      {Area of is given as=12|x1y11x2y21x3y31|}

=3025=35

Hence, the area of PQR=35.



Q 15 :

If P(h,k) be a point on the parabola x=4y2, which is nearest to the point Q(0, 33), then the distance of P from the directrix of the parabola y2=4(x+y) is equal to        [2023]

  • 4

     

  • 2

     

  • 8

     

  • 6

     

(4)

We have x=4y2

Equation of normal at P(h,k)

y-k=-k(18)(x-h)

y-k=-8k(x-h)

This line passes through the point (0,33).

33-k=8kh                                                  ...(i)

P(h,k) will satisfy x=4y2, so h=4k2.      ...(ii)

Solving equations (i) and (ii), we get  

         k=1 and h=4

  (h,k)(4,1)

Now we have equation of parabola, y2=4(x+y)(y-2)2=4(x+1)

The directrix of this parabola is x=-2

This is the line parallel to the y-axis.

So, distance of P(4,1) from the line x=-2 is 6.



Q 16 :

The parabolas : ax2+2bx+cy=0 and dx2+2ex+fy=0 intersect on the line y=1. If a,b,c,d,e,f are positive real numbers and a,b,c are in G.P., then      [2023]

  • da,eb,fc are in G.P.

     

  • d,e,f are in A.P.

     

  • d,e,f are in G.P.

     

  • da,eb,fc are in A.P.

     

(4)

ax2+2bx+cy=0                         ...(i)

and dx2+2ex+fy=0                 ...(ii)

Equation (i) and (ii) intersect at (α,1).

   aα2+2bα+c=0                  ...(iii)

        dα2+2eα+f=0                    ...(iv)

Roots of equation (iii):

α=-2b±4b2-4ac2a=-ba    [a,b,c are in G.P.b2=ac]

α is also a root of equation (iv)

  d(-ba)2+2e(-ba)+f=0da+fc=2eb

  da,eb and fc are in A.P.



Q 17 :

Let A be a point on the x-axis. Common tangents are drawn from A to the curves x2+y2=8 and y2=16x. If one of these tangents touches the two curves at Q and R, then (QR)2 is equal to             [2023]

  • 76

     

  • 81

     

  • 72

     

  • 64

     

(3)

Given curves are x2+y2=8 and y2=16x

Equation of tangent to the parabola in slope form is:

     y=mx+axy=mx+4m

Length of perpendicular from (0, 0) to the point of tangency is equal to the length of radius of circle.

  |4m1+m2|=8  |1m1+m2|=12

m2(1+m2)=2 m4+m2-2=0

(m2+2)(m2-1)=0     m=±1

Point of contact on parabola = (am2,2am)=(4,±8)

Point of contact on circle is (-2,2) or (2,-2)

Distance between Q(-2,2) and R(4,8) is QRQR2=72



Q 18 :

Let y=f(x) represent a parabola with focus (-12,0) and directrix y=-12. Then S={x: tan-1(f(x))+sin-1(f(x)+1)=π2}:        [2023]

  • is an empty set

     

  • contains exactly one element

     

  • contains exactly two elements

     

  • is an infinite set

     

(3)

Equations of parabola, (x+12)2+(y-0)2=|y+12| x2+14+x+y2=y2+14+y

x2+x=y=f(x)  (let)

Now, tan-1f(x)+sin-1f(x)+1=π2

cos-111+f(x)+sin-1f(x)+1=π2

We know, sin-1x+cos-1x=π2,

11+f(x)=f(x)+1f(x)+1=1f(x)=0

x2+x=0x(x+1)=0x=0,-1S={-1,0}



Q 19 :

Let the tangent to the curve x2+2x-4y+9=0 at the point P(1, 3) on it meet the y-axis at A. Let the line passing through P and parallel to the line x-3y=6 meet the parabola y2=4x at B. If B lies on the line 2x-3y=8, then (AB)2 is equal to ________ .          [2023]



(292)

Equation of tangent at P(1, 3) to the curve x2+2x-4y+9=0 is

x·1+2(1+x2)-4(3+y2)+9=0

x+1+x-6-2y+9=0

2x-2y+4=0y-x=2

So, the point A is (0,2).

Equation of line passing through P and parallel to x-3y=6 is x-3y=d.

Point (1,3) lies on this line.

  1-9=dd=-8

So, equation of line is x-3y=-8.

Now, x-3y=-8 meets the parabola y2=4x at B.

So, x+83=y(x+83)2=4xx2+16x+64=36x

x2-16x-4x+64=0x=4,16           y=4,8

So, the possible coordinates of B are (4,4) or (16,8).

But (4,4) doesn't lie on the line 2x-3y=8.

Thus, point B is (16,8).

  AB2=(16-0)2+(8-2)2=256+36=292



Q 20 :

The ordinates of the points P and Q on the parabola with focus (3, 0) and directrix x=-3 are in the ratio 3 : 1. If R(α,β) is the point of intersection of the tangents to the parabola at P and Q, then β2α is equal to _______.           [2023]



(16)

Parabola is y2=12x

Let Q(3t2,6t)

So, P(27t2,18t)

Now, R(α,β)=(at1t2, a(t1+t2))

=(3t·3t, 3(t+3t))=(9t2,12t)

R(α,β)=(9t2,12t)β2α=(12t)29t2=1449=16



Q 21 :

Let a common tangent to the curves y2=4x and (x-4)2+y2=16 touch the curves at the points P and Q. Then (PQ)2 is equal to _________ .      [2023]



(32)

Equation of tangent to parabola y2=4x is given by  

     y=mx+1m                                ...(i)

and equation of tangent to the circle (x-4)2+y2=16 is given by  

      y=m(x-4)±41+m2              ...(ii)

From (i) and (ii), we get  

     1m=-4m±41+m2

1+4m2=±4m1+m21+16m4+8m2=16m2+16m4

8m2=1m=±122

Point of contact of parabola is (1(122)2,2122)                      [ Point of contact is (am2,2am)]

i.e., (8,42)

Now, length of tangent PQ=(8-4)2+(42)2-16=32

   PQ2=32



Q 22 :

If the x-intercept of a focal chord of the parabola y2=8x+4y+4 is 3, then the length of this chord is equal to _______.         [2023]



(16)

Given, the x-intercept of a focal chord of the parabola y2=8x+4y+4 is 3.

y2=8x+4y+4(y-2)2=8(x+1)Y2=4·2·X

Y=y-2, X=x+1, a=2

For focus x+1=2 and y-2=0x=1 and y=2

 Focus is (1,2)

Equation of line passing through (1,2) is  

y-2=m(x-1), (3,0) lies on the above line,  

  -2=m(3-1)m=-1

  Equation of line is y-2=-1(x-1)

i.e., x+y-3=0     y=3-x

Now put the value of y in equation of parabola (3-x)2=8x+4(3-x)+4

9+x2-6x=8x+12-4x+4    x=5+42, 5-42

So, y=-2-42,  42-2

So, length of focal chord is  

=(42+42)2+(-2-42-42+2)2

=128+128=256=16



Q 23 :

Let S be the set of all aN such that the area of the triangle formed by the tangent at the point P(b,c), b,cN, on the parabola y2=2ax and the lines x=b,y=0 is 16 unit2, then aSa is equal to _________ .              [2023]



(146)

As P(b,c) lies on parabola  

So,  c2=2ab                                           ...(i)

Now, equation of tangent to parabola y2=2ax in point form is yy1=2a(x+x12), where (x1,y1)(b,c)

yc=a(x+b)

For point B, put y=0, we get x=-b

So, area of PBA12×AB×AP=16    (Given)

12×2b×c=16bc=16

As b and c are natural numbers, possible values of (b,c) are (1,16),(2,8),(4,4),(8,2),(16,1)

Now from equation (i) a=c22b and aN, So value of (b,c) are (1,16), (2,8) and (4,4). Now, values of a are 128,16 and 2.

Hence, sum of values of a is 146.



Q 24 :

Let O be the vertex of the parabola x2 = 4y and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2 : 3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1, 2), is:   [2026]

  • 4x-5y+6=0

     

  • x-2y+3=0

     

  • 5x-4y+3=0

     

  • 5x-y-3=0

     

(3)

h=4t5

k=2t25=25(5h4)2

8k=5h2

5x2=8y

T=S1

5(xx1)-4(y+y1)=5x12-8y1

5x-4(y+2)=5-8·2

5x-4y+3=0



Q 25 :

If the chord joining the points P1(x1,y1) and P2(x2,y2) on the parabola y2=12x subtends a right angle at the vertex of the parabola, then x1x2-y1y2 is equal to        [2026]

  • 284

     

  • 292

     

  • 280

     

  • 288

     

(4)

(x1y1)=(3t12,6t1)  and  (x2y2)=(3t22,6t2)

t1t2=-4

x1x2=9(t1t2)2,    y1y2=36t1t2

x1x2-y1y2=9(16)-36(-4)

=144+144

=288



Q 26 :

Let A be the focus of the parabola y2=8x. Let the line y=mx+c intersect the parabola at two distinct points B and C. If the centroid of the triangle ABC is (73,43), then (BC)2 is equal to :  [2026]

  • 41

     

  • 32

     

  • 80

     

  • 89

     

(3)

Coordinates of centroid of triangle ABC are

23(t12+t22+1)=73t12+t22=52

43(t1+t2)=43t1+t2=1

(t1+t2)2=t12+t22+2t1t2t1t2=-34

(t1-t2)2=(t1+t2)2-4t1t2=4

(BC)2=4(t12-t22)2+16(t1-t2)2

(BC)2=80



Q 27 :

Let the image of parabola x2=4y, in the line x-y=1 be (y+a)2=b(x-c), a,b,c. Then a+b+c is equal to             [2026]

  • 4

     

  • 12

     

  • 8

     

  • 6

     

(4)

Parametric point P on x2=4y is P(2t,t2)

  mirror image of P in x-y=1 is

Q(2t-2.1.(2t-t2-1)2, t2+2.2(1).(2t-t2-1)2)

Q(t2+1, 2t-1)(h,k)

  locus of Q is x=(y+1)24+1  which is the required parabola.

 (y+1)2=4(x-1)

 a=1, b=4, c=1

 a+b+c=6



Q 28 :

An equilateral triangle OAB is inscribed in the parabola y2=4x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is   [2026]

  • 4(6+3)

     

  • 4(3-3)

     

  • 2(3+3)

     

  • 2(8-33)

     

(2)

MOA=2t-0t2-0=2t

2t=tan30°

t=23

Req. Circle: (x-12)2+y2=(43)2

Least distance =|CP-R|

=|2-43|=4(3-3)



Q 29 :

Let one end of a focal chord of the parabola y2=16x be (16,16). If P(α,β) divides this focal chord internally in the ratio 5:2, then the minimum value of α+β is equal to : [2026]

  • 7

     

  • 5

     

  • 16

     

  • 22

     

(1)

y2=16x

 parameter of point A is t=2

Parameter of point B is t=-12

Coordinates of B is (1,-4)

Case 1:

α=5+327=377

β=-20+327=127

α+β=7

Case 2:

α=2+807,    β=-8+807

α+β=22

So minimum value of α+β=7



Q 30 :

Let y2=12x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that OPA=90°. Then the locus of the centroid of such triangles OPA is :      [2026]

  • y2-4x+8=0

     

  • y2-6x+4=0

     

  • y2-2x+8=0

     

  • y2-9x+6=0

     

(3)

mAP=-t2

Equation of AP is

y-6t=-t2(x-3t2)

Put y=0

x=12+3t2

A(12+3t2,0)

Let centroid of OPA be G(h,k)

3h=0+3t2+12+3t2

      3k=0+6t+0

t=k2,    h=2t2+4

h=2(k24)+4

Locus of (h,k) is

y2=2x-8



Q 31 :

Let the locus of the mid-point of the chord through the origin O of the parabola y2=4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is:    [2026]

  • 3x2=2y

     

  • 2y2=3x

     

  • 2x2=3y

     

  • 3y2=2x

     

(2)

y2=4x

Locus of mid point of OP

M(h,k)h=t22, k=t

k2=2hy2=2x

S: y2=2x

h=3t224, k=3t4

t2=8h3, t=4k3

16k29=8h32k2=3h

Locus of R: 2y2=3x