Question 1

If the focus of parabola (y-k)^2=4(x-h) always lies between the lines x+y=1 and x+y =3 then

Accepted Answer

Coordinate of focus will be (h+1, k) Now focus should lie to the opposite side of origin with respect to line x+y-1=0 and same side as origin with respect to line x+y-3=0 Hence h+k 0 and h+k 2.

Question 2

Let L_1 be the length of the common chord of the curves x^2+y^2=9 and y^2=8 x, and L_2 be the length of the latus rectum of y^2=8 x, then:

Accepted Answer

We have : x^2+(8 x)=9 x^2+9 x-x-9=0 x(x+9)-1(x+9)=0 (x+9)(x-1)=0 x=-9,1 for x=1, y= 2 2 x = 2 2 (2 2 +2 2 )^2+(1-1)^2 =4 2 L _n= Length of latus rectum =4 a=4 2=8 L _1 L _2

Question 3

If vertex of a parabola is (2,-1) and the equation of its directrix is 4 x-3 y=21, then the length of its latus rectum is

Accepted Answer

We are given that directrix of the parabola is 4 x-3 y=21 and vertex of the parabola is (2,-1) Now a= |8+3-21| 5 = 10 5 =2 latus rectum of the parabola =4 a=8

Question 4

The equation of a common tangent to the parabolas y=x^2 and y=-(x-2)^2 is

Accepted Answer

Equation of tangent of parabola y=x^2 be t x=y+a t^2 ...(i) y=t x- t^2 4 Solve with y =-( x -2)^2 t x- t^2 4 =-(x-2)^2 x^2+x(t-4)- t^2 4 +4=0 Here, Discriminant =0. (t-4)^2-4 (4- t^2 4 )=0 t^2-4 t=0 t=0 or t =4 Put value of t in eq. (i), then y=4(x-1).

Question 5

If the straight line y=m x+c touches the parabola y^2-4 a x+4 a^3=0, then c is

Accepted Answer

Solving the given equations, aligned &(m x+c)^2=4 a x-4 a^3 \ & m^2 x^2+2 m c x+c^2=4 a x-4 a^3 \ & m^2 x^2+(2 m c-4 a) x+c^2+4 a^3=0 aligned Since the straight line touches the parabola at a point, so, the discriminant =0 aligned & (2 m c-4 a)^2-4 m^2 (c^2+4 a^3 )=0 \ & 4 m^2 c^2-16 a m c+16 a^2-4 m^2 c^2-16 a^3 m^2=0 \ & -m c+a-a^2 m^2=0 \ & m c=a-a^2 m^2 c= a m -a^2 m aligned

Question 6

A normal is drawn at the point P to the parabola y^2=8 x, which is inclined at 60^ with the straight line y=8. Then the point P lies on the straight line

Accepted Answer

For the parabola y^2=4 a x, the equation of normal at P (a m^2,-2 a m ) is y=m x-2 a m-a m^3. Here, m= 60^ = 3 P (a( 3 )^2,-2 a( 3 ) ) (6,-4 3 ) ( a=2) Thus, P satisfies 2 x-y-12-4 3 =0

Question 7

The length of the latus rectum of the parabola which has focus at (-1,1) and the directrix is 4 x +3 y-24=0 is

Accepted Answer

Length of latus rectum =2 distance of focus from directrix =2 |1-4+3-24| 5 =10

Parabola — Previous Year Questions & Solutions