The common tangents to the circle `x^2 + y^2 =2` and the parabola `y^2 = 8x` touch the circle at `P,Q` andthe parabola at `R,S`. Then area of quadrilateral `PQRS` is

Options

A. 3
B. 6
C. 9
D. 15

Answer

D

Explanation

Let TPR and TQS be common tangents to the parabola `y^(2)=8x` and circle `x^(2)+y^(2)=2`.
The equation of any tangent to `y^(2)=8x` is
`y=mx+(2)/(m)` at `((2)/(m^(2)), (4)/(m))`
If it touches the circle `x^(2)+y^(2)=(sqrt(2))^(2)`, then
`((2)/(m))^(2)=2(1+m^(2)) rArr m^(4)+m^(2)-2=0 rArr m = pm 1`
Putting ` m=pm 1` in `((2)/(m^(2)), (4)/(m))`, we obtain
The points of contact of two tangents are R(2, 4) and S (2, -4).
Putting `a=sqrt(2) and m = pm 1 ` in `(pm (am)/(sqrt(1+m^(2))), pm (a)/(sqrt(1+m^(2))))` , we obtain the points of contact of the tangents to the circle are P(-1,1)

Clearly, PQRS is a trapezium. So, its area is given by
`Delta= (1)/(2)(PQ+RS) (LM)=(1)/(2)(2+8)xx3=15`sq. units