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The common tangents to the circle $\mathrm{x^2+y^2=2}$ and the parabola $\mathrm{y^2=8 x}$ touch the circle at the point $\mathrm{P, Q}$ and the parabola at the points $\mathrm{R, \mathrm{~S}}$ . Then the area of the quadrilateral $\mathrm{P Q R S}$ is

Option: 1
$3$

Option: 2
$6$

Option: 3
$9$

Option: 4
$15$

Answers (1)

best_answer

Let $\mathrm{y=m x+\frac{2}{m}}$ be a tangent to $\mathrm{y^2=8 x}$ Then it is tangent to circle as well, implying

$\mathrm{ \left|\frac{2 / m}{\sqrt{1+m^2}}\right|=\sqrt{2} \Rightarrow m^2\left(1+m^2\right)=2 }$

$\mathrm{ \Rightarrow m^2=1 \quad \therefore m=1,-1 }$

The points of contact being $\mathrm{ (-1,1),(2,4),(2,-4) \: and \: (-1,-1) }$

respectively.

The area of quadrilateral $\mathrm{P Q R S=\left[\frac{1}{2}(1+4) \times 3\right] \times 2=15 }$

Hence option 4 is correct.

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