JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Circle

For the circle $x^2+y^2=2$, a tangent with slope $m$ is $y=mx+\sqrt{2}\sqrt{1+m^2}$.

For the parabola $y^2=8x$, a tangent with slope $m$ is $y=mx+\frac{2}{m}$.

Equating: $\sqrt{2}\sqrt{1+m^2}=\frac{2}{m}$.

Squaring: $2(1+m^2)=\frac{4}{m^2}$, giving $m^4+m^2-2=0$.

This factors as $(m^2+2)(m^2-1)=0$, so $m^2=1$ (since $m^2=-2$ is impossible).

Thus $m=\pm 1$, giving tangents $y=x+2$ and $y=-x-2$.

Solving with the circle: $P=(-1,1)$ and $Q=(-1,-1)$.

Solving with the parabola: $R=(2,4)$ and $S=(2,-4)$.

The quadrilateral has vertices at $(-1,1)$, $(-1,-1)$, $(2,4)$, $(2,-4)$.

Area = $\frac{1}{2}|x_1(y_2-y_4)+x_2(y_3-y_1)+x_3(y_4-y_2)+x_4(y_1-y_3)|$ = 15.