JEE Advanced 2014 · Paper 2, Mathematics question 45

Common tangents. A tangent to $y^2=8x$ is $y=mx+\frac2m$. Its distance from the origin must be $\sqrt2$:
$$\frac{4}{m^2(1+m^2)}=2\implies m^4+m^2-2=0\implies m^2=1.$$
The tangents are $y=x+2$ and $y=-x-2$.

Points of contact.

• Circle: the feet of the perpendiculars from O, $P(-1,1)$ and $Q(-1,-1)$.
• Parabola: $\left(\frac a{m^2},\frac{2a}m\right)$ with $a=2$: $R(2,4)$ and $S(2,-4)$.

Area. PQRS is a trapezium with parallel vertical sides of length 2 (at $x=-1$) and 8 (at $x=2$), a distance 3 apart:
$$\text{Area}=\frac{2+8}2\times3=15.$$

(D)