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

The common tangents to the circle x² + y² = 2 and the parabola y² = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is $\text{Circle: } x^{2} + y^{2} = 2$ $\text{Parabola: } y^{2} = 8x \Rightarrow 4a = 8,\ a = 2$ $\text{Common tangents touch circle at } P, Q \text{ and parabola at } R, S$ $\text{To find: area of quadrilateral } PQRS$ $\text{Plan: }1.\ \text{find the common tangents}$ $2.\ \text{contact points } P, Q \text{ (circle)},\ R, S \text{ (parabola)}$ $3.\ \text{area of } PQRS$ $y^{2} = 8x \Rightarrow 4a = 8 \Rightarrow a = 2$ $\text{Vertex } (0,0),\ \text{opens right}$ $\text{Tangent to } y^{2} = 4ax \text{ with slope } m:\ y = mx + \dfrac{a}{m}$ $\Rightarrow\ y = mx + \dfrac{2}{m}\qquad \left(c = \dfrac{2}{m}\right)$ $x^{2} + y^{2} = 2 \Rightarrow \text{centre } (0,0),\ r = \sqrt{2}$ $\dfrac{|c|}{\sqrt{1+m^{2}}} = \sqrt{2} \Rightarrow c^{2} = 2(1+m^{2})$ $\left(\dfrac{2}{m}\right)^{2} = 2(1+m^{2}) \Rightarrow \dfrac{4}{m^{2}} = 2 + 2m^{2}$ $m^{4} + m^{2} - 2 = 0$ $(m^{2}+2)(m^{2}-1) = 0$ $m^{2} = 1 \Rightarrow m = \pm 1$ $c = \dfrac{2}{m} \Rightarrow c = 2 \text{ or } c = -2$ $\text{Common tangents: } y = x + 2,\quad y = -x - 2$ $\text{Parabola contact: } \left(\dfrac{a}{m^{2}},\ \dfrac{2a}{m}\right) = \left(\dfrac{2}{m^{2}},\ \dfrac{4}{m}\right)$ $R = (2, 4),\qquad S = (2, -4)$ $\text{Circle contact} = \text{foot of } \perp \text{ from } (0,0)$ $P = (-1, 1),\qquad Q = (-1, -1)$ $P(-1,1),\ Q(-1,-1),\ R(2,4),\ S(2,-4)$ $PQ = 2\ (\text{at } x=-1),\qquad RS = 8\ (\text{at } x=2)$ $\text{Trapezium: } PQ \parallel RS,\ \text{distance} = 3$ $\text{Area} = \dfrac{1}{2}(PQ + RS) \times d = \dfrac{1}{2}(2 + 8)\times 3$ $= \dfrac{1}{2}\times 10 \times 3 = 15$ $\boxed{\text{Area} = 15}\qquad \text{option (4)}$