From a point A common tangents are drawn to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=\frac{{\mathrm{a}}^2}{2}$ and the parabola y² = 4 ax. If the area of the quadrilateral formed by the common tangents, the chords of contact of the point A, w.r.t. the circle and the parabola  is $\mathrm{\lambda }$ square unit, then the value of $\frac{256}{{\mathrm{a}}^2}\mathrm{\lambda }$  must be

Here, centre of the circle is the vertex of the parabola and both circle and parabola are symmetrical about axis of parabola . in this case the point of intersection of common tangent must lie on the directrix and axis of the parabola.

i.e, $\mathrm{A}\left(-\mathrm{a},0\right)$

chord of contact of circle w,r,t   $\mathrm{A}\left(-\mathrm{a},0\right)$ is

$\mathrm{x}\left(-\mathrm{a}\right)+\mathrm{y}.0=\frac{{\mathrm{a}}^2}{2}$

$\therefore \mathrm{x}=-\frac{\mathrm{a}}{2}$

coordinates of R is $\left(-\frac{\mathrm{a}}{2},\frac{\mathrm{a}}{2}\right)$ and  chord of contact of  parabola w.r.t $\mathrm{A}\left(-\mathrm{a},0\right)$ is

$\mathrm{y}.0=2\mathrm{a}\left(\mathrm{x}-\mathrm{a}\right) \mathrm{i}.\mathrm{e} \mathrm{x}=\mathrm{a}$

coordinates of P is (a,2a)

Area of quadrilateral PQRS' 

$=2\left\{\mathrm{Area} \mathrm{of} ∆\mathrm{PAS}-\mathrm{Area} \mathrm{of} \mathrm{RAN}\right\}$

$=2\left\{\frac{1}{2}.2\mathrm{a}.2\mathrm{a}-\frac{1}{2}.\frac{\mathrm{a}}{2}\frac{\mathrm{a}}{2}\right\}$

$=4a^2-\frac{a^2}{4}=\frac{15}{4}{a}^2=\lambda$

  $\frac{256}{{\mathrm{a}}^2}\mathrm{\lambda }=960$