 Parabola
Question

# 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

Moderate
Solution

## 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}$$\therefore \mathrm{x}=-\frac{\mathrm{a}}{2}$coordinates of R is $\left(-\frac{\mathrm{a}}{2},\frac{\mathrm{a}}{2}\right)$ and  chord of contact of  parabola w.r.t $\mathrm{A}\left(-\mathrm{a},0\right)$ iscoordinates of P is (a,2a)Area of quadrilateral PQRS' $=2\left\{\frac{1}{2}.2\mathrm{a}.2\mathrm{a}-\frac{1}{2}.\frac{\mathrm{a}}{2}\frac{\mathrm{a}}{2}\right\}$$=4{a}^{2}-\frac{{a}^{2}}{4}=\frac{15}{4}{a}^{2}=\lambda$  $\frac{256}{{\mathrm{a}}^{2}}\mathrm{\lambda }=960$

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