• # question_answer A circle concentric to the ellipse $\frac{4{{x}^{2}}}{289}+\frac{{{y}^{2}}}{{{k}^{2}}}=1,$ where $k<\frac{17}{2},$ passes through foci ${{S}_{1}}$ and${{S}_{2}}$ of the ellipse and cuts ellipse at point P. If area of $\Delta P{{S}_{1}}{{S}_{2}}$ is 35 sq units, then ${{S}_{1}}{{S}_{2}}$ is equal to A)  12 units                       B)  13 units            C)  $\sqrt{149}$ units                 D)  $\sqrt{49}$ units

Since, circle is concentric to the ellipse and passes through the foci ${{S}_{1}}$ and ${{S}_{2}}$. Hece, ${{S}_{1}}{{S}_{2}}$ in the diameter of the circle. Then, area of $\Delta P{{S}_{1}}{{S}_{2}}$ $=\frac{1}{2}\,({{S}_{1}}P)\,({{S}_{2}}P)$ as ${{S}_{1}}{{S}_{2}}$ is daimeter But,      ${{S}_{1}}P+{{S}_{2}}P=2a=17,$ then Area $=\frac{1}{2}\,x\,(17-x)\,=35,$ where ${{S}_{1}}P=x$ $\Rightarrow$            ${{x}^{2}}-17x+70=0$ $\Rightarrow$            $x=7,\,\,10$ $\therefore$    ${{S}_{1}}{{S}_{2}}\,\sqrt{PS_{1}^{2}+PS_{2}^{2}}$ $=\sqrt{49+100}$ $=\sqrt{149}$            units