• question_answer The point of contact of a common tangent to the parabola ${{y}^{2}}=8x$ and an ellipse $15{{x}^{2}}+4{{y}^{2}}=60$ is A)  $\left( \frac{1}{2},\,\frac{15}{4} \right)$                  B)  $\left( -\frac{1}{2},\,\frac{15}{4} \right)$ C)  $\left( \frac{1}{2},-\,\frac{15}{4} \right)$                D)  $\left( -\frac{1}{2},-\,\frac{15}{2} \right)$

Let the point of contact with ellipse is $({{x}_{1}},\,\,{{y}_{1}})$. Then, $15x{{x}_{1}}+4y{{y}_{1}}-60=0$           ?(i) Let the tangent of ${{y}^{2}}=8x\,$ is $y=\,mx+\frac{2}{m}$ and the tangent of $\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{15}=1$ is $y=mx\pm \,\sqrt{4{{m}^{2}}+15}$ $\therefore$    $\frac{2}{m}=\pm \,\sqrt{4{{m}^{2}}+15}$ $\Rightarrow$            $4{{m}^{4}}+15{{m}^{2}}-4=0$ $\Rightarrow$            $m=\pm \frac{1}{2}$ Hence, equation of common tangent at $m=\frac{1}{2}$ $\Rightarrow$            $y=\frac{x}{2}+4$ $\Rightarrow$            $x-2y+8=0$              ?(ii) From Eqs. (i) and  (ii), we get $\frac{15{{x}_{1}}}{1}=\frac{4{{y}_{1}}}{-2}=\frac{-60}{8}$ $\Rightarrow$            ${{x}_{1}}=-\frac{1}{2}$  and      ${{y}_{1}}=\frac{15}{4}$