• # question_answer The line $2px\,+\,y\,\sqrt{1-{{p}^{2}}}\,=\,1$, $(\,\left| p \right|\,<\,1\,)$ for different values of p, touches A) an ellipse of eccentricity $\sqrt{3}\,/\,2$           B) an ellipse of eccentricity $1\,/\sqrt{3}\,$ C) a hyperbola of eccentricity 2              D) an ellipse or a hyperbola depending on p

 $2px+y\sqrt{1-{{p}^{2}}}=1$ i.e.        $y=-\frac{2p}{\sqrt{1-{{p}^{2}}}}x+\frac{1}{\sqrt{1-{{p}^{2}}}}$ ?(i) Let $m=-\frac{2p}{\sqrt{1-{{p}^{2}}}}.$ Then $1-{{p}^{2}}=\frac{4}{4+{{m}^{2}}}.$
 $\therefore$      equation of the line (i) becomes $y=mx+\frac{\sqrt{4+{{m}^{2}}}}{2}$ i.e.        $y=mx+\sqrt{\frac{{{m}^{2}}}{4}+1}$ $\therefore$      the curve is an ellipse for which ${{b}^{2}}=\frac{1}{4}$ and ${{a}^{2}}=1$ $\therefore$      ${{e}^{2}}=\frac{3}{4}$                    $\therefore$      $e=\frac{\sqrt{3}}{2}$