A normal to the hyperbola $4{\mathrm{x}}^2–9{\mathrm{y}}^2=36$ meets the coordinate axes x and y at A and B respectively. If the parallelogram OABP (O is origin) is formed, then the locus of P is

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1 \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}\left(\theta \right) \mathrm{is} \\ \Rightarrow \frac{3\mathrm{x}}{\mathrm{sec\theta }}+\frac{2\mathrm{y}}{\mathrm{tan\theta }}=13 \\ \Rightarrow 3\mathrm{x} \mathrm{cos\theta }+2\mathrm{y} \mathrm{cot\theta }=13 \\ \mathrm{It} \mathrm{meets} \mathrm{coordinate} ax\mathrm{es} \mathrm{at} \mathrm{A}=\left(\frac{13}{3}\mathrm{sec\theta }, 0\right), \mathrm{B}=\left(0, \frac{13}{2}\mathrm{tan\theta }\right) \mathrm{and} \mathrm{let} \mathrm{P}(\mathrm{x}, \mathrm{y}), o=(0, 0) \\ \mathrm{since} \mathrm{OABP} \mathrm{is} \mathrm{parallelo}gram \mathrm{then} \mathrm{diagonals} \mathrm{bisect} \mathrm{each} \mathrm{other} \\ \Rightarrow \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{OB}=\mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AP} \\ \Rightarrow \left(0, \frac{13}{4}\mathrm{tan\theta }\right)=\left(\frac{{\displaystyle \frac{13}{3}}\mathrm{sec\theta }+\mathrm{x}}{2}, \frac{\mathrm{y}}{2}\right) \end{gathered}$$

$$\begin{gathered} \overline{)\Rightarrow \frac{{\displaystyle \frac{13}{3}}\mathrm{sec\theta }+\mathrm{x}}{2}=0 \\ \Rightarrow \mathrm{x}=\frac{-13}{3}\mathrm{sec\theta }} \mathrm{and} \frac{\mathrm{y}}{2}=\frac{13}{4}\mathrm{tan\theta } \\ \overline{)\Rightarrow \mathrm{sec\theta }=\frac{-3x}{13} } \Rightarrow \mathrm{tan\theta }=\frac{4\mathrm{y}}{2\mathrm{x}13}=\frac{2\mathrm{y}}{13} \\ \mathrm{Now} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{9{\mathrm{x}}^2}{169}-\frac{4{\mathrm{y}}^2}{169}=1 \\ \Rightarrow 9{\mathrm{x}}^2-4{\mathrm{y}}^2=169 \end{gathered}$$