A normal to the hyperbola, 4 x^2-9 y^2=36 meets the co-ordinate...

Given, $4 x^2-9 y^2=36$ After differentiating w.r.t. $x$, we get $$\begin{aligned} \text{ 4.2.x-9.2.y. } \frac{d y}{d x}=0 \\ &\Rightarrow \text{ Slope\text{ of } tangent }=\frac{d y}{d x}=\frac{4 x}{9 y} \end{aligned}$$ So, slope of normal =$\frac{-9 y}{4 x} Now, equation$ of normal at point\left(x_0, y_0\right) is given by y-y_0=\frac{-9 y_0}{4 x_0}$\left(x-x_0\right)$ As normal intersects $\mathrm{X}$ axis at A$, Then A \equiv$\left(\frac{13 x_0}{9}, 0\right) and $B \equiv$\left(0, \frac{13 y_0}{4}\right)$As$O A B P$is a parallelogram \therefore$ midpoint of $O B \equiv \left(0, \frac{13 y_0}{8}\right) \equiv Midpoint$ of A P So, $P(x, y)$\equiv \left(\frac{-13 x_0}{9} , \frac{13 y_0}{4}\right)\because $\left(x_0, y_0\right) lies on hyperbola, therefore 4 \left(x_0\right)^2-9\left(y_0\right)^2=36$ From equation (i): $x_0=\frac{-9 x}{13} and y_0=\frac{4 y}{13} From equation (ii), we get 9 x²-4 y²=169 Hence, locus of pointP$ is : 9 x²-4 y²=169