Given, $4{x^2} – 9{y^2} = 36$

ater differentiating w.r.t.$x$, we get 

$4.2x – 9.2y.\frac{{dy}}{{dx}} = 0$

$ \Rightarrow $ slope of tangent $ = \frac{{dy}}{{dx}} = \frac{{4x}}{{9y}}$

so, slope of normal $ = \frac{{ – 9y}}{{4x}}$

Now, equation of normal at point $({x_0},{y_0})$ is given by

${y_0} – {y_0} = \frac{{ – 9y}}{{4x}}\left( {{x_0} – {x_0}} \right)$

As normal intersects $X$ 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 $OABP$ is a parallelogram 

$\therefore $ midpoint of $OB \equiv \left( {0,\frac{{13{y_0}}}{8}} \right) \equiv $ Midpoint of $AP$

So,$P\left( {x,y} \right) \equiv \left( {\frac{{ – 13{x_0}}}{9},\frac{{13{y_0}}}{4}} \right)\,\,\,\,\,\,……\left( i \right)$

$\therefore \,({x_0},{y_0})$ lies on hyperbola, therfore 

$4{\left( {{x_0}} \right)^2} – 9{\left( {{y_0}} \right)^2} = 36\,\,\,\,\,\,\,\,…….\left( {ii} \right)$

Feom equation $(i):$ ${x_0} = \frac{{ – 9x}}{{13}}\,$ and ${y_0} = \frac{{4y}}{{13}}$

From equation $(ii)$, we get 

$9{x^2} – 4{y^2} = 169$

Hence, locus point $P$ is :$9{x^2} – 4{y^2} = 169$