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 :$9{x^2} - 4{y^2} = 169$
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