A normal to the hyperbola, $4 x^2-9 y^2=36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$,…
A normal to the hyperbola, $4 x^2-9 y^2=36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$, respectively. If the parallelogram $O A B P(O$ being the origin) is formed, then the locus of $P$ is
$4 x^2-9 y^2=121$
$4 x^2+9 y^2=121$
$9 x^2-4 y^2=169$
$9 x^2+4 y^2=169$
Solution
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 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 $9 x^2-4 y^2=169$
Hence, locus of point $P$ is : $9 x^2-4 y^2=169$