JEE Main 2018 — Hyperbola Question with Solution

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