JEE Main 2018 — Hyperbola Question with Solution

$4y^2=x^2+1$ is $4y{y}_1=x{x}_1+1$.

This tangent intersect coordinate axes at $A$ and $B$ respectively, then $A\left(-\frac{1}{x_1}, 0\right)\&\mathrm{ }B\left(0, \frac{1}{4y_{\mathrm{1 }}}\right)$.

Let mid point is $M\left(h, k\right)$.

$2h=-\frac{1}{x_1}\Rightarrow x_1=-\frac{1}{2h} ...\left(1\right)$

$2k=\frac{1}{{4y}_1}\Rightarrow y_1=\frac{1}{8k} ...\left(2\right)$

Since, point $\left(x_1, y_1\right)$, lies on the hyperbola.

So, $4{y_1}^2={x_1}^2+1$

From equations $\left(1\right)$ and $\left(2\right)$, we get

$4{\left(\frac{1}{8k}\right)}^2={\left(-\frac{1}{2h}\right)}^2+1\Rightarrow \frac{1}{16k^2}=\frac{1}{4h^2}+1$

$\Rightarrow 4h^2=16k^2\left(1+4h^2\right)$

$\Rightarrow x^2=4y^2+16x^2{y}^2$

$\Rightarrow x^2-4y^2-16x^2{y}^2=0\Rightarrow$ Locus of $M$.