JEE Main 2023 — Hyperbola Question with Solution

Given,

Equation of hyperbola, $3x^2-4y^2=36$

And nearest line $3x+2y=1$

Now slope of the given line will be, $m=-\frac{3}{2}$

Now slope of tangent which will be parallel to given line of hyperbola $3x^2-4y^2=36$ is given by,

$m=\frac{3\sec \theta }{\sqrt{12}·\tan \theta }$

Now putting the value of $m=\frac{-3}{2}$ we get,

$\Rightarrow \frac{3}{\sqrt{12}}\times \frac{1}{\sin \theta }=\frac{-3}{2}$

$\Rightarrow \sin \theta =-\frac{1}{\sqrt{3}}$

So, point will be $\left(\sqrt{12}\sec \theta ,3\tan \theta \right)$

$=\left(\sqrt{12}·\frac{\sqrt{3}}{\sqrt{2}},-3\times \frac{1}{\sqrt{2}}\right)$

$=\left(\frac{6}{\sqrt{2}},\frac{-3}{\sqrt{2}}\right)$,

Hence, $\sqrt{2}\left(y_0-x_0\right)=\sqrt{2}\left(\frac{-3}{\sqrt{2}}-\frac{6}{\sqrt{2}}\right)=-9$