JEE Main 2022 — Hyperbola Question with Solution

Given, the tangent drawn to the parabola $y^2=24x$ at the point $\left(\alpha ,\beta \right)$ is perpendicular to the line $2x+2y=5$.

So, tangent at $\left(\alpha ,\beta \right)$ has slope $1$

And $\left(\alpha ,\beta \right)$ lies on $y^2=24x$ so ${\beta }^2=24\alpha$

Equation of tangent will be $\mathrm{y}\beta =12\left(x+\alpha \right)$ so its slope will be $\frac{12}{\beta }=1$, so $\beta =12$

$\Rightarrow \alpha =6,\beta =12$

$\therefore \left(\alpha +4,\beta +4\right)=\left(10,16\right)$

Now finding normal at $\left(10,16\right)$ to $\frac{x^2}{36}-\frac{y^2}{144}=1$,

First finding slope of the tangent $\frac{2x}{36}-\frac{{\displaystyle 2y}}{144}\frac{{\displaystyle dy}}{dx}=0$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(10,16\right)}=\frac{144\times 10}{36\times 16}=\frac{5}{2}$, so slope of the normal will be $\frac{-2}{5}$

Now equation of normal will be $y-16=\frac{-2}{5}\left(x-10\right)$

$\Rightarrow 2x+5y=100$

Now satisfying all option one by one we can see $\left(15,13\right)$ will not satisfy the given line.