JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Hyperbola

Since $(3,\alpha )$ lies on $y^2=12x$, we have ${\alpha }^2=36$, so $\alpha =\pm 6$.

For the parabola $y^2=12x$, differentiating: $2y\frac{dy}{dx}=12$, so $\frac{dy}{dx}=\frac{6}{y}$.

At $(3,\alpha )$: slope of tangent = $\frac{6}{\alpha }$.

The line $2x+2y=3$ has slope $-1$. For perpendicularity: $\frac{6}{\alpha }\times (-1)=-1$, giving $\alpha =6$.

The hyperbola becomes $\frac{x^2}{9}-\frac{y^2}{36}=1$. At point $(5,8)$, the equation of the normal is:
$\frac{9x}{5}+\frac{36y}{8}=45$, which simplifies to $2x+5y-50=0$.

Distance from $(6,-4)$ to this line:
$d=\frac{|2(6)+5(-4)-50|}{\sqrt{4+25}}=\frac{|12-20-50|}{\sqrt{29}}=\frac{58}{\sqrt{29}}$

Square of distance: $d^2=\frac{3364}{29}=116$.