Let the tangent to the parabola y ^ 2 = 12x at the point (3, α) be...

Given,

The tangent to the parabola ${y}^{2} = 12x$ at the point \left(3, \alpha\right)$be perpendicular to the line$2x + 2y = 3$</p><p>So, Slope of tangent$= 1 = \dfrac{6}{\alpha} ⇒ \alpha = 6$<br />So, equation of hyperbola will be,</p><p>$36{x}^{2} - 9{y}^{2} = 324$</p><p>$⇒ \dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{36} = 1$<br />Now equation of tangent at \left(5, 8\right)$ will be,

$\dfrac{5x}{9} - \dfrac{8y}{36} = 1$

$⇒ 5x - 2y = 9$
So, slope of normal$= \dfrac{- 2}{5}$
Now finding, equation of normal $y - 8 = \dfrac{- 2}{5}\left(x - 5\right)$
$⇒ 5y - 40 = - 2x + 10$

$⇒ 5y + 2x = 50$

Now finding,distance from \left(6, - 4\right)$, we get,<br />$= \left|\dfrac{12 - 20 - 50}{\sqrt{29}}\right| = \dfrac{58}{\sqrt{29}}$</p><p>$= 2\sqrt{29}$</p><p>Hence,${\left(2\sqrt{29}\right)}^{2} = 116$