Let the tangent drawn to the parabola y ^ 2 = 24x at the point (α,...

Given, the tangent drawn to the parabola ${y}^{2} = 24x$ at the point \left(\alpha, \beta\right)$is perpendicular to the line$2x + 2y = 5$.</p><p>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$</p><p>Equation of tangent will be$y\beta = 12 \left(x + \alpha\right) so its slope will be \dfrac{12}{\beta} =$1$, so $\beta = 12$

$⇒ \alpha = 6, \beta = 12$

$∴ \left(\alpha + 4, \beta + 4\right)$ = \left(10, 16\right)$</p><p>Now finding normal at \left(10, 16\right)$ to $\dfrac{{x}^{2}}{36} - \dfrac{{y}^{2}}{144} = 1$,

First finding slope of the tangent $\dfrac{2x}{36} - \dfrac{2y}{144}\dfrac{dy}{dx} =$0$</p><p>$⇒ {\left(\dfrac{dy}{dx}\right)}_{\left(10, 16\right)} = \dfrac{144 \times 10}{36 \times 16} = \dfrac{5}{2},$so slope of$the normal will be\dfrac{- 2}{5}$</p><p>Now equation of normal will be$y - 16 = \dfrac{- 2}{5}\left(x - 10\right)$</p><p>$⇒ 2x + 5y = 100$</p><p>Now satisfying all option one by one we can see \left(15, 13\right)$ will not satisfy the given line.