Let the normal at the point P on the parabola y ^ 2 = 6x pass...

Given ${y}^{2} = 6x$ comparing with ${y}^{2} = 4a \to a = \dfrac{3}{2}$

Now normal at $P$ is given by $P\left(\dfrac{3}{2}{t}^{2}, 3t\right)$

$y + tx = 3t + \dfrac{3}{2}{t}^{3}$

Now this normal is passing through \left(5, - 8\right)$</p><p>So$- 8 + 5t = 3t + \dfrac{3}{2}{t}^{3}$</p><p>$⇒ - 16 + 10t = 6t + 3{t}^{3}$</p><p>$⇒ 3{t}^{3} - 4t + 16 = 0$</p><p>Now solving above equation we get$t = - 2$</p><p>So tangent at$P$is given by$ty = x + a{t}^{2}$</p><p>putting$a&t$we get$- 2y = x + \dfrac{3}{2} \times 4$</p><p>$- 2y = x + 6$</p><p>Now tangent is intersecting at directrix which is$x = \dfrac{- 3}{2}$</p><p>So$- 2y = - \dfrac{3}{2} + 6 ⇒ y = \dfrac{3}{4} - 3 = - \dfrac{9}{4}$</p><p>So point of intersection is \left(\dfrac{- 3}{2}, \dfrac{- 9}{4}\right)$