Let the normal at the point P on the parabola $y^2=6x$ pass through the point (5,-8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

$$\begin{gathered} \text{ Equation of normal : }\mathrm{y}=-\mathrm{tx}+2\mathrm{at}+a{\mathrm{t}}^3\left(\mathrm{a}=\frac{3}{2}\right) \\ \text{ since the normal is passing through }(5,-8)\text{, we get }\mathrm{t}=-2 \\ \text{ Co-ordinate of P}:(6,-6) \\ \text{ Equation of tangent at P}:\mathrm{x}+2\mathrm{y}+6=0 \\ \mathrm{Equation} \mathrm{of} \mathrm{directrix} \mathrm{is} \mathrm{x}=\frac{{\displaystyle -3}}{{\displaystyle 2}} \\ \text{ Put }\mathrm{x}=\frac{-3}{2}\text{ to get Q}\left(\frac{-3}{2},\frac{-9}{4}\right) \end{gathered}$$