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

We need to find the ordinate of point $$Q$$ where the tangent at $$P$$ on the parabola $$y^2 = 6x$$ meets the directrix.

Since $$y^2 = 6x$$ implies $$4a = 6$$ and hence $$a = \frac{3}{2}$$, the directrix is $$x = -\frac{3}{2}$$ and a general point on the parabola can be written as $$P = \left(\frac{3t^2}{2},3t\right)$$ using the parametric form $$(at^2,2at)$$.

Now the equation of the normal at parameter $$t$$ is given by $$y + tx = 2at + at^3$$ which simplifies to $$y + tx = 3t + \frac{3t^3}{2}$$. Since this normal passes through $$(5,-8)$$, substituting these coordinates gives $$-8 + 5t = 3t + \frac{3t^3}{2}$$. This leads to $$-8 + 2t = \frac{3t^3}{2}$$, and multiplying by 2 yields $$-16 + 4t = 3t^3$$ or $$3t^3 - 4t + 16 = 0$$. Testing $$t = -2$$ confirms the equation, since $$3(-8) - 4(-2) + 16 = -24 + 8 + 16 = 0$$. Therefore $$t = -2$$ and $$P = (6,-6)$$.

Next, the equation of the tangent at $$P$$ follows from the general tangent $$ty = x + at^2$$ which for $$t = -2$$ becomes $$-2y = x + \frac{3\cdot4}{2} = x + 6$$, or equivalently $$x + 2y + 6 = 0$$.

From the directrix $$x = -\frac{3}{2}$$, substituting into the tangent equation gives $$-\frac{3}{2} + 2y + 6 = 0$$, which simplifies to $$2y = -\frac{9}{2}$$ and hence $$y = -\frac{9}{4}$$.

The ordinate of point $$Q$$ is $$-\frac{9}{4}$$.

The correct answer is Option A.