A normal with slope $$\frac{1}{\sqrt{6}}$$ is drawn from the point $$(0, -\alpha)$$ to the parabola $$x^2 = -4ay$$, where $$a > 0$$. Let $$L$$ be the line passing through $$(0, -\alpha)$$ and parallel to the directrix of the parabola. Suppose that $$L$$ intersects the parabola at two points $$A$$ and $$B$$. Let $$r$$ denote the length of the latus rectum and $$s$$ denote the square of the length of the line segment $$AB$$. If $$r : s = 1 : 16$$, then the value of $$24a$$ is ______.

The parabola is $$x^{2} = -4ay$$ with $$a \gt 0$$.
For this parabola:

• Vertex: $$(0,0)$$   • Focus: $$(0,-a)$$   • Directrix: $$y = a$$
• Length of the latus rectum $$r = 4a$$

1. Equation of the required normal
Let the point of contact on the parabola be $$(x_1,y_1)$$.
Write the parabola as $$y = -\dfrac{x^{2}}{4a}$$, so

$$\frac{dy}{dx} = -\frac{x}{2a}$$
Therefore, the slope of the tangent at $$(x_1,y_1)$$ is $$m_t = -\frac{x_1}{2a}$$.
The slope of the normal is the negative reciprocal: $$m_n = \frac{2a}{x_1}$$.

Given that the required normal has slope $$\dfrac{1}{\sqrt6}$$, equate:

$$\frac{2a}{x_1} = \frac{1}{\sqrt6} \;\Longrightarrow\; x_1 = 2a\sqrt6$$

Find the corresponding $$y_1$$ on the parabola:

$$y_1 = -\frac{x_1^{2}}{4a} = -\frac{(2a\sqrt6)^{2}}{4a} = -\frac{4a^{2}\,6}{4a} = -6a$$

Thus the point of contact is $$(2a\sqrt6,\,-6a)$$.

2. Locate the external point
The normal passes through the given point $$(0,-\alpha)$$ and has slope $$\dfrac{1}{\sqrt6}$$.
Equation of the normal through $$(0,-\alpha)$$:

$$y + \alpha = \frac{1}{\sqrt6}\,x \;\Longrightarrow\; y = \frac{x}{\sqrt6} - \alpha$$

Substitute $$(x_1,y_1) = (2a\sqrt6,-6a)$$ into this line:

$$-6a = \frac{2a\sqrt6}{\sqrt6} - \alpha = 2a - \alpha$$

Hence $$\alpha = 2a + 6a = 8a$$.

3. Intersection of line $$L$$ with the parabola
Line $$L$$ passes through $$(0,-\alpha)$$ and is parallel to the directrix $$y=a$$, so $$L$$ is horizontal:

$$y = -\alpha = -8a$$

Intersect with $$x^{2} = -4ay$$:

$$x^{2} = -4a(-8a) = 32a^{2}$$

Thus the two intersection points are
$$A(4a\sqrt2,\,-8a), \quad B(-4a\sqrt2,\,-8a).$$

4. Lengths $$r$$ and $$s$$
Latus-rectum length:   $$r = 4a$$.
Length $$AB = 4a\sqrt2 - (-4a\sqrt2) = 8a\sqrt2$$, so

$$s = (AB)^{2} = (8a\sqrt2)^{2} = 128a^{2}$$

Given $$\dfrac{r}{s} = \dfrac{1}{16}$$ :

$$\frac{4a}{128a^{2}} = \frac{1}{16} \;\Longrightarrow\; \frac{1}{32a} = \frac{1}{16} \;\Longrightarrow\; 16 = 32a \;\Longrightarrow\; a = \frac12$$

5. Required value
$$24a = 24 \times \frac12 = 12$$

Hence the answer is 12.