Parabola Normal Slope and Segment Ratio Analysis

To find the value of $24a$, we follow these steps:

Step 1: Parametric representation of the parabola

The given parabola is: $x^2 = -4ay \quad (a > 0)$

Any point $P$ on this parabola can be represented in parametric form as: $P = (2at, -at^2)$

Step 2: Finding the slope and equation of the normal

Differentiating the parabola equation with respect to $x$: $2x = -4a \frac{dy}{dx} \implies \frac{dy}{dx} = -\frac{x}{2a}$

At the point $P(2at, -at^2)$, the slope of the tangent is: $m_T = -\frac{2at}{2a} = -t$

Thus, the slope of the normal line $m_N$ at point $P$ is: $m_N = -\frac{1}{m_T} = \frac{1}{t}$

We are given that the slope of the normal is $\frac{1}{\sqrt{6}}$, so: $\frac{1}{t} = \frac{1}{\sqrt{6}} \implies t = \sqrt{6}$

The equation of the normal line at point $P(2at, -at^2)$ with slope $m_N = \frac{1}{t}$ is: $y - (-at^2) = \frac{1}{t}(x - 2at)$ $y + at^2 = \frac{x}{t} - 2a$ $y = \frac{x}{t} - 2a - at^2$

Substituting $t = \sqrt{6}$: $y = \frac{x}{\sqrt{6}} - 2a - a(\sqrt{6})^2$ $y = \frac{x}{\sqrt{6}} - 8a$

Step 3: Determining $\alpha$

The normal passes through the point $(0, -\alpha)$. Substituting $(0, -\alpha)$ into the normal equation: $-\alpha = \frac{0}{\sqrt{6}} - 8a \implies \alpha = 8a$

Step 4: Equation of line $L$ and intersection points

The directrix of the parabola $x^2 = -4ay$ is the line $y = a$. Since the line $L$ passes through $(0, -\alpha) = (0, -8a)$ and is parallel to the directrix, its equation is: $L: y = -8a$

To find the points of intersection $A$ and $B$ between $L$ and the parabola, substitute $y = -8a$ into $x^2 = -4ay$: $x^2 = -4a(-8a) = 32a^2 \implies x = \pm 4\sqrt{2}a$

Thus, the points of intersection are $A(-4\sqrt{2}a, -8a)$ and $B(4\sqrt{2}a, -8a)$.

Step 5: Calculating $r$ and $s$

1. The length of the line segment $AB$ is: $AB = 4\sqrt{2}a - (-4\sqrt{2}a) = 8\sqrt{2}a$ The square of the length of $AB$, denoted by $s$, is: $s = (8\sqrt{2}a)^2 = 128a^2$

2. The length of the latus rectum $r$ for $x^2 = -4ay$ is: $r = 4a$

Step 6: Using the ratio $r : s = 1 : 16$

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

Substitute the expressions for $r$ and $s$: $\frac{4a}{128a^2} = \frac{1}{16}$ $\frac{1}{32a} = \frac{1}{16} \implies 32a = 16 \implies a = \frac{1}{2}$

Step 7: Calculating the final value

$24a = 24 \times \frac{1}{2} = 12$

12