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 ________.

A normal with slope 1/√6 is drawn from the point (0,-α) to the parabola x²=-4ay, where a>0. Let L be the line passing through (0,-α) 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 ________.

$x^2 = -4ay,\quad a > 0$

$\text{normal of slope } \dfrac{1}{\sqrt{6}} \text{ from } (0,-\alpha)$

$L \parallel \text{directrix},\ L \cap \text{parabola} = \{A, B\}$

$r = \text{length of latus rectum} = 4a$

$s = AB^2$

$r : s = 1 : 16 \ \Rightarrow\ \text{find } 24a$

$1.\ \text{normal slope } 1/t \Rightarrow t$

$2.\ \text{y-intercept of normal} \Rightarrow \alpha$

$3.\ L:\ y = -\alpha \Rightarrow AB,\ s;\ \text{use } r:s$

$P(t) = (2at,\, -at^2)$

$\text{tangent slope} = -\dfrac{x}{2a} = -t$

$\text{normal slope} = \dfrac{1}{t}$

$\dfrac{1}{t} = \dfrac{1}{\sqrt{6}} \Rightarrow t = \sqrt{6}$

$\text{normal at } P:\ y + at^2 = \dfrac{1}{t}(x - 2at)$

$x = 0 \Rightarrow y = -a t^2 - 2a = -a(t^2+2)$

$\alpha = a(t^2+2) = a(6+2) = 8a$

$\text{directrix of } x^2 = -4ay:\ y = a$

$L \parallel \text{directrix} \Rightarrow L:\ y = -\alpha$

$x^2 = -4a(-\alpha) = 4a\alpha$

$A, B = \left(\pm 2\sqrt{a\alpha},\ -\alpha\right)$

$AB = 4\sqrt{a\alpha} \Rightarrow s = AB^2 = 16a\alpha$

$r = 4a$

$\dfrac{r}{s} = \dfrac{4a}{16a\alpha} = \dfrac{1}{4\alpha}$

$\dfrac{1}{4\alpha} = \dfrac{1}{16} \Rightarrow \alpha = 4$

$\alpha = 8a \Rightarrow 8a = 4 \Rightarrow a = \dfrac{1}{2}$

$24a = 24\cdot\dfrac{1}{2} = 12$

$\boxed{24a = 12}$