A normal with slope \(\frac{1}{\sqrt{6}}\) is drawn from the point \((0,-\alpha)\) to the parabola \(x^2=-4 a y\), 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 \(A B\). If \(r: s=1: 16\), then the value of 24ais ________.

\(\frac{d y}{d x}=\left.\frac{x}{-2 a} \Rightarrow \frac{d y}{d x}\right|_N=-t\)
Slope of normal \(=\frac{1}{t}=\frac{1}{\sqrt{6}} \Rightarrow t=\sqrt{6}\)
Now, \(\frac{-\mathrm{at}^2+\alpha}{2 \mathrm{at}}=\frac{1}{\mathrm{t}}\)
\(\Rightarrow-\mathrm{at}^2+\alpha=2 \mathrm{a}\)
\(\Rightarrow-6 \mathrm{a}+\alpha=2 \mathrm{a} \Rightarrow \alpha=8 \mathrm{a}\)
For A and B
\(\begin{aligned} & x^2=-4 a(-8 a) \\ & \Rightarrow x^2=32 a^2 \Rightarrow x= \pm 4 \sqrt{2} a \\ & \therefore A(-4 \sqrt{2} a,-8 a), B(4 \sqrt{2} a,-8 a) \\ & \therefore A B^2=(8 \sqrt{2} a)^2=128 a^2=s\end{aligned}\)
\(\therefore\) Length of LR \(=\mathrm{r}=4 \mathrm{a}\)
\(\begin{aligned} & \Rightarrow \frac{\mathrm{r}}{\mathrm{s}}=\frac{4 \mathrm{a}}{128 \mathrm{a}^2}=\frac{1}{16} \\ & \therefore 32 \mathrm{a}=16 \Rightarrow \mathrm{a}=\frac{1}{2} \\ & \therefore 24 \mathrm{a}=12 \text { Ans. }\end{aligned}\)