A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0,-\alpha)$ to the parabola $x^2=-4 a y$, where $a0$. 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 $r$…

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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 ________.

Aicharya · Accepted Answer

$\frac{d y}{d x}=\left.\frac{x}{-2 a} \Rightarrow \frac{d y}{d x}ight|_N=-t$ 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 $\begin{aligned} & x^2=-4 a(-8 a) \ & \Rightarrow x^2=32 a^2 \Rightarrow x= \pm 4 \sqrt{2} a \ & 	herefore A(-4 \sqrt{2} a,-8 a), B(4 \sqrt{2} a,-8 a) \ & 	herefore A B^2=(8 \sqrt{2} a)^2=128 a^2=s\end{aligned}$ $	herefore$ Length of LR $=\mathrm{r}=4 \mathrm{a}$ $\begin{aligned} & \Rightarrow \frac{\mathrm{r}}{\mathrm{s}}=\frac{4 \mathrm{a}}{128 \mathrm{a}^2}=\frac{1}{16} \ & 	herefore 32 \mathrm{a}=16 \Rightarrow \mathrm{a}=\frac{1}{2} \ & 	herefore 24 \mathrm{a}=12 	ext { Ans. }\end{aligned}$

A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0,-\alpha)$ to the parabola $x^2=-4 a y$,…

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