QJEEAdv24Ma24b6: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma24b6 Solution: A normal with slope is drawn from the point to the parabola ,…

$\frac{d y}{d x}=\left.\frac{x}{-2 a} \Rightarrow \frac{d y}{d x}\right|_N=-t$

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

$\begin{aligned}& x^2=-4 a(-8 a) \\& \Rightarrow x^2=32 a^2 \Rightarrow x= \pm 4 \sqrt{2} a \\& \therefore A(-4 \sqrt{2} a,-8 a), B(4 \sqrt{2} a,-8 a) \\& \therefore A B^2=(8 \sqrt{2} a)^2=128 a^2=s \\& \therefore \text { Length of } L R=r=4 a \\& \Rightarrow \frac{r}{s}=\frac{4 a}{128 a^2}=\frac{1}{16} \\& \therefore 32 a=16 \Rightarrow a=\frac{1}{2} \\& \therefore 24 a=12 \text { Ans. }\end{aligned}$