JEE Advanced 2024 — Paper 2 — Question 12

Rotating the axis by an angle of $90^{\circ}$ in anticlock-wise direction then slope of normal becomes $-\sqrt{6}$ and point $(0,-\alpha)$ becomes $(\alpha, 0)$ and parabola being $y^{2}=$ 4ax $\Rightarrow$ Equation of normal $\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^{3}$ passes through ( $\alpha, 0$ ) $\Rightarrow 0=-\sqrt{6} \alpha+2 \sqrt{6} a+6 \sqrt{6} a$ $\Rightarrow \alpha=8 a=\mathrm{at}^{2} \Rightarrow \mathrm{t}= \pm 2 \sqrt{2}$ $\Rightarrow A B=4 a t=8 \sqrt{2} a$ Now $A B^{2}=64 \times 2 \times a^{2}$ $\Rightarrow \frac{\mathrm{r}}{\mathrm{s}}=\frac{4 \mathrm{a}}{128 \mathrm{a}^{2}}=\frac{1}{16} \Rightarrow \mathrm{a}=\frac{1}{2} \Rightarrow 24 \mathrm{a}=12$