JEE Advanced 2024 · Paper 2, Mathematics question 12

Parabola. For $x^2=-4ay$ use the point $(2at,-at^2)$.

• The tangent slope there is $-\frac{x}{2a}=-t$, so the normal slope is $\frac1t$.
• Setting $\frac1t=\frac1{\sqrt6}$ gives $t=\sqrt6$.

The point $(0,-\alpha)$. The normal is $y+at^2=\frac1t(x-2at)$. At $x=0$:
$$y=-at^2-2a=-8a,$$
so $\alpha=8a$.

Chord AB. The directrix is $y=a$, so $L$ is $y=-8a$.

• It meets the parabola where $x^2=32a^2$, i.e. $x=\pm4\sqrt2\,a$.
• So $AB=8\sqrt2\,a$ and $s=128a^2$.

Ratio. The latus rectum is $r=4a$.
$$\frac rs=\frac{4a}{128a^2}=\frac1{32a}=\frac1{16}\implies a=\frac12.$$

So $24a=12$.