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

The correct numerical answer is 12. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Mathematics Question 12 Solution

The equation of the parabola is $x^2 = -4ay$ . A point on this parabola can be taken as $(2at, -at^2)$ .
The slope of the tangent at $(2at, -at^2)$ is $\frac{dy}{dx} = \frac{-2x}{4a} = -t$ . Thus, the slope of the normal is $m = \frac{1}{t}$ .
Given $m = \frac{1}{\sqrt{6}}$ , we have $t = \sqrt{6}$ .
The equation of the normal is $y - (-at^2) = \frac{1}{t}(x - 2at)$ , which simplifies to $y = \frac{x}{t} - 2a - at^2$ .
The normal passes through $(0, -\alpha)$ , so $-\alpha = -2a - a(6) = -8a$ , which gives $\alpha = 8a$ .
The line $L$ passes through $(0, -8a)$ and is parallel to the directrix $y = a$ . Thus, the equation of $L$ is $y = -8a$ .
Intersection of $L$ with the parabola: $x^2 = -4a(-8a) = 32a^2$ , so $x = \pm 4\sqrt{2}a$ .
The points $A$ and $B$ are $(-4\sqrt{2}a, -8a)$ and $(4\sqrt{2}a, -8a)$ . The length $AB = 8\sqrt{2}a$ .
Given $r = 4a$ (latus rectum) and $s = (AB)^2 = 128a^2$ .
From $r : s = 1 : 16$ , we have $\frac{4a}{128a^2} = \frac{1}{16} \Rightarrow \frac{1}{32a} = \frac{1}{16} \Rightarrow a = \frac{1}{2}$ .
The value of $24a = 24 \times \frac{1}{2} = 12$ .