JEE Advanced 2024 — Paper 2 — MATH-III solution

JEE Advanced2024Paper 2MATH-III

A normal with slope $\frac{1}{6}$ is drawn from the point $(0, - α)$ to the parabola $x^{2} = - 4 a y$ , where $a > 0$ . Let $L$ be the line passing through $(0, - α)$ 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 $24 a$ is ______.

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Solution

For the downward-opening parabola $x^{2} = - 4 a y$ , the normal in slope form is $y = m x - 2 a - \frac{a}{m ^{2}}$ .

With slope $m = \frac{1}{6}$ passing through $(0, - α)$ : $- α = - 2 a - \frac{a}{1/6} = - 2 a - 6 a = - 8 a$ , so $α = 8 a$ .

The directrix is $y = a$ , so $L : y = - α = - 8 a$ . Intersecting with $x^{2} = - 4 a y$ : $x^{2} = - 4 a (- 8 a) = 32 a^{2}$ , giving $x = \pm 42 a$ .

So $A = (42 a, - 8 a)$ , $B = (- 42 a, - 8 a)$ and $A B = 82 a$ , hence $s = A B^{2} = 128 a^{2}$ .

The latus rectum length is $r = 4 a$ . The ratio condition:

$\frac{r}{s} = \frac{4 a}{128 a ^{2}} = \frac{1}{32 a} = \frac{1}{16} \Rightarrow 32 a = 16 \Rightarrow a = \frac{1}{2} .$

$∴ 24 a = 12.$

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