JEE Advanced 2024 Paper 2 · Q12 · Parabola

Question

A normal with slope $\dfrac{1}{\sqrt{6}}$ 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 $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$, then the value of $24a$ is ______.

SolveKar AI · Accepted Answer

The parabola $x^{2} = -4ay$ opens downward; directrix $y = a$. A point on it is $(2at, -at^{2})$. Tangent slope $= dy/dx = -x/(2a) = -t$; normal slope $= 1/t$. Set $1/t = 1/\sqrt{6} \Rightarrow t = \sqrt{6}$. Normal at $(2at, -at^{2})$ is $y + at^{2} = (1/t)(x - 2at)$. Setting $x = 0$: $y = -at^{2} - 2a = -a(t^{2}+2) = -8a$. So $\alpha = 8a$. Line $L: y = -8a$. Intersection with parabola: $x^{2} = -4a(-8a) = 32a^{2} \Rightarrow x = \pm 4\sqrt{2}\,a$, so $|AB| = 8\sqrt{2}\,a$ and $s = 128a^{2}$. Latus rectum $r = 4a$. Ratio $r:s = 4a : 128a^{2} = 1 : 32a$. Setting $1 : 32a = 1 : 16$ gives $32a = 16 \Rightarrow a = 1/2$. Hence $24a = 12$.

JEE Advanced 2024 Paper 2 · Q12 · Parabola

Solution