JEE Advanced 2023 — Paper 1 — MATH-II solution

JEE Advanced2023Paper 1MATH-II

Let $P$ be a point on the parabola $y^{2} = 4 a x$ , where $a > 0$ . The normal to the parabola at $P$ meets the $x$ -axis at a point $Q$ . The area of the triangle $P F Q$ , where $F$ is the focus of the parabola, is $120$ . If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a, m)$ is

A
$(2, 3)$
B
$(1, 3)$
C
$(2, 4)$
D
$(3, 4)$

Solution

Parametrise $P = (a t^{2}, 2 a t)$ . The normal at $P$ has slope $m = - t$ and the equation $y = - t x + 2 a t + a t^{3}$ . Setting $y = 0$ gives $Q = (2 a + a t^{2}, 0)$ , while the focus is $F = (a, 0)$ .

Base $F Q = (2 a + a t^{2}) - a = a (1 + t^{2})$ and height from $P$ is $∣2 a t ∣$ , so

Area $= \frac{1}{2} a (1 + t^{2}) ∣2 a t ∣ = a^{2} (1 + t^{2}) ∣ t ∣ = 120$ .

With $m = - t$ , i.e. $t = - m$ and $∣ t ∣ = m$ , this becomes $a^{2} (1 + m^{2}) m = 120$ .

Testing the options: $(a, m) = (2, 3)$ gives $4 \cdot 10 \cdot 3 = 120$ . $✓$

Hence $(a, m) = (2, 3)$ .

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