JEE Advanced 2023 Paper 1 · Q07 · Parabola

Question

Let $P$ be a point on the parabola $y^2 = 4ax$, where $a > 0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $PFQ$, 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

SolveKar AI · Accepted Answer

Parametrize $P = (at^2, 2at)$. Normal at $P$ has slope $m = -t$. Equation of normal: $y - 2at = -t(x - at^2)$, meets $x$-axis at $Q = (2a + at^2, 0)$. Focus $F = (a, 0)$, so $FQ = a + at^2 = a(1+t^2)$. Height of $P$ above $x$-axis is $2at$. Area $= \dfrac{1}{2} \cdot a(1+t^2) \cdot 2at = a^2 t (1+t^2) = 120$. Testing positive integer $(a, m=t)$: $a=2, t=3$: $4 \cdot 3 \cdot 10 = 120$ ✓. So $(a, m) = (2, 3)$.

JEE Advanced 2023 Paper 1 · Q07 · Parabola

Solution