JEE Advanced 2023 — Paper 1 — Question 6

The equation of normal to $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. The normal meets the x-axis at $Q$, so set $y = 0$: $0 = mx_Q - 2am - am^3 \Rightarrow x_Q = 2a + am^2$. The focus $F = (a,0)$. The coordinates of point $P$: for slope $m$, the parameter $t = -m$, so $P = (at^2, 2at) = (am^2, -2am)$. Area of $\triangle PFQ = \frac{1}{2} \times \text{base} \times \text{height}$. Base = $x_Q - x_F = (2a + am^2) - a = a + am^2 = a(1 + m^2)$. Height = $|y_P| = | -2am | = 2am$ (since $a,m > 0$). Thus area = $\frac{1}{2} \cdot a(1+m^2) \cdot 2am = a^2m(1+m^2)$. Given area = 120, so $a^2 m (1+m^2) = 120$. Check integer options: $(2,3)$ gives $4 \cdot 3 \cdot 10 = 120$. Hence $(a,m) = (2,3)$.