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 (1) $(2,3)$ (2) $(1,3)$ (3) $(2,4)$ (4) $(3,4)$

Let P be a point on the parabola y²=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

$y^2 = 4ax, \qquad F(a, 0)$

$P = (a t^2, \ 2a t)$

Normal at P meets the x-axis at Q, area(triangle PFQ) = 120

$\text{Find } (a, m), \quad a, m \in \mathbb{Z}^{+}$

$1.\ \text{normal slope} \Rightarrow t = -m$

$2.\ Q = \text{normal's } x\text{-intercept}$

3. area(triangle PFQ) = 120 ⇒ integer (a, m)

$\text{tangent slope} = \dfrac{1}{t}, \qquad \text{normal slope} = -t = m \Rightarrow t = -m$

$y - 2a t = -t(x - a t^2)$

$\Rightarrow y = -t x + a t^3 + 2a t$

$0 = -t x + a t^3 + 2a t \Rightarrow x = a t^2 + 2a$

$\Rightarrow Q(a t^2 + 2a, \ 0)$

$FQ = a(t^2 + 1), \qquad \text{height} = |2a t|$

$\text{area} = \tfrac{1}{2}\, a(t^2+1)\,|2a t| = a^2 |t| (t^2+1)$

$\text{area} = a^2 \cdot m \cdot (m^2 + 1) = 120$

$a=2, \ m=3: \ 2^2 \cdot 3\,(3^2+1) = 4 \cdot 3 \cdot 10 = 120$

$a=1, \ m=3: \ 1^2 \cdot 3\,(3^2+1) = 30 \neq 120$

$(a, m) = (2, 3) \Rightarrow \text{option (1)}$