Find Parameters for Normal to Parabola with Given Triangle Area

To find the pair of positive integers $(a, m)$, we follow these mathematical steps:

1. Parametric Form of the Normal: The equation of a normal to the parabola $y^2 = 4ax$ in terms of its slope $m$ is given by: $y = mx - 2am - am^3$

   Since the slope of the normal at a point $P(at^2, 2at)$ is $m = -t$, we have $t = -m$. Substituting $t = -m$ into the parametric coordinates gives the point of contact $P$: $P = \left(am^2, -2am\right)$

2. Coordinates of Point $Q$: The point $Q$ is the intersection of the normal with the $x$-axis. Substituting $y = 0$ into the equation of the normal: $0 = mx - 2am - am^3$ Since $m > 0$, we divide by $m$: $x = 2a + am^2$ Thus, $Q = \left(2a + am^2, 0\right)$.

3. Area of $\triangle PFQ$: The focus of the parabola is $F = (a, 0)$. Notice that both $F(a, 0)$ and $Q(2a + am^2, 0)$ lie on the $x$-axis. Therefore, the length of the base $FQ$ is: $\text{Base } FQ = |x_Q - x_F| = (2a + am^2) - a = a(1 + m^2)$

   The height of $\triangle PFQ$ corresponds to the absolute $y$-coordinate of $P$: $\text{Height } h = |y_P| = |-2am| = 2am$

   The area of $\triangle PFQ$ is given by: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times a(1 + m^2) \times 2am = a^2 m (1 + m^2)$

4. Solving for Integer Parameters $(a, m)$: We are given that the area is $120$: $a^2 m (1 + m^2) = 120$

   Since $a$ and $m$ are positive integers, we test the options:

   • For (A) $(a, m) = (2, 3)$: $a^2 m (1 + m^2) = 2^2 \cdot 3 \cdot (1 + 3^2) = 4 \cdot 3 \cdot 10 = 120$ This satisfies the equation.

   • For (B) $(a, m) = (1, 3)$: $1^2 \cdot 3 \cdot 10 = 30 \neq 120$

   • For (C) $(a, m) = (2, 4)$: $2^2 \cdot 4 \cdot (1 + 4^2) = 16 \cdot 17 = 272 \neq 120$

   • For (D) $(a, m) = (3, 4)$: $3^2 \cdot 4 \cdot (1 + 4^2) = 36 \cdot 17 = 612 \neq 120$

Therefore, the correct pair $(a, m)$ is $(2, 3)$, which corresponds to Option A.