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

For the parabola $$y^{2}=4ax$$ (with $$a\gt 0$$) every point can be written in the parametric form $$P(at^{2},\,2at)$$, where $$t$$ is a real parameter.

Differentiate the parabola to obtain the tangent slope:
$$2y\dfrac{dy}{dx}=4a\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{2a}{y}$$
At $$P(at^{2},\,2at)$$, the tangent slope is $$\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$$.

Hence the normal slope (negative reciprocal of the tangent slope) is
$$m=-t$$.

Equation of the normal through $$P$$:
$$y-2at=m\bigl(x-at^{2}\bigr).\tag{-1}$$

Let $$Q(x_{Q},0)$$ be the x-intercept of this normal. Put $$y=0$$ in $$( -1 )$$:
$$-2at=m\,(x_{Q}-at^{2})$$
Substituting $$m=-t$$ and cancelling $$(-t)\neq 0$$ gives
$$2a=x_{Q}-at^{2}\;\Longrightarrow\;x_{Q}=a(t^{2}+2).$$
Thus $$Q\bigl(a(t^{2}+2),\,0\bigr).$$

The focus of the parabola is $$F(a,0).$$ The triangle under consideration has vertices
$$P(at^{2},\,2at),\;F(a,0),\;Q\bigl(a(t^{2}+2),0\bigr).$$

Because $$F$$ and $$Q$$ lie on the x-axis, segment $$FQ$$ is the base:
Base length $$|FQ|=a\bigl(t^{2}+2-a/a\bigr)=a(t^{2}+1).$$
Height of the triangle is the absolute y-coordinate of $$P$$, namely $$|2at|.$$

Area $$\triangle PFQ=\dfrac12\bigl(\text{base}\bigr)\bigl(\text{height}\bigr)$$:
$$\text{Area}= \dfrac12 \bigl[a(t^{2}+1)\bigr]\bigl[|2at|\bigr] = a^{2}(t^{2}+1)\,|t|.$$

The question states that this area equals $$120$$:
$$a^{2}(t^{2}+1)\,|t| = 120.\tag{-2}$$

We are told that the slope $$m$$ of the normal and $$a$$ are positive integers. Since $$m=-t$$, a positive integer $$m$$ implies $$t\lt 0$$, so set $$|t|=-t=m.$$ Let us write $$s=|t|=m$$ (a positive integer). Equation $$( -2 )$$ now becomes

$$a^{2}\,s\,(s^{2}+1)=120.\tag{-3}$$

List small positive integers $$s$$ and evaluate $$s(s^{2}+1)$$:

s = 1 ⇒ 2, s = 2 ⇒ 10, s = 3 ⇒ 30, s = 4 ⇒ 68, …
Insert into $$( -3 )$$ and check for a perfect square $$a^{2}$$.

Case s = 3: $$30a^{2}=120 \;\Longrightarrow\; a^{2}=4 \;\Longrightarrow\; a=2$$ (integer).
Case s = 1: $$2a^{2}=120 \Rightarrow a^{2}=60$$ (not a square).
Case s = 2: $$10a^{2}=120 \Rightarrow a^{2}=12$$ (not a square).
Case s ≥ 4: $$s(s^{2}+1)\gt 60$$, so $$a^{2}\lt 2$$, impossible for positive integer $$a$$.

The only admissible solution is $$s=3,\;a=2$$. Since $$s=m$$, we obtain $$m=3$$.

Therefore $$(a,\,m)=(2,\,3).$$

Option A which is: (2, 3)