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

The correct option is (2, 3).

Let us solve the problem step-by-step by finding the coordinates of the points involved and calculating the area of the triangle $PFQ$.

Step 1: Parametric form of the parabola and equation of normal
The equation of the parabola is given by:

$y^2=4ax$

where $a>0$.
The focus $F$ of the parabola is at $(a,0)$.
In terms of the slope $m$ of the normal, the equation of the normal to the parabola $y^2=4ax$ at a point $P$ is:

$y=mx-2am-a{m}^3$

The point of contact $P$ in terms of slope $m$ is:

$P=(a{m}^2,-2am)$

Step 2: Finding the coordinates of point Q
The normal meets the x-axis at point $Q$. Setting $y=0$ in the equation of the normal:

$0=mx-2am-a{m}^3$

Since slope $m>0$, we can divide by $m$:

$x=2a+a{m}^2$

Therefore, the coordinates of $Q$ are:

$Q=(2a+a{m}^2,0)$

Step 3: Calculating the Area of Triangle PFQ
The vertices of triangle $PFQ$ are:
$F(a,0)$
$Q(2a+a{m}^2,0)$
$P(a{m}^2,-2am)$

Notice that both $F$ and $Q$ lie on the x-axis. Therefore, the base of the triangle along the x-axis is the distance $FQ$:

$\text{Base}FQ=\left|\right(2a+a{m}^2)-a|=a+a{m}^2=a(1+m^2)$

The height of the triangle is the absolute value of the y-coordinate of point $P$:

$\text{Height}=|-2am|=2am$

Using the formula for the area of a triangle:

$\text{Area}=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\text{Area}=\frac{1}{2}\times a(1+m^2)\times 2am=a^{2}m(1+m^2)$

Step 4: Solving for positive integers a and m
We are given that the area of triangle $PFQ$ is $120$:

$a^{2}m(1+m^2)=120$

Since both $a$ and $m$ are positive integers, let us test the given options for $(a,m)$:

1. For $(a,m)=(2,3)$:
$a^{2}m(1+m^2)=\left(2^2\right)\left(3\right)(1+3^2)=4\times 3\times 10=120$
This satisfies the given condition perfectly.

2. For $(a,m)=(1,3)$:
$1^2\times 3\times (1+9)=30\ne 120$

3. For $(a,m)=(2,4)$:
$2^2\times 4\times (1+16)=16\times 17=272\ne 120$

4. For $(a,m)=(3,4)$:
$3^2\times 4\times (1+16)=36\times 17=612\ne 120$

Thus, the pair $(a,m)$ is (2, 3).