Problem Statement:

We are given:

• A parabola with equation $y^2=4ax$ where $a>0$.

• A point $P$ on this parabola.

• The normal to the parabola at $P$ meets the x-axis at point $Q$.

• The area of triangle $PFQ$ is 120, where $F$ is the focus of the parabola.

• The slope $m$ of the normal and $a$ are both positive integers.

• We need to find the correct pair $(a,m)$ from the given options.

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Step 1: Understand the Parabola and Focus

The standard parabola $y^2=4ax$ has:

• Vertex at $(0,0)$.

• Focus $F$ at $(a,0)$.

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Step 2: Parametrize Point $P$ on the Parabola

Let $P$ be a point on the parabola. We can parametrize $P$ as:

$$P=(a{t}^2,2at),$$

where $t$ is a parameter.

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Step 3: Find the Equation of the Normal at $P$

The slope of the tangent to the parabola $y^2=4ax$ at $P$ is:

$$\frac{dy}{dx}=\frac{2a}{y}=\frac{1}{t}\mathrm{.}$$

Thus, the slope of the normal (perpendicular to the tangent) is:

$$m=-t\mathrm{.}$$

The equation of the normal at $P$ is:

$$y-2at=-t(x-a{t}^2)\mathrm{.}$$

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Step 4: Find Point $Q$ where the Normal Meets the x-axis

At $Q$, $y=0$. Substituting into the normal equation:

$$0-2at=-t(x-a{t}^2)⟹-2at=-tx+a{t}^3⟹x=a{t}^2+2a\mathrm{.}$$

Thus, $Q=(a{t}^2+2a,0)$.

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Step 5: Compute the Area of Triangle $PFQ$

The vertices of the triangle are:

• $P=(a{t}^2,2at)$,

• $F=(a,0)$,

• $Q=(a{t}^2+2a,0)$.

Using the formula for the area of a triangle:

$$\text{Area}=\frac{1}{2}\mid x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mid \mathrm{.}$$

Substituting the coordinates:

$$\text{Area}=\frac{1}{2}\mid a{t}^2(0-0)+a(0-2at)+(a{t}^2+2a)(2at-0)\mid =\frac{1}{2}\mid -2a^{2}t+2a^{2}t(t^2+2)\mid \mathrm{.}$$

Simplify:

$$\text{Area}=\frac{1}{2}\mid -2a^{2}t+2a^2{t}^3+4a^{2}t\mid =\frac{1}{2}\mid 2a^2{t}^3+2a^{2}t\mid =a^2\mathrm{\mid }{t}^3+t\mathrm{\mid }\mathrm{.}$$

Given that the area is 120:

$$a^2\mathrm{\mid }{t}^3+t\mathrm{\mid }=120.$$

Since $a>0$ and $t>0$ (as $m=-t$ is a positive integer), we have:

$$a^2(t^3+t)=120.$$

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Step 6: Find Integer Solutions $(a,t)$

We are given that $a$ and $m=-t$ are positive integers. Since $m$ is positive, $t$ must be negative, but from the problem's context, we consider $t$ as a positive integer (as area is positive). Let's re-express $m$ as $m=t$ (assuming the slope is positive).

Now, solve:

$$a^2(t^3+t)=120.$$

Factorize 120:

$$120=2^3\times 3\times 5.$$

Test integer values of $t$:

• For $t=1$:

  $$a^2(1+1)=2a^2=120⟹a^2=60⟹a\mathrm{\notin }\mathbb{Z}\mathrm{.}$$

• For $t=2$:

  $$a^2(8+2)=10a^2=120⟹a^2=12⟹a\mathrm{\notin }\mathbb{Z}\mathrm{.}$$

• For $t=3$:

  $$a^2(27+3)=30a^2=120⟹a^2=4⟹a=2.$$

  This gives $(a,m)=(2,3)$.

• For $t=4$:

  $$a^2(64+4)=68a^2=120⟹a^2\approx 1.76⟹a\mathrm{\notin }\mathbb{Z}\mathrm{.}$$

Thus, the only valid integer solution is $(a,m)=(2,3)$.

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Step 7: Verify the Options

The options are:

• (A) $(2,3)$ – Matches our solution.

• (B) $(1,3)$ – Does not satisfy the area condition.

• (C) $(2,4)$ – Not valid.

• (D) $(1,4)$ – Not valid.

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Final Answer:

The correct pair is $(2,3)$, which corresponds to option (A).

$$\boxed{{\displaystyle A}}$$