Let P be a point on the parabola y² = 4ax, where a 0. The normal to the parabola at P meets the x-

Q: Wait, why is the slope of the normal $-t$?

Great question! The slope of the tangent at $P(at^2, 2at)$ is $\frac{1}{t}$. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal, which is $-t$. This is a fundamental property of parabolas that simplifies many problems!

Q: Why is the base $FQ = a(1+t^2)$?

The x-coordinate of the focus $F$ is $a$. The x-coordinate of $Q$ is $2a + at^2$. The distance $FQ$ is the difference between these x-coordinates: $(2a + at^2) - a = a + at^2 = a(1+t^2)$. It is just the horizontal distance between the two points on the x-axis.

Q: How do I know which values of $a$ and $m$ to test?

Since $a$ and $m$ are positive integers, the equation $a^2 m (1+m^2) = 120$ is a Diophantine equation. Prime factorization of $120$ is $2^3 \cdot 3 \cdot 5$. You can test small integer values for $a$ (like $1, 2, 3$) and see if the resulting equation for $m$ yields an integer solution. For $a=2$, $a^2=4$, so $m(1+m^2) = 30$, which is easily solved by $m=3$.

Question

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

Options:

(A) $(2, 3)$

(B) $(1, 3)$

(C) $(2, 4)$

(D) $(3, 4)$

Sigma Percentile · Accepted Answer

### The Geometry of the Parabola: A Journey into $y^2 = 4ax$

Imagine you are standing on a coordinate plane, looking at the elegant curve of a parabola defined by $y^2 = 4ax$. This is not just an equation; it is the path of a projectile, the shape of a satellite dish, and a fundamental building block of conic sections.

Today, we are going to explore the relationship between a point $P$ on this parabola, its normal, and the focus $F$. We are given a point $P$ on the parabola, and for any point on a parabola, the most powerful tool in our arsenal is the parametric coordinate system.

By setting $P(at^2, 2at)$, we reduce the complexity of the problem to a single variable, $t$. This is the secret to unlocking the geometry of the parabola.

### The Normal and the Intersection $Q$

When we draw a normal to the parabola at point $P$, we are essentially drawing a line perpendicular to the tangent at that point. The equation of this normal is a standard result:

$$y = -tx + 2at + at^3$$

This line cuts through the x-axis at a point $Q$. To find $Q$, we simply set $y = 0$ in our normal equation.

Solving $0 = -tx + 2at + at^3$ gives us $tx = 2at + at^3$, which simplifies to $x = 2a + at^2$. Thus, the coordinates of $Q$ are $(2a + at^2, 0)$. We have successfully located our points $P$, $F$, and $Q$.

### The Area of Triangle $PFQ$

Now, we turn our attention to the triangle $PFQ$. The focus $F$ is at $(a, 0)$. The base of our triangle, $FQ$, lies entirely on the x-axis.

The length of this base is the difference between the x-coordinates of $Q$ and $F$:

$$FQ = |(2a + at^2) - a| = |a + at^2| = a(1+t^2)$$

The height of the triangle is the perpendicular distance from $P$ to the x-axis, which is simply the absolute value of the y-coordinate of $P$: $h = |2at|$.

The area of a triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Substituting our values, we get:

$$	ext{Area} = \frac{1}{2} \cdot a(1+t^2) \cdot |2at| = a^2 |t| (1+t^2)$$

We are told this area is $120$. So, $a^2 |t| (1+t^2) = 120$.

### The Diophantine Challenge

We are given that the slope $m$ of the normal is a positive integer. The slope of the normal is $-t$, so $m = -t$. Since $m > 0$, $t$ must be negative, and $|t| = m$.

Substituting this into our area equation, we get:

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

This is a Diophantine equation, meaning we are looking for integer solutions. Let's test values for $a$.

If $a=1$, $m(1+m^2) = 120$. Testing values for $m$: if $m=4$, $4(17) = 68$; if $m=5$, $5(26) = 130$. No integer solution exists here.

If $a=2$, $a^2=4$, so $4m(1+m^2) = 120$, which simplifies to $m(1+m^2) = 30$. Testing $m=3$: $3(1+9) = 3(10) = 30$. It works!

The pair **$(a, m) = (2, 3)$** is our solution. Through this journey, we have seen how parametric coordinates and algebraic manipulation can turn a daunting geometric problem into a beautiful, solvable puzzle.

Select Answer:

Visualized Solution

The Parabola and Focus

Point $P$ on the Parabola

Equation of the Normal

Finding Point $Q$

Coordinates of $Q$

Triangle $P F Q$

Base of the Triangle

Height of the Triangle

Area of Triangle $P F Q$

Simplifying the Area

Slope of the Normal

Substituting the Slope

Integer Constraints

Testing Values

The Final Solution

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Parabola

The Normal and the Intersection $Q$

The Area of Triangle $P F Q$

The Diophantine Challenge

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