Let a parabola P be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4

Q: Wait, why can't I just use the standard $y^2 = 4ax$ formula? The vertex is at $(2,0)$, not $(0,0)$!

That is a great observation! You are absolutely right. The standard form $y^2 = 4ax$ assumes the vertex is at the origin. Because our vertex is shifted to $(2, 0)$, we must use the shifted form $(y - k)^2 = 4a(x - h)$. If you use the origin-based formula, you will be describing a different parabola entirely. Always check the vertex position first!

Q: How do I know the tangent equation $y = mx + \frac{a}{m}$ works for a shifted parabola?

It actually doesn't! That specific form is only for $y^2 = 4ax$. For a shifted parabola $(y - k)^2 = 4a(x - h)$, the tangent equation becomes $y - k = m(x - h) + \frac{a}{m}$. Think of it as a transformation: we are just replacing $y$ with $(y - k)$ and $x$ with $(x - h)$ in the standard tangent equation. It keeps the geometry consistent.

Q: Is there a faster way to find the area without calculating the coordinates of $S$ and $R$?

While you could use properties of the parabola, calculating the coordinates is actually quite elegant here. Since the tangents are symmetric with respect to the axis of the parabola, the points of contact $S$ and $R$ will share the same $x$-coordinate. This makes the base of the triangle vertical and the height easy to determine. Sometimes, the direct path is the most reliable one!

Question

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O(0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$, then the area (in sq. units) of $\Delta SOR$ is equal to

Options:

(A) $16\sqrt{2}$

(B) 16

(C) 32

(D) $8\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, let us embark on a journey through the elegant world of coordinate geometry. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a parabola.

Imagine you are standing on the Cartesian plane. You see a parabola $P$ whose heart—the vertex $V$—is resting at $(2, 0)$, and its focus $F$ is gazing from $(4, 0)$. Both lie on the positive $x$-axis.

The distance between the vertex and the focus is the fundamental parameter $a$. By simply calculating $a = 4 - 2 = 2$, we have unlocked the DNA of this curve. It tells us exactly how the parabola bends and breathes.

### Constructing the Equation

With the vertex $(h, k) = (2, 0)$ and our parameter $a = 2$, we can write the equation of our parabola. The standard form for a right-opening parabola is:

$$(y - k)^2 = 4a(x - h)$$

Substituting our values, we get $(y - 0)^2 = 4(2)(x - 2)$, which simplifies beautifully to:

$$y^2 = 8(x - 2)$$

This equation is the mathematical blueprint of our curve. It is the path that every point on this parabola must follow.

### The Tangent Dance

Now, we introduce the tangents. We are drawing them from the origin $O(0, 0)$.

The general equation of a tangent to a shifted parabola $(y - k)^2 = 4a(x - h)$ with slope $m$ is:

$$y - k = m(x - h) + \frac{a}{m}$$

Plugging in our values, we get $y = m(x - 2) + \frac{2}{m}$. But wait! These tangents must pass through the origin.

This is our constraint. By substituting $x = 0$ and $y = 0$ into our tangent equation, we get:

$$0 = m(0 - 2) + \frac{2}{m}$$

This simplifies to $0 = -2m + \frac{2}{m}$.

### Solving the Mystery

This is where the algebra meets the geometry. Solving $2m = \frac{2}{m}$ gives us $m^2 = 1$, which means $m = 1$ or $m = -1$.

We have found our two slopes! These represent the two distinct tangents reaching out from the origin to kiss the parabola.

Now, where do they touch? Using the point of contact formula $(h + \frac{a}{m^2}, k + \frac{2a}{m})$, we find the points $S$ and $R$.

For $m = 1$, we get:

$$S = \left(2 + \frac{2}{1^2}, 0 + \frac{2(2)}{1}ight) = (4, 4)$$

For $m = -1$, we get:

$$R = \left(2 + \frac{2}{(-1)^2}, 0 + \frac{2(2)}{-1}ight) = (4, -4)$$

### The Geometric Climax

Look at the points $S(4, 4)$ and $R(4, -4)$. They share the same $x$-coordinate! This means the segment $SR$ is a vertical line.

The length of this base is the difference in their $y$-coordinates: $4 - (-4) = 8$. The height of the triangle $\Delta SOR$ from the origin to this vertical line is simply the $x$-coordinate, which is $4$.

The area of our triangle is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 4 = 16$$

We have arrived at our destination: **16 square units**. A perfect, clean, and satisfying result. You have mastered the parabola today!

Select Answer:

Visualized Solution

Visualizing Vertex and Focus

Finding the Parabola Parameter $a$

Constructing the Parabola Equation

General Equation of the Tangent

Applying the Origin Constraint

Solving for Tangent Slopes

Determining Points of Contact Formula

Coordinates of Point $R$

Visualizing the Triangle $S O R$

Calculating the Final Area

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Constructing the Equation

The Tangent Dance

Solving the Mystery

The Geometric Climax

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