Let the tangent to the parabola S : y² = 2x at the point P(2, 2) meet the x-axis at Q and normal at

Q: Wait, why is the slope of the normal $-2$? I thought the slope of the tangent was $1/2$.

You are absolutely right! The slope of the tangent is indeed $1/2$. Since the normal is perpendicular to the tangent, their slopes must satisfy the condition $m_1 \cdot m_2 = -1$. Therefore, $m_{\text{normal}} = -1 / (1/2) = -2$. It is a fundamental property of perpendicular lines that you should always keep in your back pocket!

Q: When solving for point $R$, I got two values for $x$: $2$ and $9/2$. Why do we reject $x=2$?

Great observation! The value $x=2$ corresponds to the point $P(2, 2)$, which is where the normal *starts*. Since the problem asks for the point $R$ where the normal *meets* the parabola again, we are looking for the *other* intersection point. That is why we choose $x = 9/2$.

Q: Is there a shortcut to find the area of $\triangle PQR$ without using the coordinate formula?

While there are some specific properties for parabolas, using the coordinate geometry formula (the determinant method) is the most robust and error-proof way to solve this. It prevents you from getting lost in complex geometric theorems. Stick to the coordinates; they never lie!

Question

Let the tangent to the parabola $S : y^2 = 2x$ at the point $P(2, 2)$ meet the $x$-axis at $Q$ and normal at it meet the parabola $S$ at the point $R$. Then the area (in sq. units) of the triangle $PQR$ is equal to:

Options:

(A) $\frac{25}{2}$

(B) $\frac{35}{2}$

(C) $\frac{15}{2}$

(D) 25

Sigma Percentile · Accepted Answer

### The Dance of the Parabola: A Journey Through Geometry

Welcome, future engineer. Today, we are not just solving a problem; we are choreographing a dance between a curve and its lines. We are looking at the parabola $S: y^2 = 2x$.

Imagine this curve as a smooth, infinite slide. Our journey begins at a specific point, $P(2, 2)$, where we will draw a tangent and a normal, creating a triangle that holds the secret to our answer.

### Phase 1: The Tangent—The Grazing Line

First, let us ground ourselves. The standard form of a parabola is $y^2 = 4ax$. Comparing this to our equation $y^2 = 2x$, we find that $4a = 2$, which gives us $a = \frac{1}{2}$.

This value of $a$ is the DNA of our parabola; it dictates its shape and focus.

Now, we need the tangent at $P(2, 2)$. The formula for a tangent at any point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.

Substituting our known values, $y_1 = 2$, $x_1 = 2$, and $a = \frac{1}{2}$, we get:

$$y(2) = 2\left(\frac{1}{2}ight)(x + 2)$$

The math simplifies beautifully to $2y = x + 2$. This is our tangent line.

When this line kisses the $x$-axis, we find point $Q$. Setting $y = 0$, we get $0 = x + 2$, so $x = -2$. Thus, $Q$ is at $(-2, 0)$.

### Phase 2: The Normal—The Perpendicular Path

Now, let us pivot to the normal. The normal is the line perpendicular to the tangent at point $P$.

If we rewrite our tangent equation as $y = \frac{1}{2}x + 1$, we see the slope is $\frac{1}{2}$. The slope of the normal, $m_n$, must be the negative reciprocal, so $m_n = -2$.

Using the point-slope form, $y - y_1 = m(x - x_1)$, we write the equation of the normal:

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

Simplifying this, we get $y = 6 - 2x$. This line cuts through the parabola, and we need to find where it lands—point $R$.

### Phase 3: The Intersection—Finding Point $R$

To find $R$, we solve the system of equations: $y^2 = 2x$ and $y = 6 - 2x$. Substituting the normal equation into the parabola, we get:

$$(6 - 2x)^2 = 2x$$

Expanding this, we arrive at $36 + 4x^2 - 24x = 2x$, which simplifies to $4x^2 - 26x + 36 = 0$. Dividing by 2, we get the quadratic:

$$2x^2 - 13x + 18 = 0$$

Factoring this, we find $(2x - 9)(x - 2) = 0$. We know $x = 2$ is point $P$, so $R$ must be at $x = \frac{9}{2}$.

Plugging this back into our normal equation, $y = 6 - 2(\frac{9}{2}) = -3$. So, $R$ is $(\frac{9}{2}, -3)$.

### Phase 4: The Final Area

We have our vertices: $P(2, 2)$, $Q(-2, 0)$, and $R(\frac{9}{2}, -3)$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:

$$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting our values:

$$	ext{Area} = \frac{1}{2} \left| 2(0 - (-3)) + (-2)(-3 - 2) + \frac{9}{2}(2 - 0) ight|$$

This simplifies to:

$$	ext{Area} = \frac{1}{2} |2(3) + (-2)(-5) + 9| = \frac{1}{2} |6 + 10 + 9| = \frac{25}{2}$$

And there it is! The area is **$\frac{25}{2}$ square units**. You have successfully navigated the geometry of the parabola.

Animated Solution for Mathematics - Conic Sections: Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 2 x$

Equation of Tangent at $P (2, 2)$

Simplifying the Tangent Equation

Finding Point $Q$

Slope of the Normal at $P$

Equation of the Normal at $P (2, 2)$

Simplifying the Normal Equation

Finding Intersection Point $R$

Solving the Quadratic Equation

Finding the Coordinates of $R$

Area of Triangle $P QR$

Substituting the Coordinates

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Dance of the Parabola

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=2x

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,2)