Let P be a variable point on the parabola y = 4x² + 1. Then, the locus of the mid-point of the point

Q: Why is the slope of $PQ$ equal to $-1$?

The line $y=x$ has a slope of $1$. For any line perpendicular to it, the product of the slopes must be $-1$. Therefore, the slope of $PQ$ is $-1$.

Q: Why did we define $Q$ as $(c, c)$?

Since $Q$ lies on the line $y=x$, every point on that line satisfies the condition $y=x$. Thus, if the x-coordinate is $c$, the y-coordinate must also be $c$.

Q: Can we just replace $(h, k)$ with $(x, y)$ at the end?

Yes, because $(h, k)$ represents any arbitrary point on the locus. By replacing them with $(x, y)$, we define the general equation of the curve that the midpoint traces.

Question

Let $P$ be a variable point on the parabola $y = 4x^2 + 1$. Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

Options:

(A) $(3x - y)^2 + (x - 3y) + 2 = 0$

(B) $2(3x - y)^2 + (x - 3y) + 2 = 0$

(C) $(3x - y)^2 + 2(x - 3y) + 2 = 0$

(D) $2(x - 3y)^2 + (3x - y) + 2 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. Before you lies a beautiful, upward-opening parabola, defined by the equation $y = 4x^2 + 1$. It sits gracefully, its vertex at $(0, 1)$, never touching the $x$-axis.

Cutting across this scene is the line $y = x$, a simple, diagonal path stretching infinitely in both directions. Our problem is to track a point $P$ as it travels along this parabola.

From $P$, we drop a perpendicular to the line $y = x$, hitting it at a point $Q$. We are interested in the midpoint $R$ of the segment $PQ$. As $P$ moves, $R$ traces a path—a locus—that we must uncover.

### The Perpendicularity Insight

To begin, let $P$ have coordinates $(x_1, y_1)$. Since $P$ is on the parabola, we know that $y_1 = 4x_1^2 + 1$.

Now, consider the point $Q$, the foot of the perpendicular from $P$ to the line $y = x$. Because $Q$ lies on the line $y = x$, its coordinates must be of the form $(c, c)$.

The line segment $PQ$ is perpendicular to the line $y = x$. Since the slope of $y = x$ is $1$, the slope of our perpendicular segment $PQ$ must be $-1$. Using the slope formula:

$$\frac{y_1 - c}{x_1 - c} = -1$$

Solving this gives us $y_1 - c = -x_1 + c$, which simplifies to $2c = x_1 + y_1$, or:

$$c = \frac{x_1 + y_1}{2}$$

We have now locked $Q$ in terms of $P$.

### The Midpoint Bridge

Now, let $R(h, k)$ be the midpoint of $PQ$. By the midpoint formula:

$$h = \frac{x_1 + c}{2}, \quad k = \frac{y_1 + c}{2}$$

Substituting our expression for $c$, we get:

$$h = \frac{x_1 + \frac{x_1 + y_1}{2}}{2} = \frac{3x_1 + y_1}{4}$$

$$k = \frac{y_1 + \frac{x_1 + y_1}{2}}{2} = \frac{x_1 + 3y_1}{4}$$

These equations connect the coordinates of our moving midpoint $R$ to the original point $P$. To find the locus, we must express $x_1$ and $y_1$ in terms of $h$ and $k$.

Solving the system $3x_1 + y_1 = 4h$ and $x_1 + 3y_1 = 4k$, we find:

$$x_1 = \frac{3h - k}{2}, \quad y_1 = \frac{3k - h}{2}$$

### The Final Transformation

We know $P(x_1, y_1)$ must satisfy the parabola's equation $y_1 = 4x_1^2 + 1$. Substituting our expressions for $x_1$ and $y_1$:

$$\frac{3k - h}{2} = 4 \left( \frac{3h - k}{2} \right)^2 + 1$$

Simplifying this, we have:

$$\frac{3k - h}{2} = (3h - k)^2 + 1$$

Multiplying by $2$, we obtain $3k - h = 2(3h - k)^2 + 2$. Rearranging, we get:

$$2(3h - k)^2 + (h - 3k) + 2 = 0$$

Replacing $(h, k)$ with $(x, y)$, we arrive at the final locus:

**$$2(3x - y)^2 + (x - 3y) + 2 = 0$$**

This equation describes the path of the midpoint, a beautiful result derived from simple geometric constraints.

Animated Solution for Mathematics - Conic Sections: Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

Select Answer:

Visualized Solution

Visualize the Setup

Define Points $P$ and $Q$

Perpendicularity Condition

Equation for Slope of $P Q$

Solve for $c$

Define Midpoint $R$

Midpoint Formula for $R$

Substitute $c$ into $h$ and $k$

Solve for $x_{1}$ and $y_{1}$

Apply the Parabola Constraint

Substitute into Parabola Equation

Simplify the Equation

Final Locus Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Perpendicularity Insight

The Midpoint Bridge

The Final Transformation

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Select Answer:

Visualized Solution

Visualize the Setup

Define Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Perpendicularity Condition

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Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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