A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent t

Q: Wait, why is the x-coordinate of the midpoint zero? I thought the y-axis was just a line.

That is a great observation! The y-axis is indeed a line, but it is defined by the equation $X = 0$. If the y-axis bisects the segment $PQ$, then the midpoint of $PQ$ must lie on this line. Therefore, the x-coordinate of the midpoint must be zero. This is a classic geometric condition that translates directly into a powerful algebraic constraint.

Q: How do we know the curve is a parabola? We just solved a differential equation.

Excellent question! When we solved the differential equation $\frac{dy}{y} = \frac{dx}{2x}$, we arrived at $\ln|y| = \frac{1}{2}\ln|x| + C$, which simplifies to $y = c\sqrt{x}$. Squaring both sides gives $y^2 = kx$. This is the standard form of a parabola opening along the positive x-axis. The differential equation naturally led us to this specific family of curves.

Q: Is there a trap with the focus? I got $a = 3/4$, but option C says $(4/3, 0)$.

You have spotted the trap perfectly! The standard form of a parabola is $y^2 = 4ax$. In our case, $y^2 = 3x$, so $4a = 3$, which means $a = 3/4$. The focus is at $(a, 0)$, which is $(3/4, 0)$. Option C is indeed a distractor designed to catch students who might accidentally invert the fraction or miscalculate the focus. Always double-check your standard form!

Question

A particle is moving in the xy-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the x-axis at $Q$. If the y-axis bisects the segment $PQ$, then $C$ is a parabola with

Options:

(A) length of latus rectum 3

(B) length of latus rectum 6

(C) focus $(\frac{4}{3}, 0)$

(D) focus $(0, \frac{3}{4})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the $xy$-plane, watching a particle trace an invisible path. We are given a curve $C$ that passes through the point $(3, 3)$.

At any point $P(x, y)$ on this curve, a tangent is drawn, and it meets the $x$-axis at a point $Q$. The problem states that the $y$-axis acts as a perfect mirror, bisecting the segment $PQ$. This geometric constraint dictates the very shape of the curve.

### The Tangent and the Intercept

For any point $P(x, y)$ on the curve, the slope of the tangent is given by the derivative $y' = \frac{dy}{dx}$. The equation of this tangent line, using the point-slope form, is:

$$Y - y = y'(X - x)$$

Here, $(X, Y)$ are the coordinates of any point on the line. We are interested in the point $Q$ where this tangent meets the $x$-axis. At the $x$-axis, the $y$-coordinate is zero, so we set $Y = 0$.

This gives us $-y = y'(X - x)$. Solving for $X$, we find:

$$X = x - \frac{y}{y'}$$

Thus, the coordinates of $Q$ are $\left(x - \frac{y}{y'}, 0\right)$.

### The Midpoint Magic

Now, we apply the bisection property. The $y$-axis bisects the segment $PQ$, meaning the midpoint of $PQ$ lies on the $y$-axis. The $x$-coordinate of the midpoint is the average of the $x$-coordinates of $P$ and $Q$.

Mathematically, this is:

$$\frac{x_P + x_Q}{2} = 0$$

Substituting $x_P = x$ and $x_Q = x - \frac{y}{y'}$, we get:

$$\frac{x + (x - \frac{y}{y'})}{2} = 0$$

Simplifying this, we obtain $2x - \frac{y}{y'} = 0$, or:

$$2x = \frac{y}{y'}$$

This is the differential equation that governs our curve.

### Solving the Differential Equation

We can rewrite $y'$ as $\frac{dy}{dx}$, giving us $2x = y \cdot \frac{dx}{dy}$. Rearranging to separate the variables, we get:

$$\frac{dy}{y} = \frac{dx}{2x}$$

Integrating both sides, we have:

$$\int \frac{1}{y} dy = \int \frac{1}{2x} dx$$

This yields $\ln|y| = \frac{1}{2}\ln|x| + C$. By setting $C = \ln c$, we can combine the logarithms to get $\ln y = \ln(c\sqrt{x})$, which simplifies to $y = c\sqrt{x}$. Squaring both sides, we find the family of curves:

$$y^2 = kx$$

where $k = c^2$. This is indeed a parabola.

### The Final Reveal

We know the curve passes through $(3, 3)$. Substituting these values into $y^2 = kx$, we get:

$$3^2 = k(3) \Rightarrow 9 = 3k \Rightarrow k = 3$$

The equation of our curve is **$y^2 = 3x$**.

Comparing this to the standard form $y^2 = 4ax$, we see that $4a = 3$. The length of the latus rectum is $3$, and the focus is at $(3/4, 0)$. We have successfully navigated the geometry and calculus to reveal the true nature of the curve.

Animated Solution for Mathematics - Differential Equations: A particle is moving in the xy-plane along a curve $C$ passing through the point $(3, 3)$ . The tangent to the curve $C$ at the point $P$ meets the x-axis at $Q$ . If the y-axis bisects the segment $P Q$ , then $C$ is a parabola with

Select Answer:

Visualized Solution

Visualizing the Curve and Point $P$

Equation of the Tangent at $P$

Finding Point $Q$ on the $x$ -axis

Coordinates of Point $Q$

The Bisection Property

Applying the Midpoint Formula

Forming the Differential Equation

Separating the Variables

Integrating Both Sides

Simplifying the Equation

Finding the Constant $k$

The Final Equation of the Curve

Analyzing the Parabola

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Tangent and the Intercept

The Midpoint Magic

Solving the Differential Equation

The Final Reveal

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