Key Concepts and Formulas

This problem requires a strong understanding of coordinate geometry, differential calculus, and the properties of conic sections, specifically parabolas. We will use these concepts to translate a geometric condition into a differential equation, solve it, and then identify the resulting curve.

1. Equation of a Tangent Line: For a curve $y = f(x)$, the equation of the tangent at a specific point $P(x_1, y_1)$ is given by the point-slope form: $Y - y_1 = m(X - x_1)$ where $m = \frac{dy}{dx}\Big|_{(x_1, y_1)}$ is the slope of the tangent at that point. Here, $(X, Y)$ are the general coordinates of any point on the tangent line, distinct from the fixed point $P(x_1, y_1)$ on the curve.

2. Midpoint Formula: The coordinates of the midpoint $M$ of a line segment with endpoints $A(x_A, y_A)$ and $B(x_B, y_B)$ are: $M = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right)$

3. Condition for a Point on the y-axis: A point lies on the y-axis if and only if its x-coordinate is zero.

4. Solving Differential Equations by Separation of Variables: A first-order differential equation where variables can be separated, typically of the form $\frac{dy}{dx} = g(x)h(y)$ or $\frac{dx}{dy} = g(x)h(y)$, can be solved by rearranging terms to get all $x$ terms with $dx$ and all $y$ terms with $dy$ on opposite sides, and then integrating both sides. For example, $\frac{1}{h(y)} \, dy = g(x) \, dx$.

5. Standard Form and Properties of a Parabola:

   • The equation $y^2 = 4ax$ represents a parabola with its vertex at the origin $(0,0)$. Its axis of symmetry is the x-axis.
     • Its focus is at $(a, 0)$.
     • The length of its latus rectum is $|4a|$.
   • Similarly, $x^2 = 4ay$ represents a parabola with its vertex at the origin $(0,0)$. Its axis of symmetry is the y-axis.
     • Its focus is at $(0, a)$.
     • The length of its latus rectum is $|4a|$.

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Step-by-Step Solution

1. Define the Point P and Initial Setup

• Why this step? The problem describes a curve $C$ and a property related to its tangent at an arbitrary point. To derive a general differential equation that describes this curve, we must first define an arbitrary point on it.
• Let $P(x_1, y_1)$ be an arbitrary point on the curve $C$.
• Our goal is to find a relationship between $x_1$, $y_1$, and $\frac{dy}{dx}\Big|_{(x_1, y_1)}$ based on the given condition. Once this relationship is found, we will generalize it by replacing $(x_1, y_1)$ with $(x, y)$ to obtain the differential equation of the curve.

2. Formulate the Equation of the Tangent Line at P

• Why this step? The problem explicitly mentions the tangent to the curve, so the first crucial step is to write down its equation. This equation will allow us to find the coordinates of point $Q$.
• The slope of the tangent to the curve $C$ at point $P(x_1, y_1)$ is given by the derivative of $y$ with respect to $x$, evaluated at $P$. Let's denote this slope as $m = \frac{dy}{dx}\Big|_{(x_1, y_1)}$.
• Using the point-slope form, the equation of the tangent line to the curve $C$ at $P(x_1, y_1)$ is: $Y - y_1 = m(X - x_1)$ $Y - y_1 = \frac{dy}{dx}\Big|_{(x_1, y_1)} (X - x_1)$ Here, $(X, Y)$ represent the running coordinates of any point on the tangent line, while $(x_1, y_1)$ is the fixed point of tangency on the curve.

3. Find the Coordinates of Point Q (x-intercept of the Tangent)

• Why this step? Point $Q$ is explicitly mentioned as an endpoint of the segment $PQ$. To apply the midpoint formula later, we need the coordinates of $Q$.
• Point $Q$ is where the tangent line meets the x-axis. Any point on the x-axis has a y-coordinate of $0$. Let the x-coordinate of $Q$ be $X_Q$. So, $Q = (X_Q, 0)$.
• Substitute $Y=0$ into the tangent equation: $0 - y_1 = \frac{dy}{dx}\Big|_{(x_1, y_1)} (X_Q - x_1)$
• Now, we solve for $X_Q$. To make the algebra simpler, especially when the derivative is in the denominator, it's often convenient to think of $\frac{dy}{dx}$ as the reciprocal of $\frac{dx}{dy}$. $-y_1 = \frac{1}{\frac{dx}{dy}\Big|_{(x_1, y_1)}} (X_Q - x_1)$ $X_Q - x_1 = -y_1 \left( \frac{dx}{dy}\Big|_{(x_1, y_1)} \right)$ $X_Q = x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)}$
• Therefore, the coordinates of point $Q$ are $\left( x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)}, 0 \right)$.

4. Apply the Midpoint Condition

• Why this step? This is the core geometric condition provided in the problem statement. Applying it will allow us to establish a relationship between $x_1$, $y_1$, and their derivatives, leading directly to the differential equation of the curve.
• We are given that the y-axis bisects the segment $PQ$. This means the midpoint of $PQ$ lies on the y-axis. For a point to be on the y-axis, its x-coordinate must be zero.
• Let $P = (x_1, y_1)$ and $Q = \left( x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)}, 0 \right)$.
• Using the midpoint formula, the coordinates of the midpoint $M$ of $PQ$ are: $M = \left( \frac{x_1 + \left( x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)} \right)}{2}, \frac{y_1 + 0}{2} \right)$ $M = \left( \frac{2x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)}}{2}, \frac{y_1}{2} \right)$
• Since $M$ lies on the y-axis, its x-coordinate must be 0: $\frac{2x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)}}{2} = 0$ $2x_1 - y_1 \frac{dx}{dy}\Big|_{(x_1, y_1)} = 0$
• Now, we generalize this for any point $(x, y)$ on the curve $C$. This gives us the differential equation that describes the curve: $2x - y \frac{dx}{dy} = 0$

5. Solve the Differential Equation

• Why this step? The differential equation defines the curve implicitly. Solving it will yield the explicit algebraic equation of the curve $C$.
• We have the differential equation: $2x - y \frac{dx}{dy} = 0$
• Rearrange the equation to separate the variables $x$ and $y$. We want all terms involving $x$ with $dx$ on one side, and all terms involving $y$ with $dy$ on the other side. $2x = y \frac{dx}{dy}$ Assuming $x \neq 0$ and $y \neq 0$ (these are typically considered boundary cases that don't affect the general solution derived from integration): $\frac{2}{y} \, dy = \frac{1}{x} \, dx$
• Now, integrate both sides: $\int \frac{2}{y} \, dy = \int \frac{1}{x} \, dx$ $2 \ln|y| = \ln|x| + \ln|C_{int}|$
  • Explanation: We use $\ln|C_{int}|$ as the constant of integration.