To eliminate the arbitrary constant in the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can differentiate the equation with respect to $x$.

Differentiation:

Given the equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Differentiate both sides with respect to $x$: $\frac{d}{dx} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{d}{dx} (1)$

The right side is zero since the derivative of a constant is zero: $\frac{d}{dx} \left(\frac{x^2}{a^2}\right) + \frac{d}{dx} \left(\frac{y^2}{b^2}\right) = 0$

Differentiation of each term:

1. $\frac{d}{dx} \left(\frac{x^2}{a^2}\right) = \frac{2x}{a^2}$
2. $\frac{d}{dx} \left(\frac{y^2}{b^2}\right)$ requires the chain rule: $\frac{d}{dx} \left(\frac{y^2}{b^2}\right) = \frac{2y}{b^2} \cdot \frac{dy}{dx}$

Substituting into the equation:

$\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0$

Simplifying:

$\frac{x}{a^2} + \frac{y}{b^2} \cdot \frac{dy}{dx} = 0$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y}$

This differential equation no longer contains the arbitrary constants $a$ and $b$.

Summary:

The differential equation $\frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y}$ is the result of eliminating the arbitrary constants from the original equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Would you like a deeper explanation, or do you have any specific questions?

Relative Questions:

1. What is the significance of eliminating arbitrary constants in an equation?
2. How would you solve the differential equation $\frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y}$?
3. What are some common techniques for eliminating arbitrary constants?
4. Can arbitrary constants always be eliminated from any equation?
5. How do the parameters $a$ and $b$ affect the geometry of the original ellipse equation?

Tip:

When differentiating implicitly, always apply the chain rule carefully, especially when dealing with functions of multiple variables.