Question

The differential equation corresponding to the family of ellipses \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \), where 'a' is an arbitrary constant is

• A. \( xy \frac{dy}{dx} = 4-y^2 \)
• B. \( xy \frac{dy}{dx} = 4-x^2 \)
• C. \( xy \frac{dy}{dx} = x^2-4 \)
• D. \( xy \frac{dy}{dx} = y^2-4 \)

Hint:

To form a differential equation from a family of curves with 'n' arbitrary constants, you need to differentiate the original equation 'n' times. Then, you will have a system of n+1 equations from which you can eliminate the 'n' constants.

Answer 1

The Correct Option is D

We start with the equation of the family of ellipses:
\( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \).
To form a differential equation, we need to eliminate the arbitrary constant 'a'.
First, differentiate the equation with respect to x.
\( \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \).
\( \frac{2x}{a^2} + \frac{y}{2} \frac{dy}{dx} = 0 \).
Now, we solve this for \( \frac{1}{a^2} \).
\( \frac{2x}{a^2} = -\frac{y}{2} \frac{dy}{dx} \implies \frac{1}{a^2} = -\frac{y}{4x} \frac{dy}{dx} \).
Substitute this expression for \( \frac{1}{a^2} \) back into the original equation.
\( x^2 \left( -\frac{y}{4x} \frac{dy}{dx} \right) + \frac{y^2}{4} = 1 \).
\( -\frac{xy}{4} \frac{dy}{dx} + \frac{y^2}{4} = 1 \).
Multiply the entire equation by 4 to clear the denominators.
\( -xy \frac{dy}{dx} + y^2 = 4 \).
Rearrange the terms to match the options.
\( y^2 - 4 = xy \frac{dy}{dx} \).
This is the required differential equation. It matches option (D).