Question

If $x = a \cos \theta + b \sin \theta$, $y = a \sin \theta - b \cos \theta$, then which one is true?

• A. $y^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y = 0$
• B. $y^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0$
• C. $y^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - y = 0$
• D. $y^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - y = 0$

Hint:

For parametric equations, use chain rule carefully: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

Answer 1

The Correct Option is A

Differentiate using chain rule: \[ \frac{dy}{d\theta} = a \cos \theta - b \sin \theta, \frac{dx}{d\theta} = -a \sin \theta + b \cos \theta. \implies \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. \] Then compute $\frac{d^2 y}{dx^2}$ using \[ \frac{d^2 y}{dx^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{1}{dx/d\theta} \cdot \frac{d}{d\theta} \left( \frac{dy}{dx} \right). \] Using parametric differentiation, you get \[ y^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y = 0. \]