Find \( \frac{dy}{dx} \), Given: \[ x = a \cos t \quad \text{and} \quad y = b \sin t \] Differentiate \( x \) and \( y \) with respect to \( t \): \[ \frac{dx}{dt} = -a \sin t \quad \text{and} \quad \frac{dy}{dt} = b \cos t \] Using the chain rule, \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a} \cot t \] Find \( \frac{d^2y}{dx^2} \) Differentiate \( \frac{dy}{dx} \) with respect to \( t \): \[ \frac{d}{dt}\left(-\frac{b}{a} \cot t\right) = -\frac{b}{a} (-\csc^2 t) = \frac{b}{a} \csc^2 t \] Using the chain rule again: \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{b}{a} \csc^2 t}{-a \sin t} = -\frac{b}{a^2 \sin^3 t} \] Express in terms of \( x \) and \( y \) From \( x = a \cos t \) and \( y = b \sin t \): \[ \cos t = \frac{x}{a}, \quad \sin t = \frac{y}{b}, \quad \text{and} \quad \csc^2 t = \frac{1}{\sin^2 t} = \frac{b^2}{y^2} \] Substitute \( \sin t = \frac{y}{b} \): \[ \frac{d^2y}{dx^2} = -\frac{b}{a^2 \left(\frac{y}{b}\right)^3} = -\frac{b^4}{a^2 y^3} \]