Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, find $\frac{dy}{dx}$.

Answer 1

Given,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Now differentiating both sides with respect to $x$ both sides we get,

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

or, $\frac{dy}{dx}=-\frac{x{b}^2}{y{a}^2}$