Question

Find $\frac{dy}{dx}$, when

$x=b{\sin }^2\mathrm{\theta }\mathrm{and}y=a{\cos }^2\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x=b{\sin }^2\theta \mathrm{and}y=a{\cos }^2\theta \\ \therefore \frac{dx}{d\theta }=\frac{d}{d\theta }\left(b{\sin }^2\theta \right)=2b\sin \theta \cos \theta \\ \mathrm{and}, \\ \frac{dy}{d\theta }=\frac{d}{d\theta }\left(a{\cos }^2\theta \right)=-2a\cos \theta \sin \theta \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{-2a\cos \theta \sin \theta }{2b\sin \theta \cos \theta }=-\frac{a}{b} \end{gathered}$$