Solve this

Question:

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

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

Solution:

as $x=b \sin ^{2} \theta$

Then

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}\left(\mathrm{b} \sin ^{2} \theta\right)}{\mathrm{d} \theta}=2 \mathrm{~b} \sin \theta \cos \theta$

And $y=a \cos ^{2} \theta$

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{d}\left(\operatorname{acos}^{2} \theta\right)=-2 \operatorname{acos} \theta \sin \theta$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=-\frac{2 \mathrm{a} \cos \theta \sin \theta}{2 \mathrm{~b} \sin \theta \cos \theta}=-\frac{\mathrm{a}}{\mathrm{b}}$

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