# Find the slope of the normal to the curve

Question:

Find the slope of the normal to the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$.

Solution:

It is given that $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$.

$\therefore \frac{d x}{d \theta}=3 a \cos ^{2} \theta(-\sin \theta)=-3 a \cos ^{2} \theta \sin \theta$

$\frac{d y}{d \theta}=3 a \sin ^{2} \theta(\cos \theta)$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}=-\frac{\sin \theta}{\cos \theta}=-\tan \theta$

Therefore, the slope of the tangent at $\theta=\frac{\pi}{4}$ is given by,

$\left.\left.\frac{d y}{d x}\right]_{\theta=\frac{\pi}{4}}=-\tan \theta\right]_{\theta=\frac{\pi}{4}}=-\tan \frac{\pi}{4}=-1$

Hence, the slope of the normal at $\theta=\frac{\pi}{4}$ is given by,

$\frac{1}{\text { slope of the tangent at } \theta=\frac{\pi}{4}}=\frac{-1}{-1}=1$

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