Find the slope of the normal to the curve
Question:

Find the slope of the normal to the curve $x=1-a \sin \theta, y=b \cos ^{2} \theta$ at $\theta=\frac{\pi}{2}$.

Solution:

It is given that $x=1-a \sin \theta$ and $y=b \cos ^{2} \theta$.

$\therefore \frac{d x}{d \theta}=-a \cos \theta$ and $\frac{d y}{d \theta}=2 b \cos \theta(-\sin \theta)=-2 b \sin \theta \cos \theta$

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

Therefore, the slope of the tangent at $\theta=\frac{\pi}{2}$ is given by,

$\left.\left.\frac{d y}{d x}\right]_{\theta-\frac{\pi}{2}}=\frac{2 b}{a} \sin \theta\right]_{\theta-\frac{\pi}{2}}=\frac{2 b}{a} \sin \frac{\pi}{2}=\frac{2 b}{a}$

Hence, the slope of the normal at $\theta=\frac{\pi}{2}$ is given by,

$\frac{1}{\text { slope of the tangent at } \theta=\frac{\pi}{4}}=\frac{-1}{\left(\frac{2 b}{a}\right)}=-\frac{a}{2 b}$

 

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