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# Find the slope of the normal to the curve x = 1 minus a sin theta , y = b cos theta at theta is equal to pi by 2

6) Find the slope of the normal to the curve $x = 1- a \sin \theta , y = b \cos ^ 2 \theta \: \: at \: \: \theta = \pi /2$

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The slope of the tangent at a point on given curves is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{2}} = -a(\cos \theta)$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{2}} = 2b\cos \theta(-\sin \theta)$
$\left ( \frac{dy}{dx} \right )_{x=\frac{\pi}{2}} = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{-2b\cos \theta \sin \theta}{-a\cos \theta} = \frac{2b\sin \theta}{a} = \frac{2b\times1}{a} = \frac{2b}{a}$
Hence, the slope of the tangent at  $\theta = \frac{\pi}{2}$ is $\frac{2b}{a}$
Now,
Slope of normal =  $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{\frac{2b}{a}} = -\frac{a}{2b}$
Hence, the slope of normal at $\theta = \frac{\pi}{2}$   is  $-\frac{a}{2b}$

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