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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}

 

Posted by

Gautam harsolia

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