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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the slope of the normal to the curve \( x = 1 - a sin\theta, y = bcos^2\theta\)$\; at\;$ \( \theta = \Large {\frac{\pi}{2}}\)

$\begin{array}{1 1}(A)\;\frac{2b}{a} \\(B)\;\frac{-2b}{a} \\ (C)\;\frac{-a}{2b} \\ (D)\;\frac{a}{2b} \end{array} $

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  • $\large\frac{1}{\big(\Large\frac{dy}{dx}\big)_P}$=Slope of the normal to $y=f(x)$ at point $P$.
Step 1:
Given : $x=1-a\sin \theta$ and $y=b\cos^2\theta$
Consider $x=1-a\sin \theta$
Differentiating w.r.t $\theta$,we get
$\large\frac{dx}{d\theta}$$=-a\cos \theta$
Step 2:
Consider $y=b\cos^2\theta$
$\large\frac{dy}{dx}$$=2b\cos \theta(-\sin \theta)$
$\quad\;=-2b\sin \theta\cos \theta$
Step 3:
$\large\frac{dy}{dx}=\frac{dy}{d\theta}\times \frac{d\theta}{dx}$
$\qquad=\large\frac{-2b\sin \theta\cos \theta}{a\cos \theta}$
$\qquad=\large\frac{2b}{a}$$\sin \theta$
Step 4:
Slope of the normal at $\theta=\large\frac{\pi}{2}$,
$\large\frac{1}{\big(\Large\frac{dy}{dx}\big)_{\theta=\Large\frac{\pi}{4}}}=\large\frac{-1}{2b/a-\sin \big(\Large\frac{\pi}{2}\big)}$
But $\sin\large\frac{\pi}{2}$$=1$
Therefore slope of the normal is $\large\frac{-1}{2b/a}$
$\Rightarrow \large\frac{-a}{2b}$
answered Jul 10, 2013 by sreemathi.v

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