A) \[\frac{a}{2b}\]
B) \[\frac{2a}{b}\]
C) \[\frac{a}{b}\]
D) \[\frac{-a}{2b}\]
Correct Answer: D
Solution :
Given, \[x=1-a\sin \theta \]and \[y=b{{\cos }^{2}}\theta \] On differentiating w. r. t. \[\theta \], we get \[\frac{dx}{d\theta }=-a\cos \theta \] and \[\frac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\text{=2b cos }\!\!\theta\!\!\text{ (-sin }\!\!\theta\!\!\text{ )}\] Then, \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{2b}{a}\sin \theta \] \[\therefore \] Slope of normal at the point \[\theta =\frac{\pi }{2}\] is \[-\frac{dx}{dy}=-\frac{1}{dy/dx}\] \[=-\frac{1}{\frac{2b}{a}\sin \left( \frac{\pi }{2} \right)}=-\frac{a}{2b}\]
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