# Find the slope of normal to the curve x = 1 – $$a sin\theta$$, y = $$b cos^2\theta$$ at $$\theta$$ = $$\pi\over 2$$.

## Solution :

We have, x = 1 – $$a sin\theta$$, y = $$b cos^2\theta$$

$$\implies$$  $$dx\over d\theta$$ = $$-a cos\theta$$  and $$dy\over d\theta$$ = $$-2b cos\theta sin\theta$$

$$\therefore$$  $$dy\over dx$$ = $$dy/d\theta\over dx/d\theta$$ = $$2b\over a$$ $$sin\theta$$

$$\implies$$  $$dy\over dx$$  at $$\pi\over 2$$ = $$2b\over a$$

Hence, Slope of normal at $$\theta$$ = $$\pi\over 2$$ = $$-a\over 2b$$

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