# Find the slope of the normal to the curve x = $$a cos^3\theta$$, y = $$a sin^3\theta$$ at $$\theta$$ = $$\pi\over 4$$.

## Solution :

We have, x = $$a cos^3\theta$$, y = $$a sin^3\theta$$

$$\implies$$  $$dx\over d\theta$$ = $$-3 a cos^2\theta sin\theta$$,  $$dy\over d\theta$$ = $$3 a sin^2\theta cos\theta$$

Now, $$dy\over dx$$  = $$dy/d\theta\over dx/d\theta$$

$$\implies$$  $$dy\over dx$$ = -$$tan\theta$$

$$\therefore$$  Slope of the normal at any point on the curve = $$-1\over dy/dx$$ = $$cot\theta$$

Hence, the slope of the normal at $$\theta$$ = $$\pi\over 4$$ = $$cot\pi\over 4$$ = 1.

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