# Find the equations of the tangent and the normal at the point ‘t’ on the curve x = $$a sin^3 t$$, y = $$b cos^3 t$$.

## Solution :

We have, x = $$a sin^3 t$$, y = $$b cos^3 t$$

$$\implies$$  $$dx\over dt$$ = $$3a sin^2t cos t$$  and, $$dy\over dt$$  = $$-3b cos^2t sin t$$

$$\therefore$$   $$dy\over dx$$ = $$dy/dt\over dx/dt$$ = $$-b\over a$$ $$cos t\over sin t$$

So, the equation of the tangent at the point ‘t’ is

y – $$b cos^3 t$$ = ($$dy\over dx$$)(x – $$a sin^3 t$$)

or, y – $$b cos^3 t$$ = $$-b\over a$$ $$cos t\over sin t$$(x – $$a sin^3 t$$)

or, bx cos t + ay sin t = ab sin t cos t

The equation of the normal at the point ‘t’ is

y – $$b cos^3 t$$ = $$(-1\over ({dy\over dx})$$(x – $$a sin^3 t$$)

or, y – $$b cos^3 t$$ = $$(-1\over ({-b\over a}{cos t\over sin t})$$(x – $$a sin^3 t$$)

or, ax sin t – by cos t = $$a^2 sin^4 t – b^2 cos^4 t$$

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