Solution :
We have, x = asin3t, y = bcos3t
⟹ dxdt = 3asin2tcost and, dydt = −3bcos2tsint
∴ 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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