Solution :
We have, x = asin3t, y = bcos3t
⟹ dxdt = 3asin2tcost and, dydt = −3bcos2tsint
∴ dydx = dy/dtdx/dt = −ba costsint
So, the equation of the tangent at the point ‘t’ is
y – bcos3t = (dydx)(x – asin3t)
or, y – bcos3t = −ba costsint(x – asin3t)
or, bx cos t + ay sin t = ab sin t cos t
The equation of the normal at the point ‘t’ is
y – bcos3t = (−1(dydx)(x – asin3t)
or, y – bcos3t = (−1(−bacostsint)(x – asin3t)
or, ax sin t – by cos t = a2sin4t–b2cos4t
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