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Find the slope of the normal to the curve x = acos3θ, y = asin3θ at θ = π4.

Solution :

We have, x = acos3θ, y = asin3θ

  dxdθ = 3acos2θsinθdydθ = 3asin2θcosθ

Now, dydx  = dy/dθdx/dθ

  dydx = -tanθ

  Slope of the normal at any point on the curve = 1dy/dx = cotθ

Hence, the slope of the normal at θ = π4 = cotπ4 = 1.


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