Solution :
We have, x = acos3θ, y = asin3θ
⟹ dxdθ = −3acos2θsinθ, dydθ = 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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