Question

Find the slope of the normal to the curve x = acos^3 θ, y = asin^3θ at θ=π/4

Answer 1

you should know ,
slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal × slope of tangent = -1
but we know, slope of tangent of curve y = f(x) at a point (a,b) is 1st order derivatives at that given point. e.g., $\textbf{slope of tangent}=\frac{dy}{dx}|_{(a,b)}$
so, slope of normal = -1/slope of tangent = $\bf{-\frac{1}{\frac{dy}{dx}|_{(a,b)}}}$

so, let's start to solve :
x = acos^3 θ
differentiate with respect to θ
dx/dθ = 3acos²θ.(-sinθ) = -3acos²θ.sinθ---(1)

similarly, differentiate y with respect to θ
dy/dθ = 3asin²θ.cosθ -----(2)

dividing equations (2) by (1),
dy/dx = {3asin²θ.cosθ}/{-3acos²θ.sinθ} = -tanθ
at, θ = π/4 , dy/dx = -tanπ/4 = -1

so, slope of normal = -1/{dy/dx} = -1/-1 = 1
hence,slope of normal of the curve is 1