Question

Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos^2 θ at θ=π/2.

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 = $\bf{-\frac{1}{\frac{dy}{dx}|_{(a,b)}}}$

x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)

dividing equations (2) by (1),

dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b

Answer 2

Answer:

Step-by-step explanation: