Question

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

Answer 1

Find the slope of the normal to the curve x = 1− a sinθ, y = bcos²θ at θ = π/2

solution :- x = 1 - asinθ
differentiate x with respect to θ,
dx/dθ = 0 - acosθ
dx/dθ = - acosθ -------(1)

y = bcos²θ
differentiate y with respect to θ,
dy/dθ = 2bcosθ(-sinθ)
dy/dθ = -2bsinθ.cosθ ------(2)

dividing equation (2) by (1),
$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{-2bsin\theta cos\theta}{-acos\theta}\\\\\frac{dy}{dx}|_{\theta=\frac{\pi}{2}}=\frac{2bsin\theta}{a}\\\\\frac{dy}{dx}|_{\theta=\frac{\pi}{2}}=\frac{2bsin\frac{\pi}{2}}{a}=\frac{2b}{a}$

hence, slope of tangent = 2b/a
we know, slope of tangent × slope of normal= -1
slope of normal = -1/(slope of tangent)
= -1/(2b/a) = -a/2b

hence, slope of normal = $\frac{-a}{2b}$