Question

Find the slope of the normal to the curve $x = 1 − a\, sin θ,\, y = b\, cos^2 θ$ at θ=$\frac{π}{2}$.

Answer 1

It is given that x =1−a sin θ and y = b cos2 θ.

$\frac{dx}{dθ}$=-acosθ and $\frac{dy}{dθ}$=2b cosθ(-sinθ)=-2bsinθcosθ

$\frac{dy}{dx}$=($\frac{(\frac{dy}{dθ}) }{ (\frac{dx}{dθ}) }$=$\frac{-2b\, sinθ\,cosθ}{-a\, cosθ}$=$\frac {2b}{b}$ sinθ

Therefore, the slope of the tangent at $θ=\frac{π}{2}$ is given by,

$(\frac{dy}{dx}) \bigg]_{θ=\frac{π}{2}}$=$(\frac {2b}{b}) sinθ\bigg]_{θ=\frac{π}{2}}$= $\frac {2b}{b}$ sin $\frac{π}{2}$=$\frac {2b}{b}$

Hence, the slope of the normal at $θ=\frac{π}{2}$ is given by,

$\frac{1}{\text{slope of the tangent at} θ=\frac{π}{4}}$=-$\frac1{\frac {2b}{b} }$=$\frac {a}{2b}$