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,

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