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Question

Find the slope of the normal to the curve x=1-a sin θ,y=b cos2 θ at θ=π2

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Solution

It is given that x=1-a sin θandy=bcos2θ

On differentiating x and y both w.r.t. θ, we get

dxdθ=ddθ[1a sin θ]=a cos θ and

dydθ=ddθ[b cos2 θ]=2b cos θ(sin θ)=2b cosθ sin θ

dydx=dydθdxdθ=2b cos θ sin θa cos θ=2basin θ

Slope of normal at the point θ=π2,dxdy=1dy/dx

1(dydx)(θ=π/2)=12basin(π2)=a2b


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