Question
The tangent to the curve x = a√cos2θcosθ,y=a√cos2θsinθ at the point corresponding to θ=π6 is
Answer: Option A
:
A
dxdθ=−a√cos2θsinθ+−acosθsinθ√cos2θ=−a(cos2θsinθ+cosθsin2θ)√cos2θ=−asin3θ√cos2θ=dydθ=a√cos2θcosθ−asinθsin2θ√cos2θ=acos3θ√cos2θ
Hence dydx=−cot3θ⇒dydx|θ=π6 = 0
So the tangent to the curve at θ=π6 is parallel to the x-axis.
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:
A
dxdθ=−a√cos2θsinθ+−acosθsinθ√cos2θ=−a(cos2θsinθ+cosθsin2θ)√cos2θ=−asin3θ√cos2θ=dydθ=a√cos2θcosθ−asinθsin2θ√cos2θ=acos3θ√cos2θ
Hence dydx=−cot3θ⇒dydx|θ=π6 = 0
So the tangent to the curve at θ=π6 is parallel to the x-axis.
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