The tangent to the curve x = a √ c o s 2 θ c o s θ , y = a √ c o s 2 θ s i n θ at the point corresponding to θ = π 6 is Options: A . &nbsp parallel to the x-axis B . &nbsp parallel to the y-axis C . &nbsp parallel to line y = x D . &nbsp 3X-4Y+2=0

Answer: Option A : A d x d θ = − a √ c o s 2 θ s i n θ + − a c o s θ s i n θ √ c o s 2 θ = − a ( c o s 2 θ s i n θ + c o s θ s i n 2 θ ) √ c o s 2 θ = − a s i n 3 θ √ c o s 2 θ = d y d θ = a √ c o s 2 θ c o s θ − a s i n θ s i n 2 θ √ c o s 2 θ = a c o s 3 θ √ c o s 2 θ Hence d y d x = − c o t 3 θ ⇒ d y d x | θ = π 6 = 0 So the tangent to the curve at θ = π 6 is parallel to the x-axis.

The tangent to the curve x = a√cos2θcosθ,y=a√c

Discussion Forum : Application Of Derivatives

Question -

The tangent to the curve x = a $\sqrt{c o s 2 θ} c o s θ, y = a \sqrt{c o s 2 θ} s i n θ$ at the point corresponding to $θ = \frac{π}{6}$ is

Options:

A . parallel to the x-axis

B . parallel to the y-axis

C . parallel to line y = x

D . 3X-4Y+2=0

Answer: Option A
:
A

\frac{d x}{d θ} = - a \sqrt{c o s 2 θ} s i n θ + \frac{- a c o s θ s i n θ}{\sqrt{c o s 2 θ}} = - a \frac{(c o s 2 θ s i n θ + c o s θ s i n 2 θ)}{\sqrt{c o s 2 θ}} = \frac{- a s i n 3 θ}{\sqrt{c o s 2 θ}} = \frac{d y}{d θ} = a \sqrt{c o s 2 θ} c o s θ - a \frac{s i n θ s i n 2 θ}{\sqrt{c o s 2 θ}} = \frac{a c o s 3 θ}{\sqrt{c o s 2 θ}}

Hence

\frac{d y}{d x} = - c o t 3 θ \Rightarrow \frac{d y}{d x} |_{θ = \frac{π}{6}}

= 0
So the tangent to the curve at

θ = \frac{π}{6}

is parallel to the x-axis.

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Discussion Forum : Application Of Derivatives

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