For the curve represented parametrically by the equation $x = 2 \log (\cot t) + 1$ and $y = \tan t + \cot t$ , then

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Normal at $t = \frac{π}{4}$ is parallel to Y-axis

Tangent at $t = \frac{π}{4}$ is parallel to X-axis

Normal at $t = \frac{π}{4}$ is parallel to y=x

Tangent at $t = \frac{π}{4}$ is parallel to y = x

answer is A, B.

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Detailed Solution

$\begin{array}{r} \frac{d y}{d x} = (\frac{\sec^{2} t - {cosec}^{2} t}{\frac{2}{\cot t} (- cosec^{2} t)}) = 0 \\ \Rightarrow t = \frac{π}{4} \end{array}$

Normal at $t = \frac{π}{4}$ is parallel to Y-axis

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