Question

If the curve represented parametrically by the equations $x=2\ln \cot t+1$ and $y=\tan t+\cot t$

• A. tangent and normal intersect at the point $(2,1)$
• B. normal at $t=\frac{\pi }{4}$ is parallel to the $y$ axis
• C. tangent at $t=\frac{\pi }{4}$ is parallel to the line $y=x$
• D. tangent at $t=\frac{\pi }{4}$ is parallel to the $x$ axis

Answer 1

The correct option is D tangent at $t=\frac{\pi }{4}$ is parallel to the $x$ axis

$x=2\ln \cot t+1$
$\frac{dx}{dt}=-\frac{2cose{c}^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$
$y=\tan t+\cot t$
$\frac{dy}{dt}=se{c}^{2}t-cose{c}^{2}t=\frac{si{n}^{2}t-co{s}^{2}t}{si{n}^{2}tco{s}^{2}t}$
$\Rightarrow \frac{dy}{dx}=-\frac{si{n}^{2}t-co{s}^{2}t}{sin2t}$
$\Rightarrow \frac{dy}{dx}=\cot 2t$
Slope of tangent at $t=\frac{\pi }{4}$ is 0.
Hence, tangent is parallel to x-axis.