Question

For the curve represented parametrically by the equations, $x=2ln\cot \left(t\right)+1$ & $y=\tan \left(t\right)+\cot \left(t\right)$

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

Answer 1

Answer: (A), (B)

The correct options are
A normal at $t=\pi /4$ is parallel to y - axis
B tangent at $t=\pi /4$ is parallel to x - axis

$x=2lncot\left(t\right)+1$

$\frac{dx}{dt}=\frac{2}{cot\left(t\right)}(-cose{c}^{2}t)=\frac{-2}{si{n}^2\left(t\right)cot\left(t\right)}$

${\left(\frac{dx}{dt}\right)}_{t=\frac{\pi }{4}}=-4$

$y=tan\left(t\right)+cot\left(t\right)$

$\frac{dy}{dt}=se{c}^{2}t-cose{c}^{2}t=\frac{si{n}^{2}t-co{s}^{2}t}{si{n}^{2}tco{s}^{2}t}$

${\left(\frac{dy}{dt}\right)}_{t=\frac{\pi }{4}}=0$

$\Rightarrow \frac{dy}{dx}=0$

Slope of tangent $=0$

$\Rightarrow$ Tangent is parallel to x-axis.

Slope of normal $=-\frac{dx}{dy}=-\infty$

$\Rightarrow$ Normal is parallel to y-axis.