Question

For the curve represented parametrically by the equations, $x=\frac{2}{cott}+1$ & $y=tant+cott$

• A. tangent at $t=\frac{\pi }{4}$ is parallel to x - axis
• B. normal at $t=\frac{\pi }{4}$ is parallel to y - axis
• C. tangent at $t=\frac{\pi }{4}$ 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 tangent at $t=\frac{\pi }{4}$ is parallel to x - axis
B normal at $t=\frac{\pi }{4}$ is parallel to y - axis

$dy=se{c}^{2}t-cose{c}^{2}t.dt$ ...(i)
$dx=2se{c}^{2}t.dt$ ...(ii)

$\frac{dy}{dx}$$=\frac{se{c}^{2}t-cose{c}^{2}t.dt}{2se{c}^{2}t.dt}$
$=\frac{se{c}^{2}t-cose{c}^{2}t}{2se{c}^{2}t}$.

Hence
${\frac{dy}{dx}}_{t=\frac{\pi }{4}}$ $=0$ slope of tangent

Slope of normal
$=\frac{-dx}{dy}$
$=\frac{-2se{c}^{2}t}{se{c}^{2}t-cose{c}^{2}t}$

Hence $={\frac{-dx}{dy}}_{t=\frac{\pi }{4}}=\infty$

Hence the normal is perpendicular to x axis and thus parallel to y axis.

and tangent is parallel to x - axis.