finding t for parametric equation tangent line

Question

Here is the problem I'm trying to solve:

Find the tangent line at the point (0,2)

$$x=2 \, \cot(t)$$ $$y=2 \, \sin^2(t)$$ $$\frac{dy}{dx} = -2 \, > \sin^3(t) \, \cos(t)$$ The tangent line is: $$y-2=m(x-0)$$

$t$ isn't given, I have to find it. I've tried setting $x$ and $y$ equal to each other, and using $\cot^{-1}$ to find what sets $x$ equal to $0$, but I get the wrong number.

How do I find t? A photo is attached here:

Should I set $x = y$ or use $\cot^{-1}$ on x? How do I find $t$ given $x$ and $y$ so I can get the tangent line equation?

Answer 1

For a parametric curve, a directing vector of the tangent line at a point $(x(t_0),y(t_0))$ is the first non-zero derived vector. Here the point $(0,2)$ is obtained for $t_0=\dfrac\pi2$, hence a directing vector for the tangent line is: $$\Bigl(-\frac2{\sin^2t_0},4\sin t_0\cos t_0\Bigr)=(-2,0).$$

Hence the equation of the tangent line is $\;y=2$.

Answer 2

At the point $(0,2)$, $\;\;x=2\cot t=0\implies\cos t=0$, so

$\;\;\;\displaystyle\frac{dy}{dx}=-2\sin^3 t\cos t =0$ and therefore $y=2$ is the tangent line.