<s>[INST] Parametric Equations: Finding Given Tangent Lines At Points The parametric equations are: $x=2\cot\theta$ and $y=2\sin^2\theta$
$$\frac{dy}{dx}=-2\sin^3\theta\cdot \cos\theta$$
And the coordinates are: $(-\frac{2}{\sqrt{3}},3/2)$, $(0,2)$, and $(2\sqrt{3}, 1/2)$
I wasn't quite sure what to do with those coordinates, could someone please make sense of what I am to do, and why it should be done that way?
 [/INST] For example, with the point $\,\displaystyle{\left(-\frac{2}{\sqrt 3}\,,\,\frac{3}{2}\right)}\,$ , and with $\,t=\theta\,$ , for simplicity:
$$2\cot t=x=-\frac{2}{\sqrt 3}\Longrightarrow \tan t=-\sqrt 3\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$$
$$\frac{3}{2}=y=2\sin^2t\Longrightarrow \sin t=\pm\frac{\sqrt 3}{2}\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,\,,\,\,k\in\Bbb Z$$
Well, choose one of the infinite ammount of possible $\,t'$s above, say 
$$t=\frac{\pi}{3}\,\,(k=0\,)\Longrightarrow\,\left.\frac{dy}{dx}\right|_{t=\pi/3}=-2\sin^3\frac{\pi}{3}\cos\frac{\pi}{3}=-2\left(\frac{\sqrt 3}{2}\right)^3\frac{1}{2}=-\frac{3\sqrt 3}{8}$$
so the tangent line to the curve at this point is
$$y-\frac{3}{2}=-\frac{3\sqrt 3}{8}\left(x+\frac{2}{\sqrt 3}\right)$$
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