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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?

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    I honestly can't see what the problem is working with t directly, but if you want $\,x,y\,$ is just a very little algebra and trigonometry: $\frac{dy}{dx}=-2\sin^3t\cos t=-2\sin^4t\cot t=-\frac{xy^2}{4}$2012-11-27

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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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    I didn't have to: I just wrote as an example the simplest one. This, of course, may depend heavily on the given curve's domain or whatever.2012-11-27