Math Help - Help with cartesian equation

1. Help with cartesian equation

x = 2 cot t
y = 2 sin^2 t

I need to find the cartesian equation in the form y=f(x). Can somebody pls help

2. Hi

Use x² = cot²t = cos²t / sin²t = (1-sin²t)/sin²t
substitute sin²t=y/2
and finally rearrange to get y=f(x)

3. thanks you're such a lifesaver

4. Hello, threerose!

Here's another approach.
. . It should surprise/impress/terrify your professor.

$\begin{array}{c}x \:=\: 2\cot\theta \\
y \:=\: 2\sin^2\theta \end{array}$

Find the cartesian equation in the form $y\,=\,f(x)$

We have: . $\begin{array}{cccccccc}
x \;=\;2\cot\theta &\Longrightarrow& \cot\theta \;=\; \dfrac{x}{2} &\Longrightarrow& \cot^2\theta \;=\; \dfrac{x^2}{4} & [1] \\ \\[-3mm]
y \;=\; 2\sin^2\theta &\Longrightarrow& y \;=\; \dfrac{2}{\csc^2\theta} &\Longrightarrow & \csc^2\theta \;=\; \dfrac{2}{y} & [2] \end{array}$

$\text{Subtract }[2]\text{ - }[1]\!:\;\;\underbrace{\csc^2\!\theta - \cot^2\!\theta}_{\text{This is 1}} \;=\;\frac{2}{y} - \frac{x^2}{4}$

. . So we have: . $1 \:=\:\frac{2}{y} - \frac{x^2}{4}$

Multiply by $4y\!:\;\;4y \;=\;8 - x^2y \quad\Rightarrow\quad x^2y+4y \:=\:8 \quad\Rightarrow\quad y(x^2+4)\:=\:8$

Therefore: . $y \;=\;\frac{8}{x^2+4}$