Parametric to Implicit ( {x(t),y(t)} --> P(x,y) == 0 ) I have this parametric equations:
$x(\theta) = r Cos(\theta) - \frac{v_{0}^2Cos(\theta)Sin(\theta)}{g}$
$y(\theta) = \frac{v_{0}^{2}Cos^2(\theta)}{2g} + r Sin(\theta)$
This is for $\theta \in (-\frac{\pi}{2},\frac{\pi}{2})$
The last part of the exercise I am doing, ask me to find a relation between the previous $x(\theta)$ and $y(\theta)$ and specifically tells me that it exist a 4th degree polynomial $P(x,y)$ so that the previous x and y satisfy $P(x(\theta),y(\theta)) = 0$ $\forall \theta \in (-\frac{\pi}{2},\frac{\pi}{2})$
I have tried using Solve with mathematica with no good results, and I don't really know how to face this problem.
 A: In the second equation you can use $\cos(\theta)^2=1-\sin(\theta)^2$ and get a quadratic in $\sin(\theta)$, solve that for $\sin(\theta)$ in terms of $y(\theta)$. Sub your expression for $\sin(\theta)$ into the equation for $x(\theta)$ and also use the fact that $\cos \left( \theta \right) =\sqrt {1- \sin(\theta)^2}$ over your domain. This will give you a relation between $x(\theta)$ and $y(\theta)$ involving square roots. Square that, isolate the remaining square root on one side and square again to get a quartic in $x(\theta),y(\theta)$. It seems to work, let me know if it goes wrong. For comparison I got: 
$-16\,{g}^{2}{v_{{0}}}^{8} y ^{4}+16\,
{v_{{0}}}^{6}g \left( {r}^{2}{g}^{2}+{v_{{0}}}^{4} \right)  y ^{3}-4\,{v_{{0}}}^{4} \left( {v_{{0}}}^{8}+
2\,{v_{{0}}}^{4}{g}^{2} x^{2}-2\,{r}
^{2}{g}^{2}{v_{{0}}}^{4}+{r}^{4}{g}^{4} \right)  y ^{2}+ \left( 4\,{v_{{0}}}^{6}g \left( {v_{{0}}}^{4}+
3\,{r}^{2}{g}^{2} \right)  x-16
\,{r}^{2}g{v_{{0}}}^{6} \left( {r}^{2}{g}^{2}+{v_{{0}}}^{4} \right) 
 \right) y -{g}^{2}{v_{{0}}}^{8} x ^{4}-4\, \left( {r}^{2}{g}^{2}+{v_{{0}}}^{4}
 \right) {r}^{2}{g}^{2}{v_{{0}}}^{4} x^{2}+4\,{r}^{2}{v_{{0}}}^{4} \left( {r}^{2}{g}^{2}+{v_{{0}}}^
{4} \right) ^{2}=0$
Which I checked is true by subbing in the original equations; hopefully I have copied it across correctly.
