<s>[INST] eliminate the parameters Given:
$x = \frac{1}{2} \cos(\theta)$ and $y = 2\sin(\theta)$
Part a) 
solving the first one for theta:
1) multiply both sides by $2$:
$$2x = \cos(\theta)$$
2) divide both sides by $\cos (\arccos)$:
$$\cos^{-1}(2x) = \theta$$
Part b)
Now plugging this into the second equation for theta:
1) plug in theta:
$$y = 2\sin(\cos^{-1}(2x))$$
2) Simplify:
$$y = 2 \sqrt{1-x^2}$$
$$\frac{1}{2} y=\sqrt{1-x^2}$$
3) Squaring both sides gives:
$$\frac{1}{4} y^2 = 1-x^2$$
$$\frac{1}{4} y^2 +x^2 = 1$$
I have laid it out in parts and subsections of each part please refer to the appropriate part and subsection where a mistake was made. One area where I notice there may be a mistake is in part b section 2 where I substitute $2\sin(\cos^{-1}(2x))$ with the square root. $\sqrt{1-x^{2}}$ is for $\sin(\cos^{-1}(x))$ I am not sure how the $2$ in $2\sin$ and the $2$ in $\cos^{-1}(2x)$ affect the substitution.
 [/INST] This is an idea:
$$\begin{cases}x=\frac12\cos\theta\\{}\\y=2\sin\theta\end{cases}\implies \cos^2\theta+\sin^2\theta=4x^2+\left(\frac y2\right)^2=1$$
and you have an ellipse
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