eliminate the parameters

Question

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.

Answer 1

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

Answer 2

Easy solution:

$$\cos\theta=2x,\\\sin\theta=\frac y2.$$

Then

$$(2x)^2+\left(\frac y2\right)=\cos^2\theta+\sin^2\theta=1.$$

This is the equation of a centered, axis-aligned ellipse of axis lenghts $1$ and $4$.