Finding cartesian equation from parametric trigonometric equations

Question

I'm trying to find the cartesian equation of the curve which is defined parametrically by:

$$ x = 2\sin\theta, y = \cos^2\theta $$

Both approaches I take result in the same answer:

$$ y = 1 - \sin^2\theta\\ \sin \theta = \sqrt{y-1} \\ x = 2\sqrt{y-1} \\ x^2 = 4(y- 1) \\ x^2 + 4 = 4y $$ Method 2: $$ \sin^2 = y - 1 \\ \sin\theta = \frac{x}{2} \\ \sin^2\theta = \frac{x^2}{4} \\ x^2 + 4 = 4y $$

But the answer listed is $x^2 + 4y^2 = 4$. Are my calculations wrong?

Answer 1

You have a mistake in your calculation. Note that $$\sin^2\theta\not=y-1$$ and that $$\sin^2\theta=1-y.$$

By the way, I think you have a typo in your question.

If $x=2\sin\theta,y=\cos^2\theta$ are correct, then since $$\sin^2\theta=\left(\frac x2\right)^2,\ \ \cos^2\theta=y,$$ we have $$\cos^2\theta+\sin^2\theta=1\Rightarrow y+\left(\frac{x}{2}\right)^2=1\Rightarrow y=-\frac{x^2}{4}+1.$$

Note that this is not $x^2+4y^2=4$.

Answer 2

Note that $\frac{x}{2}=\sin \theta.$ Now,

$$1=\sin^2\theta+\cos^2\theta=\left(\frac{x}{2}\right)^2+y,$$

from where

$$x^2+4y=4.$$

Note that $y=\cos^2\theta \le 1.$ So it doesn't makes sense $\sqrt{y-1}$ unless $y=1$ (or you are working on complex numbers, what is not the case).

Answer 3

Answer: Squring each equation and multiplying the last on by 4 we get $$ \begin{align} x^2&=4sin^2\theta\\ 4y^2&=4cos^2\theta\\ \end{align} $$ Adding the above two equations we get $$ \begin{align} x^2+4y^2&=4(cos^2\theta+sin^2\theta)\\ &=4 \end{align} $$ Where we used the identity $cos^2\theta+sin^2\theta=1$.