# Thread: Parametric Equation of Parabola

1. ## Parametric Equation of Parabola

Hello, just a question.

Find the cartesian equation of the curves whose parametric equations are:

a) x = 2cos(theta), y = 2sin(theta), 0 ≤ theta ≤ pi

the answer is: y = sqrt( 4 - x^2), -2 ≤ x ≤ 2

I understand how to do it up to x^2 + y^2 = 4, and I can arrive at the answer, but I do not know WHY the solution is y = sqrt( 4 - x^2), -2 ≤ x ≤ 2. Something to do with the domain, but I dont understand.

Thanks.

2. Originally Posted by noobonastick
I understand how to do it up to x^2 + y^2 = 4, and I can arrive at the answer, but I do not know WHY the solution is y = sqrt( 4 - x^2), -2 ≤ x ≤ 2. Something to do with the domain, but I dont understand.
I am not too sure which part you are having trouble with, so I will go through my own solution.

$\displaystyle x=2\cos\theta,\;y=2\sin\theta\quad0\leq\theta\leq\ pi$

We have

$\displaystyle x^2+y^2=(2\cos\theta)^2+(2\sin\theta)^2=4\cos^2\th eta+4\sin^2\theta$ (just substituting for $\displaystyle x$ and $\displaystyle y$)

$\displaystyle \Rightarrow x^2+y^2=4.$ (Pythagorean identity)

But note that $\displaystyle y$ must be nonnegative (since $\displaystyle 2\sin\theta$ is nonnegative when $\displaystyle 0\leq\theta\leq\pi$). Similarly, $\displaystyle -2\leq x\leq2.$ So, we can solve for $\displaystyle y,$ but we must throw out the negative root as it is not part of the original curve. So we get

$\displaystyle y^2=4-x^2$

$\displaystyle \Rightarrow y=\pm\sqrt{4-x^2}$

$\displaystyle \Rightarrow y=\sqrt{4-x^2}$ (because $\displaystyle y\geq0$)

$\displaystyle y=\sqrt{4-x^2},\quad-2\leq x\leq2.$
After you got $\displaystyle x^2+ y^2= 4$, you need to recognise that $\displaystyle y= sin(\theta)$ is non-negative for $\displaystyle 0\le \theta\le \pi$ and do you can solve for y taking the positive root.