Parametrization of this curve $$ z = x^2 + \frac{y^2}{4}$$
I found out that the level curves of this function at z = 1 is an ellipse with major axis length $2$ and minor axis length $2\sqrt{2}$.
How would I parametrize this curve in the form
$x = x(t), y = y(t)$ ?
 A: $x(t) = \cos(t)$
$y(t) = 2\sin(t)$
This fulfills $(\forall t) \ 1 = x(t)^2 + y(t)^2/4$. You could also flip the $\sin$ and $\cos$, which changes the inital coordinates $x(0)$ and $y(0)$ but achieves the same thing.

To answer filip's question where the trigonometric functions come from, let me explain it with the equation that every point $(x,y)$ on a circle (to keep things simple) fulfills:
$$x^2 + y^2 = r^2$$
$$\left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1$$
Geometrically, the fractions relate to the cosine and sine of the angle of our point $(x,y)$ on the circle (intuition tells us that a point on a circle is best described by an angle). This is how trigonometric functions are usually defined in school:
$$\frac{x}{r} = \cos(\phi) = \frac{x}{\sqrt{x^2+y^2}}$$
$$\frac{y}{r} = \sin(\phi) = \frac{y}{\sqrt{x^2+y^2}}$$
And of course, this still fulfills the equation
$$\cos^2(\phi) + \sin^2(\phi) = 1$$
That's why circle and ellipses equations are closely related to trigonometric functions. Not the best explanation but I hope it helps.
A: Ellipses of the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
are typically parameterised by the equations
$$x(t)=a\sin t,\, y(t)=b\cos t.$$
This works, because
$$\frac{(a\sin t)^2}{a^2}+\frac{(b\cos t)^2}{b^2}=\sin^2 t+\cos^2 t=1.$$
In your case, the parameterisation would be $x(t)=\sin t,\,y(t)=2\cos t$.
