Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane.
Here is what I did.
Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\frac25=1$ to solve for $x^2$. I got $x^2=1.8$ and then got $y=0$. How should I parametrize the intersection from here onward?
$\mathbf Error$ in my calculation, $y^2=\frac 35-\frac{x^2}{3}$

You started out correctly, by plugging $z = 2$ into the equation and finding $\frac{x^2}{3} + y^2 + \frac{2}{5} = 1$, or
$$\frac{x^2}{3} + y^2 = \frac{3}{5}.$$
The next step is to put the equation in the standard form for an ellipse. (Recall that the standard form is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.)
\begin{align}
\frac{5}{3} \cdot \frac{x^2}{3} + \frac{5}{3} \cdot y^2 &= \frac{5}{3} \cdot \frac{3}{5} \\
\frac{x^2}{\frac{9}{5}} + \frac{y^2}{\frac{3}{5}} &= 1
\end{align}
Here we have $h = k = 0$, $a = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$, and $b = \sqrt{\frac{3}{5}}$. The next step is to parametrize the ellipse, and recall that the parametrization for the $z$ coordinate is $z(t) = 2$.
