<s>[INST] 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, $y^2=\frac 35-\frac{x^2}{3}$
 [/INST] I'm afraid you have make something wrong.Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. You should get $y^2=\frac{3}{5}-\frac{x^2}{3}$,  rather than  $y^2=9-5x^2$ this is the equation of the curve while $z=2$
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