Find curve at the intersection of two level surfaces. Let C be the curve at the intersection of two level surfaces M(x, y, z) = 5 and N(x,y,z) = 0, passing through a point P (1,1,1). Let $M(x, y, z) = 2x^2 + y^2 + 2z^2$ and $N(x, y, z) = xy-z$. Find curve C in the parametric form <x(t), y(t), z(t)>.
I am quite stuck on this one. I could let $z = xy$ and then plugging that into M I'd have $2x^2 + y^2 + 2x^2y^2 = 5$, and I have no idea what would be good ways to parametrize that.
Another way is to let $y=z/x$, in which case I end up with $2x^4 + (2z^2-5)x^2+z^2 = 0$, but again, I'm quite stuck. Any help would be really appreciated!
Or, if it is impossible to find the curve C, the question I'm working on actually asks for the tangent line to curve C at P if x'(0) = 3. If this makes things easier, how exactly is using this tangent more useful?
 A: It often helps to visualize the situation:

First, eliminate $z$ from your two equations (quite simple... just look at your second equation and plug $z$ into your first equation). Solve that first equation to get $y(x) = \pm \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}}$.  (It is quadratic, so you'll get a + and a - solution.)
Now you have your three-dimensional parameterization:  $(x, y(x), z(x,y)) = (x, \pm \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}}, \pm x \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}})$.
Note especially that this curve includes the point $(1,1,1)$, as you can verify by simple substitutions of $x=1$.
(Note that $-2 \leq x \leq 2$.)
A: Eliminate $2z^2.$ We have $5-2x^2-y^2=2x^2y^2\Rightarrow y^2=\frac{5-2x^2}{1+2x^2}.$ We take the positive square root since in a neighborhood of $(1,1,1),\ y>0.$  Then we may take $(t, \sqrt{\frac{5-2t^2}{1+2t^2}},t\sqrt{\frac{5-2t^2}{1+2t^2}}).$ This paramterization is valid for $t\in [-\sqrt {5/2},\sqrt {5/2}].$
A: In the equation $2x^2 + y^2 + 2 x^2 y^2 = 5$  put $x = r \cos t$, $y = \sqrt{2} \,r \sin t$, get $r = \frac{\sqrt{5} }{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } }$, and  the parametrization
$$\left( \frac{\sqrt{5} \cos t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } , \frac{\sqrt{10}\sin t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } \right)$$
Note that $t= \tan^{-1} (\frac{1}{\sqrt{2}})$ maps to $(1,1)$.
