True of False

The trace of x^2+2y^2 +10z^2=5 in the plane z=2 is an ellipse. True or False?

First I substituted z=2 in the equation and got x^2+2y^2=-35. I say False because x^2+2y^2=-35 is not an ellipse. But what is it if it is not any ellipse?

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- Nov 2nd 2009, 01:28 PMandreaspohleTrue or False traces of functions
True of False

The trace of x^2+2y^2 +10z^2=5 in the plane z=2 is an ellipse. True or False?

First I substituted z=2 in the equation and got x^2+2y^2=-35. I say False because x^2+2y^2=-35 is not an ellipse. But what is it if it is not any ellipse? - Nov 2nd 2009, 03:02 PMshawsend
You can put it into the form $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ which is an ellipsoid right. What then do the values a, b, c represent and where is z=2 in all of that?

- Nov 2nd 2009, 03:21 PMandreaspohle
so a b c are the lengths in the x y z directions. and when z=2 it will be a trace of an ellipse?

- Nov 2nd 2009, 03:54 PMshawsend
No. I get:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{5/2}+\frac{z^2}{1/2}=1$

That's an ellipsoid that crosses over the axes at $\displaystyle x\pm \sqrt{5}$, $\displaystyle y=\pm\sqrt{5/2}$ and more importantly, at $\displaystyle z=\pm \sqrt{1/2}$ which means the ellipsoid has a z-height of $\displaystyle 1/\sqrt{2}$. What then does that mean when we ask, where does the ellipsoid cut through the plane z=2 then? No where right? So false. But you can see that since the equation $\displaystyle x^2+2y^2=-35$ has no real solution.