TITLE: Is there a way to cut a an ellipsoid with a plane such that it gives a circle?
QUESTION [5 upvotes]: I'm trying to answer this

In $\Bbb {R^3} $ consider the ellipsoid:
  $2x^2+3y^2+4z^2=1$ It exists a subspace of dimension 2 which intersection with the ellipsoid is a circle. Justify any answer.

I know that I must search for a plane since these are the only linear subspaces of $\Bbb {R^3} $ of dimension 2, but I don't know what else to do to check if such a plane exists. any hint would be much appreciated. Thanks!

REPLY [3 votes]: Let $a,b,c$ be any 3 positive numbers such that $0 < a < b < c$. 
Let $\mathcal{E}$ and $\mathcal{S}$ be the ellipsoid and sphere defined by
$$\mathcal{E} = \{ (x,y,z) : ax^2+by^2 + cz^2 = 1 \}
\quad\text{ and }\quad
\mathcal{S} = \{ (x,y,z) : b(x^2+y^2+z^2) = 1 \}$$
For any point $(x,y,z) \in \mathcal{E} \cap \mathcal{S}$, we have
$$ax^2 + by^2 + cz^2 = 1 = b(x^2+y^2+z^2)
\implies (c-b)z^2 = (b-a)x^2
\iff z = \pm k x
$$
where $k = \sqrt{\frac{b-a}{c-b}}$. The RHS is the equation for a pair of planes
$P_{\pm} = \{ (x,y,z) : z = \pm k x \}$.
This implies
$$\mathcal{S} \cap \mathcal{E} \subset \mathcal{P}_{+} \cup \mathcal{P}_{-}$$
Converse, for any point $(x,y,z) \in \mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-})$, a similar argument tell us $ax^2 + by^2 + cz^2 = 1$.
This means we also has
$$\mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-}) \subset \mathcal{E}$$
Combine these two piece of information, we get
$$\mathcal{S} \cap \mathcal{E}
= \mathcal{S} \cap ( \mathcal{P}_{+} \cup \mathcal{P}_{-} )
= ( \mathcal{S} \cap \mathcal{P}_{+} ) \cup (\mathcal{S} \cap \mathcal{P}_{-})$$
Since the intersection of a sphere and a plane is a circle, we find 
$\mathcal{S} \cap \mathcal{E}$ is the union of two circles 
$\mathcal{S} \cap \mathcal{P}_{+}$ and $\mathcal{S} \cap \mathcal{P}_{-}$.
We can repeat essentially the same argument to other combinations of $\mathcal{S}$, $\mathcal{E}$ and $\mathcal{P}_{\pm}$. At the end, we have something like
$$\mathcal{E} \cap \mathcal{P}_{+} = \mathcal{S} \cap \mathcal{P}_{+}
\quad\text{ and }\quad
\mathcal{E} \cap \mathcal{P}_{-} = \mathcal{S} \cap \mathcal{P}_{-}$$
Substitute $(a,b,c)$ by $(2,3,4)$, we find the intersection of
the ellipsoid $$\{ (x,y,z) : 2x^2 + 3y^2 + 4z^2 = 1 \}$$ with the two planes $z = \pm x$ are two circles of radius $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{3}}$.