Show this set forms an ellipse in R2 I'm trying to show that $\{ (x, y) : \|x\vec v +y \vec w \| = 1 \}$ in $\mathbb{R}^2$, with $\vec v,\vec w$ elements of a real inner product space, is the equation of an ellipse centered at $\vec 0$. 
Knowing that $x^2{/}a+y^2/b =1$ is the equation of an ellipse centered at $\vec 0$, I'm guessing I need to show that all pairs in the above set satisfy this equation for fixed scalars $a$ and $b$.
I've tried simplifying $\|x\vec v +y \vec w \|$ and arrived at $x^2\|v\|^2+2xy\langle v,w\rangle +y^2\|w\|^2$. Is there a trick to factoring this that I'm overlooking? Any nudges are appreciated.
 A: You are forgetting about rotated ellipses. The equation 
$$
ax^2+b\,xy+cy^2=1,
$$
where $a=\|v\|^2$, $b=2\langle v,w\rangle$, $c=\|w\|^2$, is the equation of an ellipse, provided that $b^2\leq4ac$, which in your case is given precisely by the Cauchy-Schwarz Inequality. 
A: You have the equation
$$x^2\|{\bf v}\|^2+2xy\langle {\bf v},{\bf w}\rangle+y^2\|{\bf w}\|^2=1$$
which would be in standard form for an ellipse if it were not for the $xy$ term.  This should give you the hint that diagonalisation is the way to go.  The equation can be written as
$${\bf x}^TA{\bf x}=1$$
where
$$A=\pmatrix{\langle {\bf v},{\bf v}\rangle&\langle {\bf v},{\bf w}\rangle\cr
             \langle {\bf v},{\bf w}\rangle&\langle {\bf w},{\bf w}\rangle\cr}\ .$$
For an ellipse, this matrix needs to have two positive eigenvalues.  The characteristic equation is
$$\lambda^2-\bigl[\langle {\bf v},{\bf v}\rangle+\langle {\bf w},{\bf w}\rangle\bigr]\lambda
  +\bigl[\langle {\bf v},{\bf v}\rangle\langle {\bf w},{\bf w}\rangle-\langle {\bf v},{\bf w}\rangle^2\bigr]=0\ .$$
If you can explain why both terms in square brackets are positive, you should be able to finish the question.  Good luck!
Comment.  To be completely precise, $A$ could have a zero eigenvalue for certain ${\bf v}$ and ${\bf w}$.  This will give degenerate cases.
