Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the 
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$

Rearranging B. Goddard's system of equations:
$$\begin{align}12x-4y-\lambda x=0\end{align}\tag1$$
$$\begin{align}x=\lambda y\end{align}\tag2$$
$$\begin{align}x^2-4y^2=4\end{align}\tag3$$
Inserting $(2)$ into $(1)$ yields
$$\begin{align}12\lambda y - 4y - \lambda^2y=0\end{align}$$
$\Leftrightarrow$
$$\begin{align}(\lambda^2-12\lambda + 4)y=0\end{align}.$$
The two solutions to the quadratic equation are 
$$\begin{align}\lambda_1=-4\sqrt 2 + 6\end{align}$$
and
$$\begin{align}\lambda_2=+4\sqrt 2 + 6\end{align}.$$
Inserting $(2)$ into $(3)$ gives us
$$\begin{align}(\lambda^2-4)y^2=4\end{align}.$$
For $\lambda_1$ with
$$\begin{align}\lambda_1^2=-48\sqrt 2+68\end{align}$$
and
$$\begin{align}\lambda_1^2-4=16(-3\sqrt 2+4)\end{align}$$
so that
$$\begin{align}\dfrac{4}{\lambda_1^2-4}=\dfrac{1}{4(-3\sqrt 2+4)}=\dfrac{3\sqrt 2+4}{4\cdot-2}\end{align}$$
it follows that
$$\begin{align}-8y^2=+3\sqrt{2}+4\end{align}.$$
Unfortunately this is not solvable over the reals.
Likewise for $\lambda_2$ we have
$$\begin{align}-8y^2=-3\sqrt{2}+4\end{align}$$
which is solvable over the reals since $3\sqrt{2}\geq 4$.
The minimum is 
$$\begin{align}3x^2-2xy=4\sqrt 2 + 6\end{align}.$$
