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
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$

Let $3x^2-2xy=k$.
Since $x^2=4y^2+4$, we obtain:
$$k=3x^2-2xy\geq3x^2-(x^2+y^2)=2(4+4y^2)-y^2=8+7y^2\geq8.$$
Hence, 
$$3x^2-2xy=k\left(\frac{x^2}{4}-1\right)$$ or
$$\left(3-\frac{k}{4}\right)x^2-2xy+ky^2=0,$$
which gives 
$$1-k\left(3-\frac{k}{4}\right)\geq0$$ or
$$\frac{k^2}{4}-3k+1\geq0,$$ which is
$k\geq2(3+2\sqrt2)$ or $k\leq2(3-2\sqrt2)$ and since $k\geq8$, we obtain:
$$k\geq2(3+2\sqrt2).$$
The equality occurs for $k=2(3+2\sqrt{2})$, $\frac{x^4}{4}-y^2=1$ and $\frac{x}{y}=\frac{1}{3-\frac{2(3+2\sqrt2)}{4}}=2(3+2\sqrt2),$
which says that the equality occurs.
Thus, we got the answer:
$$\min_{\frac{x^2}{4}-y^2=1}(3x^2-2xy)=2(3+2\sqrt2).$$
Done!
