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 $x=2(u+v)$ and $y = u-v$. Then $\frac{x^2}{4}-y^2 = 1 \Rightarrow uv = \frac{1}{4}$
Now $3x^2 - 2xy = 8u^2+16v^2+24uv = 8\left(u^2+\frac{1}{8u^2} \right)+6 \ge 4\sqrt 2+6$ by AM-GM Inequality
