<s>[INST] 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
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
 [/INST] This sure looks like a Lagrange multiplier problem to me.  Let $f(x,y) = 3x^2-2xy$ and $g(x,y) = \frac{x^2}{4}-y^2.$  Then $\nabla f = (6x-2y,-2x)$ and $\nabla g = (\frac{x}{2}, -2y)$.  We get the system of 3 equations in three variables:
$$6x-2y = \lambda \frac{x}{2}$$
$$-2x = \lambda (-2y)$$
$$\frac{x^2}{4}-y^2 = 1$$
After some algebraic juggling, there are $8$ solutions:
$$x= \pm\frac{1}{3}\sqrt{18\pm 6\sqrt{10}}, y=  \mp \frac{1}{6}\sqrt{18\pm 6\sqrt{10}}(-3\pm\sqrt{10}),$$
where there are $4$ choices for $x$ and for each one, two corresponding choices for $y$.  
If all $8$ points are plugged into $f$, there are $4$ results: $$ \pm 2\sqrt{10}\pm\frac{20}{3},$$
and the max and min are obvious.
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