Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$ Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$
Prove that:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$

$36 + 9y^2 = 4x^2$ Suggests we could use $y = 2 \tan A$ to get $x = \pm 3\sec A$.  Given the objective, it is sufficient to consider $x = 3 \sec A, ~~ A \in [0, \frac{\pi}{2})$
Then $(2x - y - 4 )^2 = 4(3 \sec A -  \tan A - 2)^2 $
It is easily checked that the minimum occurs when $3 \tan A = \sec A$ or when $\sin A = \frac{1}{3}$. So the minimum value is $4(3 \sec A -  \tan A - 2)^2 = 4(3\frac{3}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}-2)^2 = \left(4\sqrt{2}-4\right)^2$
