Different methods for finding the minimum of $|x-2y|$ when $x^2+1=2y^2$.

Question

For $x, y \in \Bbb R$, $x^2 + 1 = 2y^2$, find the minimum of $|x - 2y|$.

At a glance I found that the point $(x, y)$ lies on a hyperbola and $|x - 2y|$ is just the distance between the point and the line $x - 2y = 0$ times $\sqrt 5$. Now this problem can be solved by using derivatives.

My question is: are there any other solutions? I want to find some other beautiful solutions to this interesting problem. All methods are accepted as long as they are rigorous.

Answer 1

Calculations give us $$(x-2y)^2=2(x-y)^2+1 \ge 1$$

Thus we have $$|x-2y| \ge 1$$ With the equality holding when $x=y=\pm 1$.

Answer 2

We have $$2y^2-x^2=1$$

WLOG $\sqrt2y=\sec2t,x=\tan2t$

$$u=x-2y=\dfrac{\sin2t-\sqrt2}{\cos2t}$$

$$\sqrt2=\sin2t-u\cos2t\iff\sqrt{\dfrac2{u^2+1}}=\sin\left(2t-\arccos\dfrac1{1+u^2}\right)\le1$$

$$u^2+1\ge2\iff|u|\ge1$$