$A=\{(x,y) : x = u+v,y = v,u^2+v^2 ≤ 1\}$ Derive the length of the longest line segment that can be enclosed inside the region $A$. 
QUESTION: Let $A$ be the region in the $xy$-plane given by $$A=\{(x,y) : x = u+v,y = v,u^2+v^2 ≤ 1\}$$Derive the length of the longest line segment that can be enclosed inside the region $A$.


MY APPROACH: Firstly, I didn't quite get the meaning of the question. I mean, we are considering the rectangular coordinates with normal meaning of symbols. So $x$ stands for the $x$ axis and $y$ stands for the $y$ axis. In that case, what are $u$ and $v$ here?
I tried to formulate an equation in $x$ and $y$. This is what I found out-
$$x^2+2y^2-2xy-1≤0$$ Now I do not know how to proceed from here.. does $u$ and $v$ stand for a separate coordinate system altogether? If so, how?

Pardon me, this question has been asked before.  But the solution was too brief and incomplete. I didn't get that.

Any help will be much appreciated. Thank you so much.
 A: The region is described by the inequality you wrote in terms of $x,y$. However the formulation in terms of $u,v$ helps to see it as a circle whose points have been transformed. 
The farthest points in the new region will originally have been diametrically opposite points on the circle, so they'd be of the form $(u,v)$ and $(-u,-v)$. The corresponding points for the region $A$ are $(u+v,v)$ and $(-u-v,-v)$. 
If the distance between them is $D$, then $D^2=4[(u+v)^2+v^2]$ which we want to maximize subject to $u^2+v^2=1$. I haven't checked but I'd guess this happens for $u=0,v=1$, giving $D=2\sqrt{2}$.
A: $x^2 + 2y^2 - 2xy -1 = 0$ is some curve, and the region is enclosed in this curve. Certainly our two point  in $A$ should be at the boundary to maximize their distance. 
Let $(a, b)$ and $(c, d)$ be two points. Then under condition $$a^2 + 2b^2 - 2ab - 1 = c^2 + 2d^2 - 2cd -1 =0$$ we have to maximize $M = (a-b)^2 + (c-d)^2$. 
By adding two condition, we have $a^2 + c^2 + 2(b^2 + d^2) - 2(ab + cd) = 2$ or $$a^2 + b^2 + c^2 + d^2 - 2(ab+cd) = 2 + b^2 + d^2 .$$
Note that the LHS is $M$. So we have $M = 2+ b^2 + d^2$. This is maximized where $b^2$ and $d^2$ are maximized. Since $(x-y)^2 + y^2 \le 1$, we have $|y| =\pm1$ attains the max value of $y^2$ with $x = y$. i.e. $(a, b) = (-1, 1)$, $(c, d) = (1, -1)$. The distance is maximum and is $2\sqrt{2}$. 
Here is a plot of the region. It is an ellipse. 

