Midpoint locus

If the end-points of one of the line segments are $(u,u)$ and $(v,2v)$ , then we must have $2u^2 - 6uv + 5v^2 \; = \; (u - v)^2 + (u - 2v)^2 \; = \; 4^2 \; = \; 16$ and the midpoint of this line-segment has coordinates $(x,y)$ where $x \; = \; \tfrac12(u+v) \hspace{1cm} y \; = \; \tfrac12(u+2v)$ But then $\begin{aligned} 25x^2 - 36xy + 13y^2 & = \; \tfrac14\big[25(u+v)^2 - 36(u+v)(u+2v) + 13(u+2v)^2\big] \; = \; \tfrac14\big[(25 - 36 + 13)u^2 + (50 - 108 + 52)uv + (25 - 72 + 52)v^2\big] \\ & = \; \tfrac14(2u^2 - 6uv + 5v^2) \; = \; 4 \end{aligned}$ so the midpoints of all these lines lie on the ellipse $\boxed{25x^2 - 36xy + 13y^2 = 4}$ .