Actually x'=kx, y'=hy, then x=x'/k, y = y'/h.

The equation x2-y2-2x = 0: x' 2/k2-y' 2/H2-2x'/k = 0.

X' 2-(k 2/h 2) y' 2-(2k) x' = 0 When the left and right sides of the equation are multiplied by k 2 at the same time.

The coefficients of all terms correspond to the transformed equation one by one, and it can be known that they should be satisfied at the same time: k 2/h 2 = 16, and 2k = 4.

Solve the equation: k=2, h=0.5 (excluding negative values)