Let A run the first lap at the speed of X and the total length of the runway is Y. According to the meaning of the question, A runs the second lap at the speed of 4X/3, B runs the first lap at 2X/3 and B runs the second lap at 4X/5.

When we meet for the first time, the length of A run is 3Y/5, and the length of B run is 2Y/5.

A finished a lap, B ran 2Y/3, then A turned around and accelerated.

B It takes (Y/3)/(2X/3)=Y/2X to run the remaining distance. In the meantime, A ran away.

(Y/2X)(4X/3)= 2Y/3；

B starts to run the second lap. When they meet, they need (Y/3)/(4X/3)+(4X/5)=5Y/32X, during which B runs (5y/32x) (4x/5) = y/8;

And because the starting distance a of the first meeting is 3Y/5, they all get (3Y/5)-Y/8= 190.

The solution is Y=400, so the length of the elliptical runway is 400 meters.