is an ellipse centered at (0,0) with major axis parallel to the x-axis of length 2a. The distance from the center to the foci is a2−(a2−16)​=4, so the foci are F1​=(−4,0) and F2​=(4,0).\
Similarly, the graph of
b2−1(x−20)2​+b2(y−11)2​=1
is an ellipse centered at (20,11) with major axis parallel to the y-axis of length 2b. The distance from the center to the foci is b2−(b2−1)​=1, so the foci are G1​=(20,10) and G2​=(20,12).
Let P(x,y) be a point that lies on both ellipses, so by a property of ellipses,
2a=PF1​+PF2​ and 2b=PG1​+PG2​
Adding these equations and using the triangle inequality yields
Thus a+b≥23. Equality is achieved by running this argument in reverse: that is, by taking P to be the intersection of F1​G1​​ and F2​G2​​, and then setting a=21​(PF1​+PF2​) and b=21​(PG1​+PG2​). In this case, a=16,b=7, and P=(14,7.5). Therefore the least possible value of a+b is 23​ .
