Problem:
Let a,b,x, and y be real numbers with a>4 and b>1 such that
a2x2+a2−16y2=b2−1(x−20)2+b2(y−11)2=1
Find the least possible value of a+b.
Solution:
The graph of
a2x2+a2−16y2=1
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 F1G1 and F2G2, 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 .
