jmc

Question

The foci of the ellipse x^216 + y^2b^2 = 1 and the foci of the hyperbola x^2144 - y^281 = 125coincide. Find b^2.

Accepted Answer

We can write the equation of the hyperbola as

\frac{x ^{2}}{144/25} - \frac{y ^{2}}{81/25} = 1,

so for the hyperbola,

a = \frac{12}{5}

and

b = \frac{9}{5} .

Then

c = a^{2} + b^{2} = \frac{144}{25} + \frac{81}{25} = 3.

Thus, the foci are at

(\pm 3, 0) .

Then for the ellipse,

a^{2} = 16,

so

b^{2} = a^{2} - c^{2} = 16 - 9 = 7 .