Question

the hyperbola is (x-5)^2/64 - (y+6)^2/b^2 = 1 find b such that the eccentricity is 5/4. how would i find b?

Answer 1

If a is half the length of the major axis and b is half the length of the minor axis then the eccentricity is given by:
\[\epsilon=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{1-\frac{b^2}{a^2}}\]
So plug in epsilon, b, and a then solve for b.

Answer 2

i thought eccentricity is e = c/a ...making 5/4 c = 5 a = 4 ?