If the foci of ellipse x^2/16 + y^2/7 = 1 coincide with the foci of hyperbola x^2/144 - y^2/α = 1/125 then the eccentricity of hyperbola is: ​

Question

If the foci of ellipse \(\frac{x^2}{16} + \frac{y^2}7 = 1\) coincide with the foci of hyperbola \(\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac1{125}\) then the eccentricity of hyperbola is:

(1) \(\frac{5\sqrt5}{3}\)

(2) \(\frac{5\sqrt5}{4}\)

(3) \(\frac{5\sqrt5}{6}\)

(4) \(\frac{5\sqrt7}{4}\)

Answer 1

Correct option is (2) \(\frac{5\sqrt5}{4}\)