Let the eccentricity of the hyperbola x2/a2 - y2/b2 = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then?

Given equation of ellipse can be written as x2/4 + y2/1 = 1
For ellipse x2/a2 + y2/b2 = 1, eccentricity e = √(1 - b2/a2)
Here a2 = 4, b2 = 1
e = √(1 - 1/4) = √3/2
For hyperbola x2/a2 - y2/b2 = 1, eccentricity e' = √(1 + b2/a2)
Given that e' = 1/e = 2/√3
(1 + b2/a2) = 4/3
b2/a2 = 1/3
b2 = a2/3
The foci of the ellipse are (±√3, 0)
Since the hyperbola passes through a focus of the ellipse, it passes through (√3, 0)
Substituting in the hyperbola equation:
3/a2 = 1
a2 = 3
b2 = 1
Therefore, the equation of the hyperbola is x2/3 - y2/1 = 1 or x2 - 3y2 = 3
Eccentricity of hyperbola = √(1 + 1/3) = √4/3 = 2/√3
A focus of hyperbola is (√(a2 + b2), 0) = (√4, 0) = (2, 0)