Solution :
For ellipse e = \(4\over 5\), so foci = (\(\pm\)4, 0)
for hyperbola e = 2, so a = \(ae\over e\) = \(4\over 2\) = 2, b = \(2\sqrt{4-1}\) = 2\(\sqrt{3}\)
Hence the equation of the hyperbola is \(x^2\over 4\) โ \(y^2\over 12\) = 1
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