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Question: The foci of a hyperbola coincide with the foci of the ellipse x<sup>2</sup>/25 + y<sup>2</sup>/9 = 1...

The foci of a hyperbola coincide with the foci of the ellipse x2/25 + y2/9 = 1. The equation of the hyperbola if its eccentricity is 2, is

A

x2/4 – y2/12 = 1

B

x2/12 – y2/4 = 1

C

x2/12 – y2/16 = 1

D

None of these

Answer

x2/4 – y2/12 = 1

Explanation

Solution

Foci of ellipse x2/25 + y2/9 = 1

are (± 4, 0) so foci of hyperbola are also (± 4, 0)

Ž ae = 4 Ž a × 2 = 4 Ž a = 2 …(i)

b2 = a2 (e2 – 1) = 4(4 – 1) Ž b2 = 12 ….(ii)

Equation x2/a2 – y2/b2 = 1 Ž x2/4 – y2/12 = 1