The foci of a hyperbola coincide with the foci of the ellipse x 2 /25 + y 2 /9 = 1 . If…
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The foci of a hyperbola coincide with the foci of the ellipse x2/25 + y2/9 = 1 . If eccentricity of the hyperbola is 2, then its equation is :
- x2 – 3y2 – 12 = 0
- 3x2 – y2 – 12 = 0
- x2 – y2 – 4 = 0
- none of these
Answer
(B) 3x 2 – y 2 – 12 = 0
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