The transverse axis of a hyperbola is along the x-axis and its length is 2a. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
A
6x2−y2=3a2
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B
x2−3y2=3a2
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C
x2−6y2=3a2
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D
3x2−y2=3a2
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Solution
The correct option is D3x2−y2=3a2 Let e be the eccentricity of hyperbola and length of conjugate axis be 2b. Since, vertex (a,0) bisects the join of centre (0,0) and focus (ae,0). a=ae+02⇒e=2 b2=a2(e2−1)=3a2 ∴ Required hyperbola is x2a2−y2b2=1 ⇒x2a2−y23a2=1⇒3x2−y2=3a2