WBJEE 2012 Math Solved Paper

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+0

2

⇒e=2
b2=a2(e2−1)=3a2
∴ Required hyperbola is ‌

x2

a2

−‌

y2

b2

=1
⇒‌

x2

a2

−‌

y2

3a2

=1⇒3x2−y2=3a2