For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

Q: For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

(A) $\frac{x^2}{4} – \frac{y^2}{12} = 1$

(B) $\frac{x^2}{12} – \frac{y^2}{4} = 1$

(C) $\frac{x^2}{16} – \frac{y^2}{4} = 1$

(D) $\frac{x^2}{4} – \frac{y^2}{16} = 1$

Sol. ae = 4, a = 2

⇒ e = 2

But b² = a²(e² – 1)

⇒ b² = 12

Hyperbola is $\frac{x^2}{4} – \frac{y^2}{12} = 1$

Hence (A) is the correct answer.