Question

For a hyperbola, the foci are at ($\pm$ 4, 0) and vertices at ($\pm$ 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. $\frac{x^2}{4}-\frac{y^2}{16}=1$

Answer 1

Answer: (A)

Foci are $\left(\pm 4, 0\right)$, vertices are $\left(\pm 2, 0\right)$
$ \therefore 4 = 2e $
$\Rightarrow e=2$ and $a=2 $
But $b^{2} = a^{2}\left(e^{2}-1\right)$
$ = 4\left(4-1\right) = 12$
$\therefore $ hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$