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 $\frac{x^2}{4}-\frac{y^2}{12}=1$