Question

The equation of the hyperbola whose foci are $(-2,0)$ and $(2,0)$ and eccentricity is $2$ is given by:

• A. $x^2-3y^2=3$
• B. $3x^2-y^2=3$
• C. $-x^2+3y^2=3$
• D. $-3x^2+y^2=3$

Answer 1

The correct option is B $3x^2-y^2=3$

Given $e=2$

We have $s{s}^{\prime }=2ae$

$\Rightarrow \sqrt{(2-(-2){)}^2+0}=2a\times 2$

$\Rightarrow 4=4a$

$\therefore a=1$

We have $e^2=1+\frac{b^2}{a^2}$

$\Rightarrow 4=1+\frac{b^2}{1}$

$\Rightarrow 4-1=b^2$ or $b^2=3$ or $b=\sqrt{3}$

Centre is the midpoint of s and s'

$c(\frac{2-2}{2},\frac{0-0}{2})$

$\Rightarrow c(0,0)$

$\therefore$ Equation of hyperbola is

$\frac{(x-0{)}^2}{a^2}-\frac{(y-0{)}^2}{(\sqrt{3}{)}^2}=1$

(or)

$\frac{x^2}{1}-\frac{y^2}{b}=1$ (or) $3x^2-y^2=3$

is the required equation of hyperbola.