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10-2. Parabola, Ellipse, Hyperbola
normal
The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-
A
$-3x^2 + y^2 = 3$
B
$x^2 -3y^2 = 3$
C
$3x^2 -y^2 = 3$
D
$-x^2 + 3y^2 = 3$
Solution
Given, ae $=2$
$\Rightarrow 2 \mathrm{a}=2$
or $a=1$
Also, $\mathrm{e}=\sqrt{\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{a}^{2}}}$
$\therefore 4=1+b^{2}$
$\mathrm{b}=\pm \sqrt{3}$
Regd. equation; $\frac{x^{2}}{1}-\frac{y^{2}}{3}=1$
$ \Rightarrow \boxed{3{x^2} – {y^2} = 3}$
Standard 11
Mathematics