Gujarati
Hindi
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

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