The equation of the hyperbola whose foci are | vedclass

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(C) The foci are at $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.Given eccentricity $e = 2$,we have $a(2) = 2$,which implies $a = 1$.For a hyperbola,$b^2 =

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$3x^2 - y^2 = 3$

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(C) The foci are at $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.Given eccentricity $e = 2$,we have $a(2) = 2$,which implies $a = 1$.For a hyperbola,$b^2 = a^2(e^2 - 1)$.Substituting the values,$b^2 = 1^2(2^2 - 1) = 4 - 1 = 3$.The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Substituting $a^2 = 1$ and $b^2 = 3$,we get $\frac{x^2}{1} - \frac{y^2}{3} = 1$.Multiplying by $3$,we get $3x^2 - y^2 = 3$.

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

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