The equation of the hyperbola having its ecce | vedclass

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(B) Given: Eccentricity $e = 2$ and distance between foci = $2ae = 8$.Since $2ae = 8$ and $e = 2$,we have $2a(2) = 8$,which implies $4a = 8$,so $a = 2

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

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(B) Given: Eccentricity $e = 2$ and distance between foci = $2ae = 8$.Since $2ae = 8$ and $e = 2$,we have $2a(2) = 8$,which implies $4a = 8$,so $a = 2$.Thus,$a^2 = 4$.For a hyperbola,the relation between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.Substituting the values: $b^2 = 4(2^2 - 1) = 4(4 - 1) = 4(3) = 12$.The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Substituting $a^2 = 4$ and $b^2 = 12$,we get $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

The equation of the hyperbola having its eccentricity $e = 2$ and the distance between its foci as $8$ is:

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