The distance between the foci of the hyperbol | vedclass

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(C) Given the equation of the hyperbola is $x^2 - 3y^2 - 4x - 6y - 11 = 0$. Rearranging the terms,we get $(x^2 - 4x) - 3(y^2 + 2y) = 11$. Completing t

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$8$

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(C) Given the equation of the hyperbola is $x^2 - 3y^2 - 4x - 6y - 11 = 0$. Rearranging the terms,we get $(x^2 - 4x) - 3(y^2 + 2y) = 11$. Completing the square,we have $(x^2 - 4x + 4) - 3(y^2 + 2y + 1) = 11 + 4 - 3$. This simplifies to $(x - 2)^2 - 3(y + 1)^2 = 12$. Dividing by $12$,we get $\frac{(x - 2)^2}{12} - \frac{(y + 1)^2}{4} = 1$. Here,$a^2 = 12$ and $b^2 = 4$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$. The distance between the foci is $2ae = 2 	imes \sqrt{12} 	imes \frac{2}{\sqrt{3}} = 2 	imes 2\sqrt{3} 	imes \frac{2}{\sqrt{3}} = 8$.

The distance between the foci of the hyperbola $x^2 - 3y^2 - 4x - 6y - 11 = 0$ is

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