The eccentricity of the hyperbola conjugate t | vedclass

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(C) Given,the equation of the hyperbola is ${x^2} - 3{y^2} = 2x + 8$.Rearranging the terms,we get ${x^2} - 2x - 3{y^2} = 8$.Completing the square for

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$2$

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(C) Given,the equation of the hyperbola is ${x^2} - 3{y^2} = 2x + 8$.Rearranging the terms,we get ${x^2} - 2x - 3{y^2} = 8$.Completing the square for $x$,we have ${(x - 1)^2} - 1 - 3{y^2} = 8$,which simplifies to ${(x - 1)^2} - 3{y^2} = 9$.Dividing by $9$,we get $\frac{{{{(x - 1)}^2}}}{9} - \frac{{{y^2}}}{3} = 1$.The conjugate hyperbola is given by $-\frac{{{{(x - 1)}^2}}}{9} + \frac{{{y^2}}}{3} = 1$,or $\frac{{{y^2}}}{3} - \frac{{{{(x - 1)}^2}}}{9} = 1$.For a hyperbola of the form $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{{{a^2}}}{{{b^2}}}}$.Here,${a^2} = 3$ and ${b^2} = 9$.Therefore,$e = \sqrt{1 + \frac{3}{9}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.Wait,re-evaluating the standard definition: For $\frac{{{y^2}}}{A^2} - \frac{{{x^2}}}{B^2} = 1$,$e = \sqrt{1 + \frac{B^2}{A^2}}$.Here $A^2 = 3, B^2 = 9$,so $e = \sqrt{1 + \frac{9}{3}} = \sqrt{1 + 3} = 2$.

The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is

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