If (8,2) is a point on the hyperbola whose le | vedclass

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given that the length of the transverse axis is $2a = 12$,so $a = 6$

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$\frac{2 \sqrt{2}}{\sqrt{7}}$

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given that the length of the transverse axis is $2a = 12$,so $a = 6$. Since the point $(8, 2)$ lies on the hyperbola,it must satisfy the equation: $\frac{8^2}{a^2} - \frac{2^2}{b^2} = 1$ Substituting $a = 6$: $\frac{64}{36} - \frac{4}{b^2} = 1$ $\frac{16}{9} - 1 = \frac{4}{b^2}$ $\frac{7}{9} = \frac{4}{b^2} \implies b^2 = \frac{36}{7}$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. $e = \sqrt{1 + \frac{36/7}{36}} = \sqrt{1 + \frac{1}{7}} = \sqrt{\frac{8}{7}} = \frac{2 \sqrt{2}}{\sqrt{7}}$.

If $(8,2)$ is a point on the hyperbola whose length of the transverse axis is $12$ and conjugate axis is $x=0$,then the eccentricity of that hyperbola is

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