If the eccentricity of the hyperbola frac{x^2 | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} =

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$18$

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(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$,so $a = 3$. The eccentricity $e$ is given by $e = \frac{\sqrt{13}}{3}$. We know that $e^2 = 1 + \frac{b^2}{a^2}$. Substituting the values: $(\frac{\sqrt{13}}{3})^2 = 1 + \frac{b^2}{9} \implies \frac{13}{9} = 1 + \frac{b^2}{9}$. Subtracting $1$ from both sides: $\frac{4}{9} = \frac{b^2}{9}$,which gives $b^2 = 4$. The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Since the hyperbola passes through $(k, 2)$,we substitute $x = k$ and $y = 2$: $\frac{k^2}{9} - \frac{2^2}{4} = 1$. $\frac{k^2}{9} - 1 = 1 \implies \frac{k^2}{9} = 2$. Therefore,$k^2 = 18$.

If the eccentricity of the hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ passing through the point $(k, 2)$ is $\frac{\sqrt{13}}{3}$,then the value of $k^2$ is:

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