The eccentricity of the hyperbola which passe | vedclass

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(3,0)$,we have $\frac{3^2}{a^2} - \frac{0^2}

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$\frac{\sqrt{13}}{3}$

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(3,0)$,we have $\frac{3^2}{a^2} - \frac{0^2}{b^2} = 1$,which implies $a^2 = 9$.Now,the equation is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$.Since it passes through $(3\sqrt{2}, 2)$,we substitute these coordinates:$\frac{(3\sqrt{2})^2}{9} - \frac{2^2}{b^2} = 1$$\frac{18}{9} - \frac{4}{b^2} = 1$$2 - \frac{4}{b^2} = 1$$1 = \frac{4}{b^2} \implies b^2 = 4$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.$e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.

The eccentricity of the hyperbola which passes through the points $(3,0)$ and $(3\sqrt{2}, 2)$ is

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