The eccentricity of the hyperbola 4x^2 - 9y^2 | vedclass

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(C) Given the equation of the hyperbola: $4x^2 - 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Comparing this w

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

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(C) Given the equation of the hyperbola: $4x^2 - 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$. Thus,$a = 3$ and $b = 2$. The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}}$. Substituting the values: $e = \sqrt{\frac{9 + 4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.

The eccentricity of the hyperbola $4x^2 - 9y^2 = 36$ is

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