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

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(C) The given equation of the hyperbola is $4x^2 - 9y^2 = 16$. Dividing by $16$,we get $\frac{x^2}{4} - \frac{y^2}{16/9} = 1$. Comparing this with the

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

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(C) The given equation of the hyperbola is $4x^2 - 9y^2 = 16$. Dividing by $16$,we get $\frac{x^2}{4} - \frac{y^2}{16/9} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = \frac{16}{9}$. Thus,$a = 2$ and $b = \frac{4}{3}$. The eccentricity $e$ is given by the formula $b^2 = a^2(e^2 - 1)$. Substituting the values,$\frac{16}{9} = 4(e^2 - 1)$. $\frac{4}{9} = e^2 - 1$. $e^2 = 1 + \frac{4}{9} = \frac{13}{9}$. Therefore,$e = \frac{\sqrt{13}}{3}$.

The eccentricity of the hyperbola $4x^2 - 9y^2 = 16$ is

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