The foci of the hyperbola 4x^2 - 9y^2 - 36 = | vedclass

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(A) The given equation of the hyperbola is $4x^2 - 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{4x^2}{36} - \frac{9y^2}{36} = 1$ $\frac{x^2}

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$(\pm \sqrt{13}, 0)$

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(A) The given equation of the hyperbola is $4x^2 - 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{4x^2}{36} - \frac{9y^2}{36} = 1$ $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$. Here,$a = 3$ and $b = 2$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$. The coordinates of the foci are $(\pm ae, 0)$. Substituting the values,we get $(\pm 3 	imes \frac{\sqrt{13}}{3}, 0) = (\pm \sqrt{13}, 0)$.

The foci of the hyperbola $4x^2 - 9y^2 - 36 = 0$ are:

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