The foci of the hyperbola 2x^2 - 3y^2 = 5 are | vedclass

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(A) The given equation is $2x^2 - 3y^2 = 5$. Dividing by $5$,we get $\frac{x^2}{5/2} - \frac{y^2}{5/3} = 1$. This is of the form $\frac{x^2}{a^2} - \f

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$\left( \pm \frac{5}{\sqrt{6}}, 0 \right)$

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(A) The given equation is $2x^2 - 3y^2 = 5$. Dividing by $5$,we get $\frac{x^2}{5/2} - \frac{y^2}{5/3} = 1$. This is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \frac{5}{2}$ and $b^2 = \frac{5}{3}$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5/3}{5/2}} = \sqrt{1 + \frac{2}{3}} = \sqrt{\frac{5}{3}}$. The foci are at $(\pm ae, 0)$. $ae = \sqrt{\frac{5}{2}} 	imes \sqrt{\frac{5}{3}} = \sqrt{\frac{25}{6}} = \frac{5}{\sqrt{6}}$. Thus,the foci are $\left( \pm \frac{5}{\sqrt{6}}, 0 ight)$.

The foci of the hyperbola $2x^2 - 3y^2 = 5$ are

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