Equation- of the hyperbola with foci- -0- 5- and e - 5-3-is y-z-9 - y-z-16 - 1-x-z-16 - y-z-9 - 1-x-z-16 - y-z-9 - 1-x-z-12 - y-z-13 - 1-

The correct option is B x^z/16 y^z/9 = 1 Foci = ( 0 , ±5 ) = ( 0 , ± be) b5/3 = 5 ⇒ b = 3 a^2 = 9 ( 25/9 1 ) = 16 x^2/16 y^2/9 = 1 ...

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Equation of the hyperbola with foci (0,± 5) and e = $\frac{5}{3}$ is

$\frac{y^{z}}{9} - \frac{y^{z}}{16} = 1$

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$\frac{x^{z}}{16} - \frac{y^{z}}{9} = 1$

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$\frac{x^{z}}{16} - \frac{y^{z}}{9} = 1$

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$\frac{x^{z}}{12} - \frac{y^{z}}{13} = 1$

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