The equation of a hyperbola,whose foci are (5 | vedclass

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(B) The foci are $(\pm 5, 0)$,so the hyperbola is horizontal with $ae = 5$.The length of the conjugate axis is $2b = 8$,which implies $b = 4$.For a hy

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$16x^2 - 9y^2 = 144$

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(B) The foci are $(\pm 5, 0)$,so the hyperbola is horizontal with $ae = 5$.The length of the conjugate axis is $2b = 8$,which implies $b = 4$.For a hyperbola,the relation between $a$,$b$,and $e$ is $c^2 = a^2 + b^2$,where $c = ae = 5$.Thus,$5^2 = a^2 + 4^2$,which gives $25 = a^2 + 16$.$a^2 = 25 - 16 = 9$,so $a = 3$.The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Substituting the values,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$.Multiplying by $144$,we obtain $16x^2 - 9y^2 = 144$.

The equation of a hyperbola,whose foci are $(5, 0)$ and $(-5, 0)$ and the length of whose conjugate axis is $8$,is

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