Find the equation of the hyperbola satisfying | vedclass

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(A) Given the foci are $(0, \pm 13)$,the hyperbola is vertical and centered at the origin. The standard form is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^

Vedclass · Accepted Answer

$\frac{y^{2}}{25} - \frac{x^{2}}{144} = 1$

Vedclass · Answer

(A) Given the foci are $(0, \pm 13)$,the hyperbola is vertical and centered at the origin. The standard form is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.Since the foci are $(0, \pm c)$,we have $c = 13$.The length of the conjugate axis is $2b = 24$,which implies $b = 12$.Using the relation $c^{2} = a^{2} + b^{2}$,we get $13^{2} = a^{2} + 12^{2}$.$169 = a^{2} + 144 \Rightarrow a^{2} = 169 - 144 = 25$.Substituting $a^{2} = 25$ and $b^{2} = 144$ into the standard equation,we get $\frac{y^{2}}{25} - \frac{x^{2}}{144} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm 13)$,the conjugate axis is of length $24$.

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