Find the equation of the hyperbola whose foci | vedclass

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(A) Since the foci are $(0, \pm 12)$,the hyperbola is vertical and $c = 12$.The length of the latus rectum is given by $\frac{2b^{2}}{a} = 36$,which i

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$3y^{2} - x^{2} = 108$

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(A) Since the foci are $(0, \pm 12)$,the hyperbola is vertical and $c = 12$.The length of the latus rectum is given by $\frac{2b^{2}}{a} = 36$,which implies $b^{2} = 18a$.For a hyperbola,$c^{2} = a^{2} + b^{2}$. Substituting the values,we get $144 = a^{2} + 18a$.Rearranging gives the quadratic equation $a^{2} + 18a - 144 = 0$.Factoring the quadratic,we get $(a + 24)(a - 6) = 0$,so $a = -24$ or $a = 6$.Since $a$ must be positive,we take $a = 6$. Then $b^{2} = 18(6) = 108$.The equation of the hyperbola is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,which is $\frac{y^{2}}{36} - \frac{x^{2}}{108} = 1$.Multiplying by $108$,we get $3y^{2} - x^{2} = 108$.

Find the equation of the hyperbola whose foci are $(0, \pm 12)$ and the length of the latus rectum is $36$.

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