The equation of the hyperbola whose coordinat | vedclass

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(A) Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. Given,foci are $(\pm 8, 0) = (\pm ae, 0)$,so $ae = 8$. The l

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$3x^{2} - y^{2} = 48$

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(A) Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. Given,foci are $(\pm 8, 0) = (\pm ae, 0)$,so $ae = 8$. The length of the latus rectum is $\frac{2b^{2}}{a} = 24$,which implies $b^{2} = 12a$. We know that $b^{2} = a^{2}(e^{2} - 1) = a^{2}e^{2} - a^{2}$. Substituting $ae = 8$ and $b^{2} = 12a$,we get $12a = 8^{2} - a^{2}$. $a^{2} + 12a - 64 = 0$. $(a + 16)(a - 4) = 0$. Since $a > 0$,we have $a = 4$. Then $b^{2} = 12(4) = 48$. Substituting $a^{2} = 16$ and $b^{2} = 48$ into the standard equation: $\frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$. Multiplying by $48$,we get $3x^{2} - y^{2} = 48$.

The equation of the hyperbola whose coordinates of the foci are $(\pm 8, 0)$ and the length of the latus rectum is $24$ units,is

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