Let the hyperbola H : frac{x^2}{a^2} - y^2 = | vedclass

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(A) For the hyperbola $H : \frac{x^2}{a^2} - y^2 = 1$,the length of the latus rectum is $LR_H = \frac{2b^2}{a} = \frac{2(1)^2}{a} = \frac{2}{a}$.For t

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$42$

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(A) For the hyperbola $H : \frac{x^2}{a^2} - y^2 = 1$,the length of the latus rectum is $LR_H = \frac{2b^2}{a} = \frac{2(1)^2}{a} = \frac{2}{a}$.For the ellipse $E : 3x^2 + 4y^2 = 12$,we rewrite it as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$. The length of the latus rectum is $LR_E = \frac{2b^2}{a} = \frac{2(3)}{2} = 3$.Given $LR_H = LR_E$,we have $\frac{2}{a} = 3$,which implies $a = \frac{2}{3}$.The eccentricity of the hyperbola is $e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{(2/3)^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$. Thus,$e_H^2 = \frac{13}{4}$.The eccentricity of the ellipse is $e_E = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$. Thus,$e_E^2 = \frac{1}{4}$.Finally,$12(e_H^2 + e_E^2) = 12(\frac{13}{4} + \frac{1}{4}) = 12(\frac{14}{4}) = 3 	imes 14 = 42$.

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