Let the foci of a hyperbola H coincide with t | vedclass

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(B) For the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$,we have $a^2=100$ and $b^2=75$.The eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \s

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$225$

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(B) For the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$,we have $a^2=100$ and $b^2=75$.The eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{75}{100}} = \sqrt{\frac{25}{100}} = \frac{5}{10} = \frac{1}{2}$.The foci of the ellipse are $(h \pm ae_1, k) = (1 \pm 10 	imes \frac{1}{2}, 1) = (1 \pm 5, 1)$,which are $F_1(6, 1)$ and $F_2(-4, 1)$.The distance between the foci is $2ae_1 = 10$.For the hyperbola $H$,the eccentricity $e_2 = \frac{1}{e_1} = 2$.The distance between the foci of the hyperbola is $2ae_2 = 10$,so $2a(2) = 10$,which gives $a = \frac{5}{2}$.The length of the transverse axis is $\alpha = 2a = 5$.For a hyperbola,$b^2 = a^2(e_2^2-1) = a^2(2^2-1) = 3a^2$.Thus,$b = a\sqrt{3} = \frac{5\sqrt{3}}{2}$.The length of the conjugate axis is $\beta = 2b = 5\sqrt{3}$.Therefore,$3\alpha^2 + 2\beta^2 = 3(5)^2 + 2(5\sqrt{3})^2 = 3(25) + 2(75) = 75 + 150 = 225$.

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