Let the foci of the ellipse frac{x^{2}}{16}+f | vedclass

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(D) For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$,we have $a^2=16$ and $b^2=7$.Eccentricity $e_1 = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}

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$\frac{27}{10}$

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(D) For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$,we have $a^2=16$ and $b^2=7$.Eccentricity $e_1 = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.Foci are $(\pm a_1 e_1, 0) = (\pm 4 \cdot \frac{3}{4}, 0) = (\pm 3, 0)$.For the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$,rewrite as $\frac{x^{2}}{(12/5)^2} - \frac{y^{2}}{(\sqrt{\alpha}/5)^2} = 1$.Here $a^2 = \frac{144}{25}$ and $b^2 = \frac{\alpha}{25}$.Eccentricity $e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\alpha/25}{144/25}} = \sqrt{1 + \frac{\alpha}{144}} = \frac{\sqrt{144+\alpha}}{12}$.Foci are $(\pm a_2 e_2, 0) = (\pm \frac{12}{5} \cdot \frac{\sqrt{144+\alpha}}{12}, 0) = (\pm \frac{\sqrt{144+\alpha}}{5}, 0)$.Since foci coincide,$\frac{\sqrt{144+\alpha}}{5} = 3$ $\Rightarrow \sqrt{144+\alpha} = 15$ $\Rightarrow 144+\alpha = 225$ $\Rightarrow \alpha = 81$.Thus,$b^2 = \frac{81}{25}$.Length of latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot (81/25)}{12/5} = \frac{162}{25} \cdot \frac{5}{12} = \frac{27}{10}$.

Let the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:

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