Let the foci of the hyperbola coincide with t | vedclass

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(C) For the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$, we have $a^2=36$ and $b^2=16$.Eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{

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$\frac{96}{\sqrt{5}}$

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(C) For the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$, we have $a^2=36$ and $b^2=16$.Eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{16}{36}} = \sqrt{\frac{20}{36}} = \frac{\sqrt{5}}{3}$.The foci are $(\pm ae_1, 0) = (\pm 6 	imes \frac{\sqrt{5}}{3}, 0) = (\pm 2\sqrt{5}, 0)$.For the hyperbola, let the equation be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ with eccentricity $e=5$.The foci are $(\pm Ae, 0) = (\pm 5A, 0)$.Since the foci coincide, $5A = 2\sqrt{5} \Rightarrow A = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$.Using $e^2 = 1 + \frac{B^2}{A^2}$, we have $25 = 1 + \frac{B^2}{4/5}$ $\Rightarrow 24 = \frac{5B^2}{4}$ $\Rightarrow B^2 = \frac{96}{5}$.The length of the latus rectum is $\frac{2B^2}{A} = \frac{2(96/5)}{2/\sqrt{5}} = \frac{96}{\sqrt{5}}$.

Let the foci of the hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is:

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