Let e_1 be the eccentricity of the hyperbola | vedclass

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(C) For the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. The eccentricity $e_1 = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{

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$\frac{10 \sqrt{5}}{3}$

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(C) For the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. The eccentricity $e_1 = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.Given $e_1 e_2 = 1$,we have $e_2 = \frac{1}{e_1} = \frac{4}{5}$.The foci of the hyperbola are $(\pm ae_1, 0) = (\pm 4 	imes \frac{5}{4}, 0) = (\pm 5, 0)$.Since the ellipse passes through $(\pm 5, 0)$,we have $a=5$.For the ellipse,$e_2^2 = 1 - \frac{b^2}{a^2}$ $\Rightarrow (\frac{4}{5})^2 = 1 - \frac{b^2}{25}$ $\Rightarrow \frac{16}{25} = 1 - \frac{b^2}{25}$ $\Rightarrow \frac{b^2}{25} = \frac{9}{25}$ $\Rightarrow b^2 = 9$.The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.For the chord parallel to the $x$-axis passing through $(0,2)$,we set $y=2$ in the ellipse equation:$\frac{x^2}{25} + \frac{4}{9} = 1$ $\Rightarrow \frac{x^2}{25} = 1 - \frac{4}{9} = \frac{5}{9}$ $\Rightarrow x^2 = \frac{125}{9}$ $\Rightarrow x = \pm \frac{5 \sqrt{5}}{3}$.The length of the chord is the distance between $(-\frac{5 \sqrt{5}}{3}, 2)$ and $(\frac{5 \sqrt{5}}{3}, 2)$,which is $\frac{5 \sqrt{5}}{3} - (-\frac{5 \sqrt{5}}{3}) = \frac{10 \sqrt{5}}{3}$.

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