An ellipse E: frac{x^{2}}{a^{2}}+frac{y^{2}}{ | vedclass

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(B) The hyperbola $H$ is given by $\frac{y^{2}}{64} - \frac{x^{2}}{49} = 1$. Its vertices are $(0, \pm 8)$.Since the ellipse $E: \frac{x^{2}}{a^{2}} +

Vedclass · Accepted Answer

$1552$

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(B) The hyperbola $H$ is given by $\frac{y^{2}}{64} - \frac{x^{2}}{49} = 1$. Its vertices are $(0, \pm 8)$.Since the ellipse $E: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through $(0, \pm 8)$,we have $b^2 = 64$,so $b = 8$.The eccentricity of the hyperbola $H$ is $e_H = \sqrt{1 + \frac{49}{64}} = \sqrt{\frac{113}{64}} = \frac{\sqrt{113}}{8}$.The eccentricity of the ellipse $E$ is $e_E = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{a^2}{64}}$.Given $e_E 	imes e_H = \frac{1}{2}$,we have $\sqrt{1 - \frac{a^2}{64}} 	imes \frac{\sqrt{113}}{8} = \frac{1}{2}$.Squaring both sides: $(1 - \frac{a^2}{64}) 	imes \frac{113}{64} = \frac{1}{4}$ $\Rightarrow 1 - \frac{a^2}{64} = \frac{64}{4 	imes 113} = \frac{16}{113}$.Thus,$\frac{a^2}{64} = 1 - \frac{16}{113} = \frac{97}{113}$,so $a^2 = \frac{64 	imes 97}{113}$.The length of the latus rectum $l = \frac{2a^2}{b} = \frac{2}{8} 	imes \frac{64 	imes 97}{113} = \frac{16 	imes 97}{113} = \frac{1552}{113}$.Therefore,$113l = 1552$.

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