Consider a hyperbola H having its centre at t | vedclass

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(A) Let the equation of the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $b^2 = a^2(e^2 - 1)$.Since $C_1$ is centered at the origin

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$\frac{28}{3}$

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(A) Let the equation of the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $b^2 = a^2(e^2 - 1)$.Since $C_1$ is centered at the origin and touches the hyperbola at its vertices $(\pm a, 0)$,its radius is $a$. Given the area of $C_1$ is $36 \pi$,we have $\pi a^2 = 36 \pi$,so $a = 6$.$C_2$ is centered at a focus $(ae, 0)$ and touches the hyperbola at its vertex $(a, 0)$. The distance between the focus and the vertex is $|ae - a| = a(e - 1)$. Thus,the radius of $C_2$ is $r = a(e - 1)$.Given the area of $C_2$ is $4 \pi$,we have $\pi r^2 = 4 \pi$,so $r^2 = 4$,which means $r = 2$.Substituting $a = 6$,we get $6(e - 1) = 2$,so $e - 1 = \frac{1}{3}$,which gives $e = \frac{4}{3}$.Now,$b^2 = a^2(e^2 - 1) = 36 \left( (\frac{4}{3})^2 - 1 ight) = 36 \left( \frac{16}{9} - 1 ight) = 36 \left( \frac{7}{9} ight) = 28$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 28}{6} = \frac{28}{3}$.

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