Find the eccentricity of a hyperbola whose la | vedclass

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The length of the latus rectum is $\frac{2b^2}{a} = 8$,which implies

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$\frac{2}{\sqrt{3}}$

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The length of the latus rectum is $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$. The distance between the foci is $2ae$. The length of the conjugate axis is $2b$. According to the problem,$2b = \frac{1}{2}(2ae)$,so $b = \frac{ae}{2}$,which means $b^2 = \frac{a^2e^2}{4}$. Equating the two expressions for $b^2$: $4a = \frac{a^2e^2}{4}$,so $16a = a^2e^2$,which gives $a = \frac{16}{e^2}$. We know that $b^2 = a^2(e^2 - 1)$. Substituting $b^2 = 4a$ and $a = \frac{16}{e^2}$: $4(\frac{16}{e^2}) = (\frac{16}{e^2})^2(e^2 - 1)$. $\frac{64}{e^2} = \frac{256}{e^4}(e^2 - 1)$. $1 = \frac{4}{e^2}(e^2 - 1) = 4 - \frac{4}{e^2}$. $\frac{4}{e^2} = 3$,so $e^2 = \frac{4}{3}$,which gives $e = \frac{2}{\sqrt{3}}$. Thus,the eccentricity is $\frac{2}{\sqrt{3}}$.

Find the eccentricity of a hyperbola whose latus rectum is $8$ and whose conjugate axis is half the distance between its foci.

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