The eccentricity of the hyperbola whose lengt | vedclass

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(A) Given,length of latus rectum = $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.The length of the conjugate axis is $2b$,and the distance between the

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

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(A) Given,length of latus rectum = $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.The length of the conjugate axis is $2b$,and the distance between the foci is $2ae$.According to the problem,$2b = \frac{1}{2}(2ae)$,which simplifies to $b = \frac{ae}{2}$,or $b^2 = \frac{a^2e^2}{4}$.Equating the two expressions for $b^2$: $4a = \frac{a^2e^2}{4}$,which gives $a^2e^2 = 16a$,or $ae^2 = 16$ (since $a 
eq 0$).For a hyperbola,$b^2 = a^2(e^2 - 1)$.Substituting $b^2 = 4a$: $4a = a^2(e^2 - 1)$.Since $a 
eq 0$,we have $4 = a(e^2 - 1) = ae^2 - a$.Substituting $ae^2 = 16$: $4 = 16 - a$,which gives $a = 12$.Now,$ae^2 = 16 \implies 12e^2 = 16$.$e^2 = \frac{16}{12} = \frac{4}{3}$.Therefore,$e = \frac{2}{\sqrt{3}}$.

The eccentricity of the hyperbola whose length of the latus rectum is equal to $8$ and the length of its conjugate axis is equal to half of the distance between its foci is:

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