If the eccentricity of the hyperbola frac{x^2 | vedclass

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(B) Given the equation $15(e^2 + 1) = 34e$, we rewrite it as $15e^2 - 34e + 15 = 0$.Solving this quadratic equation: $(3e - 5)(5e - 3) = 0$.Since for

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$\frac{40}{3}$

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(B) Given the equation $15(e^2 + 1) = 34e$, we rewrite it as $15e^2 - 34e + 15 = 0$.Solving this quadratic equation: $(3e - 5)(5e - 3) = 0$.Since for a hyperbola the eccentricity $e > 1$, we must have $e = 5/3$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the relation between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.Substituting $e = 5/3$, we get $b^2 = a^2((5/3)^2 - 1) = a^2(25/9 - 1) = 16a^2/9$.Since the hyperbola passes through $(6, 4\sqrt{3})$, we have $\frac{6^2}{a^2} - \frac{(4\sqrt{3})^2}{b^2} = 1$, which simplifies to $\frac{36}{a^2} - \frac{48}{b^2} = 1$.Substituting $b^2 = 16a^2/9$ into the equation: $\frac{36}{a^2} - \frac{48}{16a^2/9} = 1 \implies \frac{36}{a^2} - \frac{27}{a^2} = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9$.Then $b^2 = 16(9)/9 = 16$.Now consider the second hyperbola $\frac{x^2}{a^2} - \frac{y^2}{2(a^2 + 1)} = 1$. Here $a^2 = 9$ and $b'^2 = 2(9 + 1) = 20$.The length of the latus rectum is given by $\frac{2b'^2}{a} = \frac{2(20)}{\sqrt{9}} = \frac{40}{3}$.

If the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, passing through $(6, 4\sqrt{3})$, satisfies $15(e^2 + 1) = 34e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{2(a^2 + 1)} = 1$ is:

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