Let the eccentricity e of a hyperbola satisfy | vedclass

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(C) Given the quadratic equation $6e^2 - 11e + 3 = 0$. Factoring the equation, we get $(2e - 3)(3e - 1) = 0$. This gives $e = 3/2$ or $e = 1/3$. Since

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$15$/$2$

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(C) Given the quadratic equation $6e^2 - 11e + 3 = 0$. Factoring the equation, we get $(2e - 3)(3e - 1) = 0$. This gives $e = 3/2$ or $e = 1/3$. Since the eccentricity of a hyperbola must satisfy $e > 1$, we take $e = 3/2$. The distance between the foci is $2ae = \sqrt{(3 - 3)^2 + (5 - (-4))^2} = \sqrt{0^2 + 9^2} = 9$. Substituting $e = 3/2$, we have $2a(3/2) = 9$, which simplifies to $3a = 9$, so $a = 3$. For a hyperbola, $b^2 = a^2(e^2 - 1)$. Substituting the values, $b^2 = 3^2((3/2)^2 - 1) = 9(9/4 - 1) = 9(5/4) = 45/4$. The length of the latus rectum is given by $LR = \frac{2b^2}{a}$. $LR = \frac{2(45/4)}{3} = \frac{45/2}{3} = 15/2$.

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5)$ and $(3, -4)$, then the length of its latus rectum is:

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