Length of Latus Rectum of Hyperbola Given Foci and Eccentricity Relation

To find the length of the latus rectum of the given hyperbola, we follow these steps:

Step 1: Determine the eccentricity $e$ The eccentricity $e$ satisfies the quadratic equation: $6e^2 - 11e + 3 = 0$

Factorizing the quadratic equation: $(2e - 3)(3e - 1) = 0$

This gives two roots: $e = \frac{3}{2} \quad \text{or} \quad e = \frac{1}{3}$

Since the eccentricity of a hyperbola must be greater than $1$ ($e > 1$), we select: $e = \frac{3}{2}$

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Step 2: Find the semi-transverse axis $b$ The foci of the hyperbola are given as $S_1(3, 5)$ and $S_2(3, -4)$. Since the $x$-coordinates of both foci are equal ($x = 3$), the transverse axis of the hyperbola is parallel to the $y$-axis.

The distance between the foci is $2be$: $2be = |5 - (-4)| = 9$

Substitute $e = \frac{3}{2}$ into the equation: $2b \left(\frac{3}{2}\right) = 9 \implies 3b = 9 \implies b = 3$

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Step 3: Determine $a^2$ For a hyperbola with a vertical transverse axis, the relation between the eccentricity $e$, semi-transverse axis $b$, and semi-conjugate axis $a$ is: $e^2 = 1 + \frac{a^2}{b^2}$

Substituting $e = \frac{3}{2}$ and $b = 3$: $\left(\frac{3}{2}\right)^2 = 1 + \frac{a^2}{3^2}$ $\frac{9}{4} = 1 + \frac{a^2}{9}$ $\frac{a^2}{9} = \frac{9}{4} - 1 = \frac{5}{4}$ $a^2 = \frac{45}{4}$

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Step 4: Calculate the length of the latus rectum The length of the latus rectum for a vertical hyperbola is given by: $\text{Length of Latus Rectum} = \frac{2a^2}{b}$

Substituting $a^2 = \frac{45}{4}$ and $b = 3$: $\text{Length of Latus Rectum} = \frac{2 \times \frac{45}{4}}{3} = \frac{\frac{45}{2}}{3} = \frac{45}{6} = \frac{15}{2}$

Thus, the length of the latus rectum is $\frac{15}{2}$, which corresponds to option C.