Length of Latus Rectum of Hyperbola from Eccentricity and Point

The given hyperbola is: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The eccentricity $e$ satisfies the equation: $15(e^2 + 1) = 34e$ $15e^2 - 34e + 15 = 0$

Solving this quadratic equation for $e$: $(3e - 5)(5e - 3) = 0$ $e = \frac{5}{3} \quad \text{or} \quad e = \frac{3}{5}$

Since the eccentricity of a hyperbola must be greater than $1$ ($e > 1$), we have: $e = \frac{5}{3}$

The relation between $e$, $a$, and $b$ for the hyperbola is: $e^2 = 1 + \frac{b^2}{a^2}$ $\left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{a^2}$ $\frac{25}{9} - 1 = \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{16}{9} \implies b^2 = \frac{16}{9}a^2$

Since the hyperbola passes through the point $(6, 4\sqrt{3})$, this point satisfies its equation: $\frac{6^2}{a^2} - \frac{(4\sqrt{3})^2}{b^2} = 1$ $\frac{36}{a^2} - \frac{48}{b^2} = 1$

Substituting $b^2 = \frac{16}{9}a^2$ into the equation: $\frac{36}{a^2} - \frac{48}{\frac{16}{9}a^2} = 1$ $\frac{36}{a^2} - \frac{27}{a^2} = 1$ $\frac{9}{a^2} = 1 \implies a^2 = 9$

Now, calculating $b^2$: $b^2 = \frac{16}{9}(9) = 16$

We are required to find the length of the latus rectum of the hyperbola: $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$

Substituting $a^2 = 9$ and $b^2 = 16$: $\frac{x^2}{16} - \frac{y^2}{2(9 + 1)} = 1$ $\frac{x^2}{16} - \frac{y^2}{20} = 1$

Comparing this with the standard equation of a hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$: $A^2 = 16 \implies A = 4$ $B^2 = 20$

The length of the latus rectum is given by: $\text{Length of Latus Rectum} = \frac{2B^2}{A} = \frac{2 \times 20}{4} = 10$

Thus, the correct option is A.