If the eccentricity e of the hyperbola (x^2)/(a^2) - (y^2)/(b^2) =...

The given equation of the hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.

Since it passes through $(6, 4\sqrt{3})$, we have:

$\dfrac{36}{a^2} - \dfrac{48}{b^2} = 1$

The eccentricity $e$ satisfies $15(e^2 + 1) = 34e$.

$15e^2 - 34e + 15 = 0$

$(3e - 5)(5e - 3) = 0$

Since the eccentricity of a hyperbola is $e > 1$, we get $e = \dfrac{5}{3}$.

We know that $b^2 = a^2(e^2 - 1)$.

$b^2 = a^2 \left(\dfrac{25}{9} - 1\right) = \dfrac{16}{9}a^2$

Substituting $b^2$ into the first equation:

$\dfrac{36}{a^2} - \dfrac{48}{\dfrac{16}{9}a^2} = 1$

$\dfrac{36}{a^2} - \dfrac{27}{a^2} = 1$

$\dfrac{9}{a^2} = 1 \Rightarrow a^2 = 9$

Then, $b^2 = \dfrac{16}{9} \times 9 = 16$.

The second hyperbola is given by $\dfrac{x^2}{b^2} - \dfrac{y^2}{2(a^2+1)} = 1$.

Substituting the values of $a^2$ and $b^2$:

$\dfrac{x^2}{16} - \dfrac{y^2}{2(9+1)} = 1$

$\dfrac{x^2}{16} - \dfrac{y^2}{20} = 1$

Here, $A^2 = 16 \Rightarrow A = 4$ and $B^2 = 20$.

The length of the latus rectum is $\dfrac{2B^2}{A}$.

Length of latus rectum $= \dfrac{2 \times 20}{4} = 10$.

Answer: $10$