Let an ellipse (x^2)/(a^2) + (y^2)/(b^2) = 1 , a < b , pass through...

Given the equation of the ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $a < b$.

The eccentricity is given by $e = \dfrac{\sqrt{5}}{3}$.

Since $a < b$, the relation between $a$, $b$, and $e$ is $a^2 = b^2(1 - e^2)$.

Substituting the value of $e$:

$a^2 = b^2\left(1 - \dfrac{5}{9}\right) = \dfrac{4}{9}b^2$

The ellipse passes through the point $(4, 3)$, so substituting $x = 4$ and $y = 3$ into the equation of the ellipse gives:

$\dfrac{16}{a^2} + \dfrac{9}{b^2} = 1$

Substituting $a^2 = \dfrac{4}{9}b^2$ into the above equation:

$\dfrac{16}{\dfrac{4}{9}b^2} + \dfrac{9}{b^2} = 1$

$\dfrac{36}{b^2} + \dfrac{9}{b^2} = 1$

$\dfrac{45}{b^2} = 1 \implies b^2 = 45$

Now, finding $a^2$:

$a^2 = \dfrac{4}{9} \times 45 = 20$

For an ellipse with $a < b$, the length of the latus rectum is given by $\dfrac{2a^2}{b}$.

Substituting the values of $a^2$ and $b = \sqrt{45} = 3\sqrt{5}$:

Length of latus rectum $= \dfrac{2 \times 20}{3\sqrt{5}} = \dfrac{40}{3\sqrt{5}} = \dfrac{8\sqrt{5}}{3}$

Answer: $\dfrac{8\sqrt{5}}{3}$