Length of Latus Rectum of Vertical Ellipse

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

Step 1: Understand the orientation of the ellipse The equation of the ellipse is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Since $a < b$, the major axis lies along the $y$-axis, which means this is a vertical ellipse.

Step 2: Relate $a$ and $b$ using the eccentricity For a vertical ellipse ($a < b$), the eccentricity $e$ is related to $a$ and $b$ by the formula: $e = \sqrt{1 - \frac{a^2}{b^2}}$

Given that $e = \frac{\sqrt{5}}{3}$, we square both sides: $e^2 = 1 - \frac{a^2}{b^2}$ $\left(\frac{\sqrt{5}}{3}\right)^2 = 1 - \frac{a^2}{b^2}$ $\frac{5}{9} = 1 - \frac{a^2}{b^2}$ $\frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$

Thus, we can express $a^2$ in terms of $b^2$: $a^2 = \frac{4}{9}b^2$

Step 3: Use the given point to find $a^2$ and $b^2$ The ellipse passes through the point $(4, 3)$. Substituting $x = 4$ and $y = 3$ into the equation of the ellipse gives: $\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1$ $\frac{16}{a^2} + \frac{9}{b^2} = 1$

Substituting $a^2 = \frac{4}{9}b^2$ into this equation: $\frac{16}{\frac{4}{9}b^2} + \frac{9}{b^2} = 1$ $\frac{16 \times 9}{4b^2} + \frac{9}{b^2} = 1$ $\frac{36}{b^2} + \frac{9}{b^2} = 1$ $\frac{45}{b^2} = 1 \implies b^2 = 45$

Now, calculate $b$ and $a^2$: $b = \sqrt{45} = 3\sqrt{5}$ $a^2 = \frac{4}{9} \times 45 = 20$

Step 4: Calculate the length of the latus rectum For a vertical ellipse ($a < b$), the length of the latus rectum is given by: $\text{Length of Latus Rectum} = \frac{2a^2}{b}$

Substitute the values of $a^2$ and $b$: $\text{Length of Latus Rectum} = \frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}}$

Rationalizing the denominator: $\text{Length of Latus Rectum} = \frac{40 \sqrt{5}}{3 \times 5} = \frac{8\sqrt{5}}{3}$

Thus, the correct option is D.