If eccentricity of an ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

which passes through which passes through point (3, 4) is √5/3 , then length of latus rectum of ellipse is

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

Step 1: Write down the standard equation of the ellipse and the eccentricity formula.
The equation of the ellipse is given by:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
The eccentricity $e$ of an ellipse (assuming $a>b$) is related to its semi-major axis $a$ and semi-minor axis $b$ by the relation:
$e^2=1-\frac{b^2}{a^2}$

Step 2: Use the given eccentricity to find a relation between $b^2$ and $a^2$.
We are given that the eccentricity $e=\frac{\sqrt{5}}{3}$. Substituting this value into the eccentricity equation:
${\left(\frac{\sqrt{5}}{3}\right)}^2=1-\frac{b^2}{a^2}$
$\frac{5}{9}=1-\frac{b^2}{a^2}$
Rearranging the equation to solve for $\frac{b^2}{a^2}$:
$\frac{b^2}{a^2}=1-\frac{5}{9}=\frac{4}{9}$
This gives us the relationship:
$b^2=\frac{4}{9}{a}^2$

Step 3: Substitute the coordinates of the point (3, 4) into the ellipse equation.
Since the ellipse passes through the point $(3,4)$, these coordinates must satisfy the equation of the ellipse:
$\frac{3^2}{a^2}+\frac{4^2}{b^2}=1$
$\frac{9}{a^2}+\frac{16}{b^2}=1$

Step 4: Solve for $a^2$ and $b^2$.
Substitute $b^2=\frac{4}{9}{a}^2$ into the equation from Step 3:
$\frac{9}{a^2}+\frac{16}{\frac{4}{9}{a}^2}=1$
Simplify the second term:
$\frac{9}{a^2}+\frac{16\times 9}{4a^2}=1$
$\frac{9}{a^2}+\frac{36}{a^2}=1$
$\frac{45}{a^2}=1$
Thus, we find:
$a^2=45$
Taking the positive square root for the semi-major axis length:
$a=\sqrt{45}=3\sqrt{5}$
Now, compute $b^2$ using the relation from Step 2:
$b^2=\frac{4}{9}\times 45=20$

Step 5: Calculate the length of the latus rectum.
The formula for the length of the latus rectum of an ellipse is:
$\text{Length of latus rectum}=\frac{2b^2}{a}$
Substitute the values of $b^2$ and $a$:
$\text{Length}=\frac{2\times 20}{3\sqrt{5}}=\frac{40}{3\sqrt{5}}$
Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{5}$:
$\text{Length}=\frac{40\sqrt{5}}{3\times 5}=\frac{8\sqrt{5}}{3}$

Therefore, the length of the latus rectum of the ellipse is $\frac{8\sqrt{5}}{3}$.