Step 1: Understanding the Question:
This question comes from the topic of conic sections, specifically the properties of a hyperbola.
We are given the eccentricity and the length of the latus rectum of a hyperbola, and we need to calculate the length of its transverse axis.
Step 2: Key Formulas and Approach:
Latus Rectum Formula: $\text{L.R.} = \frac{2b^2}{a}$
Eccentricity Relation: $e^2 = 1 + \frac{b^2}{a^2}$
Transverse Axis Length: The length of the transverse axis is equal to $2a$.
Our approach will be to express $b^2$ in terms of $a$ using the latus rectum equation, substitute it into the eccentricity equation, and solve for $a$.
Step 3: Detailed Explanation:
Set up the Latus Rectum Equation:
We are given that the length of the latus rectum is $\frac{10}{3}$:
\[
\frac{2b^2}{a} = \frac{10}{3} \quad \implies \quad b^2 = \frac{5a}{3} \quad \cdots (1)
\]
Set up the Eccentricity Equation:
We are given the eccentricity $e = \frac{\sqrt{13}}{3}$. Squaring both sides:
\[
e^2 = \frac{13}{9}
\]
Using the eccentricity relationship:
\[
\frac{13}{9} = 1 + \frac{b^2}{a^2} \quad \implies \quad \frac{b^2}{a^2} = \frac{13}{9} - 1 = \frac{4}{9} \quad \cdots (2)
\]
Solve for $a$:
Substitute the expression for $b^2$ from equation (1) into equation (2):
\[
\frac{\left(\frac{5a}{3}\right)}{a^2} = \frac{4}{9}
\]
\[
\frac{5}{3a} = \frac{4}{9}
\]
\[
27 a = 45 \quad \implies \quad a = \frac{45}{12} = \frac{15}{4}
\]
Calculate the Length of the Transverse Axis:
The length of the transverse axis is $2a$:
\[
\text{Length} = 2 \times \frac{15}{4} = \frac{15}{2}\text{ units}
\]
Step 4: Final Answer:
The length of the transverse axis of the hyperbola is $15/2$ units, which corresponds to Option (C).