Question

If eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which passes through the point $(3, 4)$ is $\frac{\sqrt{5}}{3}$, then length of latus rectum of ellipse, is :

• A. $\frac{4\sqrt{5}}{3}$
• B. $\frac{8\sqrt{5}}{3}$
• C. $\frac{4\sqrt{7}}{3}$
• D. $\frac{8\sqrt{7}}{3}$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The eccentricity \( e \) relates the semi-major axis \( a \) and semi-minor axis \( b \). Using this and the given point, we can solve for \( a \) and \( b \).
Step 2: Key Formula or Approach:
Eccentricity squared: \( e^2 = 1 - \frac{b^2}{a^2} \).
Given \( e = \frac{\sqrt{5}}{3} \implies e^2 = \frac{5}{9} \).
Length of Latus Rectum \( (LR) = \frac{2b^2}{a} \).
Step 3: Detailed Explanation:
From \( e^2 \): \( \frac{5}{9} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = 1 - \frac{5}{9} = \frac{4}{9} \implies b^2 = \frac{4a^2}{9} \).
The ellipse passes through \( (3, 4) \):
\( \frac{3^2}{a^2} + \frac{4^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{16}{b^2} = 1 \).
Substitute \( b^2 = \frac{4a^2}{9} \):
\( \frac{9}{a^2} + \frac{16}{\frac{4a^2}{9}} = 1 \implies \frac{9}{a^2} + \frac{36}{a^2} = 1 \implies \frac{45}{a^2} = 1 \implies a^2 = 45 \).
Then \( a = \sqrt{45} = 3\sqrt{5} \).
Calculate \( b^2 \): \( b^2 = \frac{4(45)}{9} = 20 \).
Latus Rectum: \( LR = \frac{2b^2}{a} = \frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{40\sqrt{5}}{15} = \frac{8\sqrt{5}}{3} \).
Step 4: Final Answer:
The length of the latus rectum is \( \frac{8\sqrt{5}}{3} \).