Question

Let an ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a<b \] pass through the point \((4,3)\) and have eccentricity \( \frac{\sqrt5}{3} \). Then the length of its latus rectum is:

• A. \( \frac{4\sqrt5}{3} \)
• B. \(2\sqrt5\)
• C. \( \frac{7\sqrt5}{3} \)
• D. \( \frac{8\sqrt5}{3} \)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For an ellipse with \(a<b\), the major axis is along the y-axis.
The eccentricity relation is \(a^2 = b^2(1 - e^2)\).
Length of Latus Rectum = \(\frac{2a^2}{b}\).
Step 2: Key Formula or Approach:
1. Eccentricity: \(e^2 = 1 - a^2/b^2 \Rightarrow \frac{a^2}{b^2} = 1 - e^2\).
2. Point substitution: \(\frac{16}{a^2} + \frac{9}{b^2} = 1\).
Step 3: Detailed Explanation:
Given \(e = \frac{\sqrt{5}}{3} \Rightarrow e^2 = \frac{5}{9}\).
Using \(a<b\):
\(\frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow a^2 = \frac{4}{9}b^2\).
Substitute point \((4, 3)\) into ellipse equation:
\(\frac{16}{a^2} + \frac{9}{b^2} = 1\)
\(\frac{16}{(4b^2/9)} + \frac{9}{b^2} = 1\)
\(\frac{16 \times 9}{4b^2} + \frac{9}{b^2} = 1 \Rightarrow \frac{36}{b^2} + \frac{9}{b^2} = 1\)
\(\frac{45}{b^2} = 1 \Rightarrow b^2 = 45 \Rightarrow b = \sqrt{45} = 3\sqrt{5}\).
Now find \(a^2 = \frac{4}{9} \times 45 = 20\).
Length of Latus Rectum = \(\frac{2a^2}{b} = \frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}}\).
Rationalize the denominator: \(\frac{40\sqrt{5}}{3 \times 5} = \frac{8\sqrt{5}}{3}\).
Step 4: Final Answer:
The length of the latus rectum is \(\frac{8\sqrt{5}}{3}\).