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:
- \( \frac{4\sqrt5}{3} \)
- \(2\sqrt5\)
- \( \frac{7\sqrt5}{3} \)
- \( \frac{8\sqrt5}{3} \)
The Correct Option is D
Solution and Explanation
Given:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\quad a < b \] Point \((4,3)\) lies on ellipse and eccentricity \(e = \frac{\sqrt{5}}{3}\)
Step 1: Use eccentricity relation
For ellipse (major axis along \(y\)-axis since \(a < b\)): \[ e = \sqrt{1 - \frac{a^2}{b^2}} \] \[ \frac{\sqrt{5}}{3} = \sqrt{1 - \frac{a^2}{b^2}} \] Squaring: \[ \frac{5}{9} = 1 - \frac{a^2}{b^2} \] \[ \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9} \] \[ a^2 = \frac{4}{9} b^2 \]
Step 2: Use point (4,3)
\[ \frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \] \[ \frac{16}{a^2} + \frac{9}{b^2} = 1 \] Substitute \(a^2 = \frac{4}{9}b^2\): \[ \frac{16}{\frac{4}{9}b^2} + \frac{9}{b^2} = 1 \] \[ \frac{16 \times 9}{4b^2} + \frac{9}{b^2} = 1 \] \[ \frac{144}{4b^2} + \frac{9}{b^2} = 1 \] \[ \frac{36}{b^2} + \frac{9}{b^2} = 1 \] \[ \frac{45}{b^2} = 1 \Rightarrow b^2 = 45 \] Then, \[ a^2 = \frac{4}{9} \times 45 = 20 \]
Step 3: Length of latus rectum
For ellipse: \[ \text{Length of latus rectum} = \frac{2a^2}{b} \] \[ = \frac{2 \times 20}{\sqrt{45}} = \frac{40}{3\sqrt{5}} \] Rationalizing: \[ = \frac{40\sqrt{5}}{15} = \frac{8\sqrt{5}}{3} \]
Final Answer:
\[ \boxed{\frac{8\sqrt{5}}{3}} \]
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