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: Use the eccentricity formula.
$e^2 = 1 - \frac{b^2}{a^2}$
Given $e = \frac{\sqrt{5}}{3}$, so $e^2 = \frac{5}{9}$.
$\frac{5}{9} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = 1 - \frac{5}{9} = \frac{4}{9}$
$b^2 = \frac{4}{9} a^2 \quad \text{--- (Eq. 1)}$


Step 2: Use the point $(3, 4)$ in the ellipse equation.
$\frac{3^2}{a^2} + \frac{4^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{16}{b^2} = 1$
Substitute $b^2 = \frac{4}{9} a^2$ into this equation:
$\frac{9}{a^2} + \frac{16}{\frac{4}{9} a^2} = 1$
$\frac{9}{a^2} + \frac{16 \times 9}{4 a^2} = 1$
$\frac{9}{a^2} + \frac{36}{a^2} = 1 \implies \frac{45}{a^2} = 1 \implies a^2 = 45$


Step 3: Calculate $b^2$ and the Length of Latus Rectum.
From Eq. 1: $b^2 = \frac{4}{9} \times 45 = 20$.
Length of Latus Rectum (LR) = $\frac{2b^2}{a}$.
$a = \sqrt{45} = 3\sqrt{5}$.
LR = $\frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{40\sqrt{5}}{3 \times 5} = \frac{8\sqrt{5}}{3}$.
The answer is option (2).