The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, -3)$ and eccentricity is $\frac{4}{5}$, is
The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, -3)$ and eccentricity is $\frac{4}{5}$, is
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- $\frac{6}{5}$
- $\frac{50}{3}$
- $\frac{10}{3}$
- $\frac{18}{5}$
The Correct Option is D
Solution and Explanation
2. Calculate the distance between the foci: \[ 2c = |5 - (-3)| = 8 \implies c = 4 \]
3. Use the relationship between $a$, $b$, and $c$: \[ e = \frac{c}{a} = \frac{4}{5} \implies a = 5 \] \[ b^2 = a^2 - c^2 = 25 - 16 = 9 \implies b = 3 \]
4. Calculate the length of the latus-rectum: \[ \text{Length of latus-rectum} = \frac{2b^2}{a} = \frac{2 \cdot 3^2}{5} = \frac{18}{5} \] Therefore, the correct answer is (4) $\frac{18}{5}$.
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