The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is , is [2025]

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Solution

Q.
The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, –3) and eccentricity is $\frac{4}{5}$ , is [2025]

1 $\frac{6}{5}$

2 $\frac{10}{3}$

3 $\frac{18}{5}$

4 $\frac{50}{3}$

Ans.
(3)

We have, two foci (2, 5), (2, –3) and eccentricity =4/5

$\Rightarrow 2 b e = 8 \Rightarrow b e = 4 \Rightarrow b (\frac{4}{5}) = 4 \Rightarrow b = 5$

$∴ c^{2} = b^{2} - a^{2} \Rightarrow 16 = 25 - a^{2} \Rightarrow a = 3$

$∴$ Length of latus rectum = $\frac{2 a^{2}}{b} = \frac{18}{5}$

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