The length of the latus rectum of the ellipse $5 x^{2} + 9 y^{2} = 45$ is

$\frac{\sqrt{5}}{4}$

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$\frac{\sqrt{5}}{2}$

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$\frac{5}{3}$

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$\frac{10}{3}$

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Solution

The correct option is D
$\frac{10}{3}$

Here the ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ .

Latus rectum = $\frac{2 b^{2}}{a} = \frac{2.5}{3} = \frac{10}{3}$ .

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Similar questions

The length of the latus rectum of the ellipse $5 x^{2} + 9 y^{2} = 45$ is

4 x^{2} + 9 y^{2} = 36

4 x^{2} + 9 y^{2} = 1

5 x^{2} + 4 y^{2} = 1

4 x^{2} + 9 y^{2} + 8 x + 36 y + 4 = 0

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