If the eccentricity of an ellipse is 5/8

Question:

If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.

Solution:

We know that equation of an ellipse

$=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Also we have

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

Length of minor axis $=2 \mathrm{~b}$

$e=\frac{5}{8}$

We know that foci $=(\pm a e, 0)$

Given that distance between foci $=10$

$2 \mathrm{ae}=10$

$2 \times a \times \frac{5}{8}=10$

$a=8$

$b^{2}=a^{2}\left(1-e^{2}\right)$

$b^{2}=8^{2}\left(1-\left(\frac{5}{8}\right)^{2}\right)$

$b^{2}=8^{2}\left(1-\left(\frac{5}{8}\right)^{2}\right)$

On simplifying we get

$b^{2}=64\left(1-\frac{25}{64}\right)$

$b^{2}=64 \times \frac{39}{64}=39$

Length of Latus Rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2 \times \frac{39}{8}=9.75$

Hence, the length of latus rectum is $9.75$ units.

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