Get Answers to all your Questions

header-bg qa

Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0, 0).

Answers (1)

Let the equation of the ellipse be \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

e=\frac{2}{3}

Given that latus rectum \frac{2b^{2}}{a}=5

b^{2}=\frac{5a}{2}=a^{2}\left ( 1-e^{2} \right )

\frac{5}{2}=a\left ( 1-\left ( \frac{2}{3} \right )^{2} \right )

a=\frac{9}{2}

b^{2}=\frac{45}{4}

The required equation of the ellipse is\frac{4x^{2}}{81}+\frac{4y^{2}}{45}=1

Posted by

infoexpert22

View full answer