If the length of the latus rectum of an ellipse is 4 units and the distance between a focu…

Question

If the length of the latus rectum of an ellipse is $4$ units and the distance between a focus and its nearest vertex on the major axis is $\dfrac32$ units, then its eccentricity is?

Accepted Answer

$a(1-e)=\dfrac32$ and $\dfrac{2b^2}a=4\Rightarrow b^2=2a$.

$$e^2=1-\frac{b^2}{a^2}=1-\frac2a.$$

From $a-\dfrac32=ae$ and $e^2=1-\dfrac2a$, substituting and solving gives $a=\dfrac94$, so
$$e^2=1-\frac2{9/4}=1-\frac89=\frac19\ \Longrightarrow\ e=\frac13.$$

Solutions · 1