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 (3/2) units, then its eccentricity is :

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 (3/2) units, then its eccentricity is : (A) (1/2) (B) (1/3) (C) (2/3) (D) (1/9)

Tardigrade Admin · Accepted Answer

Given: (2 b2/a)=4 ⇒ b2=2 a b2=a2(1-e2) a(1-e)=(3/2) So 2=a(1-e)(1+e) ⇒ 2=(3/2)(1+e) ⇒ 4=3+3 e ⇒ e=(1/3)

Q. 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 $\frac{3}{2}$ units, then its eccentricity is :

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{9}$

Given: $\frac{2 b ^{2}}{a} = 4$
$\Rightarrow b^{2} = 2 a$
$b^{2} = a^{2} (1 - e^{2})$
$a (1 - e) = \frac{3}{2}$
So $2 = a (1 - e) (1 + e) \Rightarrow 2 = \frac{3}{2} (1 + e)$
$\Rightarrow 4 = 3 + 3 e$
$\Rightarrow e = \frac{1}{3}$

Q. 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 23​ units, then its eccentricity is :

21​

31​

32​

91​