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 an its nearest vertex on the major axis is $${3 \over 2}$$ units, then its eccentricity is :

Accepted Answer

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a - ae = $${3 \over 2}$$ (given)

$$ \Rightarrow $$ $$\,\,\,$$ a (1 - e) = $${3 \over 2}$$

Length of latus rectum,

$${{2{b^2}} \over a} = 4$$

$$ \Rightarrow $$ $$\,\,\,$$ b2 = 2a

$$ \Rightarrow $$ $$\,\,\,$$ a2(1 - e2) = 2a   [As b2 = a2 (1 - e2)]

$$ \Rightarrow $$ $$\,\,\,$$ a (1 - e) ( 1 + e) = 2

Putting    a (1 - e) = $${3 \over 2}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${3 \over 2}$$ (1 + e) = 2

$$ \Rightarrow $$ $$\,\,\,$$ 3 + 3e = 4

$$ \Rightarrow $$ $$\,\,\,$$ e = $${1 \over 3}$$

Solutions · 1