JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 20)
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 :
$${1 \over 2}$$
$${1 \over 3}$$
$${2 \over 3}$$
$${1 \over 9}$$
Explanation
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 = $${1 \over 3}$$
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 = $${1 \over 3}$$
Comments (0)
