JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 20)
If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12,
then the length of its latus rectum is :
$$\sqrt 3 $$
$$3\sqrt 2 $$
$${3 \over {\sqrt 2 }}$$
$$2\sqrt 3 $$
Explanation
Distance between foci = 2ae = 6
$$ \Rightarrow $$ ae = 3 .....(1)
Distance between directrices = $${{2a} \over e}$$ = 12
$$ \Rightarrow $$ $${a \over e}$$ = 6 .....(2)
from (1) and (2)
a2 = 18
also a2e2 = 9
$$ \Rightarrow $$ 18e2 = 9
$$ \Rightarrow $$ e2 = $${1 \over 2}$$
We know e2 = 1 - $${{{b^2}} \over {{a^2}}}$$
$$ \therefore $$ $${1 \over 2}$$ = 1 - $${{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow $$ b2 = 9
$$ \therefore $$ Length of latus rectum = $${{2{b^2}} \over a}$$
= $${{2 \times 9} \over {\sqrt {18} }}$$
= $$3\sqrt 2 $$
$$ \Rightarrow $$ ae = 3 .....(1)
Distance between directrices = $${{2a} \over e}$$ = 12
$$ \Rightarrow $$ $${a \over e}$$ = 6 .....(2)
from (1) and (2)
a2 = 18
also a2e2 = 9
$$ \Rightarrow $$ 18e2 = 9
$$ \Rightarrow $$ e2 = $${1 \over 2}$$
We know e2 = 1 - $${{{b^2}} \over {{a^2}}}$$
$$ \therefore $$ $${1 \over 2}$$ = 1 - $${{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow $$ b2 = 9
$$ \therefore $$ Length of latus rectum = $${{2{b^2}} \over a}$$
= $${{2 \times 9} \over {\sqrt {18} }}$$
= $$3\sqrt 2 $$
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