JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 2)
In an ellipse, with centre at the origin, if the
difference of the lengths of major axis and minor
axis is 10 and one of the foci is at (0,5$$\sqrt 3$$), then
the length of its latus rectum is :
5
8
10
6
Explanation
Focus (0, be) = (0, 5$$\sqrt 3$$)
$$ \therefore $$ be = 5$$\sqrt 3$$
$$ \Rightarrow $$ b2e2 = 75
As here b > a
so e2 = $$1 - {{{a^2}} \over {{b^2}}}$$
$$ \therefore $$ b2$$\left( {1 - {{{a^2}} \over {{b^2}}}} \right)$$ = 75
$$ \Rightarrow $$ b2 - a2 = 75 .......(1)
Given that,
difference of the lengths of major axis and minor axis is 10.
$$ \therefore $$ 2b - 2a = 10
$$ \Rightarrow $$ b - a = 5 .......(2)
From (1),
(b + a)(b - a) = 75
$$ \Rightarrow $$ (b + a)5 = 75
$$ \Rightarrow $$ (b + a) = 15 .......(3)
From (2) and (3) we get,
b = 10, a = 5
$$ \therefore $$ Length of its latus rectum = $${{{2{a^2}} \over b}}$$
= $${{2 \times 25} \over {10}}$$ = 5
$$ \therefore $$ be = 5$$\sqrt 3$$
$$ \Rightarrow $$ b2e2 = 75
As here b > a
so e2 = $$1 - {{{a^2}} \over {{b^2}}}$$
$$ \therefore $$ b2$$\left( {1 - {{{a^2}} \over {{b^2}}}} \right)$$ = 75
$$ \Rightarrow $$ b2 - a2 = 75 .......(1)
Given that,
difference of the lengths of major axis and minor axis is 10.
$$ \therefore $$ 2b - 2a = 10
$$ \Rightarrow $$ b - a = 5 .......(2)
From (1),
(b + a)(b - a) = 75
$$ \Rightarrow $$ (b + a)5 = 75
$$ \Rightarrow $$ (b + a) = 15 .......(3)
From (2) and (3) we get,
b = 10, a = 5
$$ \therefore $$ Length of its latus rectum = $${{{2{a^2}} \over b}}$$
= $${{2 \times 25} \over {10}}$$ = 5
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