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Let $x^2+y^2=20$ be the director circle of an ellipse $E$ whose major axis is $\mathrm{X}$-axis and minor axis is $\mathrm{Y}$-axis. If the length of the latus rectum of $\mathrm{E}$ is 2 , then the distance between its foci is
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The correct answer is:
$4 \sqrt{3}$
Equation of director circle ellipse is
$$
x^2+y^2=a^2+b^2 \Rightarrow a^2+b^2=20
$$
Given that lotus rectum $=\frac{2 b^2}{a}=2 \Rightarrow b^2=a$
from (i) and (ii) we get $a=4$
We have $c^2=a^2-b^2=16-4=12$
$\Rightarrow c=2 \sqrt{3}$
$\therefore$ Distance between foci $=2 c=4 \sqrt{3}$.
$$
x^2+y^2=a^2+b^2 \Rightarrow a^2+b^2=20
$$
Given that lotus rectum $=\frac{2 b^2}{a}=2 \Rightarrow b^2=a$
from (i) and (ii) we get $a=4$
We have $c^2=a^2-b^2=16-4=12$
$\Rightarrow c=2 \sqrt{3}$
$\therefore$ Distance between foci $=2 c=4 \sqrt{3}$.
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