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Let S and S^' be the foci of an ellipse and B be any one of the extrmities of its minor axis. If $\Delta {S}'BS$ is a right angled triangle with right angle at B and area $\left ( \Delta {S}'BS \right )=8\: Sq.$ units, then the length of a latus rectum of the ellipse is :
Option 1)
$4$

Option 2)
$2$
Option 3)
$2\sqrt{2}$
Option 4)
$4\sqrt{2}$

Answers (1)

best_answer

Foci of Ellipse -

The two fixed points on the ellipse.

- wherein

Sum of focal distance -

$2a$

- wherein

$=4$ $d=\sqrt{3}\left ( x-25 \right )$

$tan60=\frac{x+25}{d}$

$\Rightarrow d=\frac{x+25}{\sqrt{3}}$

$\sqrt{3}\left ( x-25 \right )=\frac{x+25}{\sqrt{3}}$

$\Rightarrow x=50m$

$\Delta BS{S}'=\frac{1}{2}\left ( S{S}' \right )\left ( OB \right )=8$

$\Rightarrow ae..b=8\Rightarrow b^{2}=8\Rightarrow b=2\sqrt{2}$

$\Delta BS{S}'=\frac{1}{2}\: \left ( BS \right )\: \left ( B{S}' \right )=8$

$=\frac{1}{2}(BS)^{2}=\frac{1}{2}a^{2}=8$

$\Rightarrow a=4$

$\therefore e=\frac{1}{\sqrt{2}}$

$length\: of \: lotus\: rectum\: =\frac{2b^{2}}{a}=4$ ???????

Option 1)
$4$

Option 2)

$2$

Option 3)

$2\sqrt{2}$

Option 4)

$4\sqrt{2}$

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