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Let S and S' be the foci of the ellipse ...

Let S and S' be the foci of the ellipse and B be any one of the extremities of its minor axis. If `DeltaS'BS=8sq.` units, then the length of a latus rectum of the ellipse is

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The correct Answer is:
4
Area of `Delta S'BS = (1)/(2) xx SB xx S'B`
`(1)/(2)(sqrt((ae)^(2)+b^(2)))^(2)=8`
`(ar)^(2) + b^(2) = 16` …(i)
`(a)^(2) (1-(b^(2))/(a^(2))) + b^(2) + 16`
`a = 4 " "because e^(2) = 1 - (b^(2))/(a^(2))`
Now as `SB _|_S'B`
`m_(SB).m_(S'B) = -1 , " "(b)/(-ae)(b)/(ae) = -1`
`b^(2) = (ae)^(2)` ...(iii)
By equation (i) and (ii) `b^(2) = 8`
Length of latus rectum `= (2b^(2))/(a) = (2 xx 8)/(4) = 4`
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