One of extremity of minor of ellipse is `B`. `S` and `S\'` are focii of ellipse then area of right angle `/_\SBS\'` is equal to 8. Then latus rectum of ellipse is equal to

Options

A. `3 sqrt(2)`
B. `4 sqrt(2)`
C. 2
D. 4

Answer

D

Explanation

Let the ellipse be `(x^(2))/(a^(2))+y^(2))/(b^(2))=1.`
then according to given information , we have the following figure .

Clearly , slope of line`SB=(b)/(-ae)` and slope of line `S'b=(b)/(ae)=-1`
`:' ` Lines SB and S'B are perpendicular , so
`((b)/(-ae)).((b)/(ae))=-1`
`[:' " product of slope of two perpendicular lines is "(-1)]`
`implies b^(2)=8`
also it is given that area of `Delta s' BS=8`
`therefore (1)/(2) a^(2)=8`
`[:' s'B=SB= alpha ` because s'b+SB = 2a and S'B =SB]
`implies a^(2)=16implies A=4`
`:' e^(2)=1-(b^(2))/(a^(2))=1-e^(2))`[ from Eq.(i)]
`implies 2e^(2)=1`
`implies 2e^(2)=1`
`impliese^(2)=(1)/(2)`
from Eqs. (i) and (iii) we, get
`b^(2)=a^(2)((1)/(2))=16((1)/(2))`[ using Eq.(ii)]
`implies b^(2)=8`
Now , length of latus rectum =`(2b^(2))/(alpha)=(2xx8)/(4)=4` units