about to only mathematics

Options

A. `1-sqrt(2//3)` sq unit
B. `sqrt(3//2)-1` sq unit
C. `1+sqrt (2//3)` sq unit
D. `sqrt(3//2)+1` sq unit

Explanation

Given equation can be rewritten as focal chord
`((x-sqrt(2))^(2))/(4)-((y+sqrt(2))^(2))/(2)=1`
For point `A(x,y), e =sqrt(1+(2)/(4))=sqrt((3)/(2))`
`rArr x -sqrt(2)=2`
`rArr x=2 +sqrt(2)`

For point `C(x,y), x -sqrt(2)=ae=sqrt(6) rArr x =sqrt(6)+sqrt(2)`
Now, `AC=sqrt(6)+sqrt(2)-2-sqrt(2)=sqrt(6)-2`
`and BC = (b^(2))/(a)=(2)/(2)=1`
` therefore " Area of " triangle ABC = (1)/(2) xx (sqrt(6)-2) xx 1=sqrt((3)/(2))-1` sq unit