Y^2 = X^2(1+X)/(1-X)

edit: post cleared.

 

ya sorry i noted the mistake...but is it really the graph u want?

No

d ans is wrong

 

check again as when x -> 1   Y -> infinity

look if you put a value less than -1 for x the graph does not exist becoz y^2 becomes -ve so the graph exists b/w 0 and -1 in the second or third qd.
now the graph is symmetrical about the x axis. so what ever be the graph in 2nd and 3rd qd. . you can put limits from -1 to 0 for dx and integrate it and then double it.
that would be the required area.