Let F and F ^1 be the foci of the ellipse x^2 4 + y^2 b^2 =1(b 2) and B is one end of the minor axis. If the area of the triangle FBF ^1 is 3 sq. units, then the eccentricity of the ellipse is

Step-by-step solution

Given equation of ellipse is x^2 4 + y^2 b^2 =1 and (b 2) so, F(c, 0), F^ (-c, 0) and B(0, b) Area of F B F^ = 1 2 2 c b 3 =b c b^2 c^2=3 c^2= 3 b^2 ....(i) We know that B F+B F^ =2 a b^2+c^2 + b^2+c^2 =2 a2 b^2+c^2 =2 a b^2+c^2 =ab^2+c^2=a^2=4 b^2+ 3 b^2 =4 b^4-4 b^2+3=0 (b^2-1 ) (b^2-3 )=0 So, b=1 c= 3 or b= 3 c=1 when c= 3 a e= 3 e= 3 2 or c=1 a e=1 e= 1 2