Let the product of the focal distances of the point ( 3 , 1 2 ) on the ellipse x^2 a^2 + y^2 b^2 =1,( a b ) , be 7 4 . Then the absolute difference of the eccentricities of two such ellipses is

Step-by-step solution

aligned & Product of focal distances = (a+e x₁ ) (a-e x₁ ) & =a^2-e^2 x₁^2=a^2-e^2(3) & =a^2-3 e^2= 7 4 a^2= 7 4 +3 e^2 & 4 a^2=7+12 e^2 & & ( 3 , 1 2 ) lines on x^2 a^2 + y^2 b^2 =1 & 3 a^2 + 1 4 b^2 =1 aligned aligned & 3 a ^2 + 1 4 ( a ^2 ) (1- e ^2 ) =1 & 12 (1- e ^2 )+1=4 a ^2 (1- e ^2 ) & 13-12 e ^2= (7+12 e ^2 ) (1- e ^2 ) & 13-12 e ^2=7-7 e ^2+12 e ^2-12 e ^4 & 12 e ^4-17 e ^2+6=0 & e ^2= 17 289-288 24 = 17 1 24 = 3 4 & 2 3 & e = 3 2 & 2 3 & difference == 3 2 - 2 3 = 3-2 2 2 3 aligned