Let the ellipses E₁: x^2 a^2 + y^2 b^2 = 1 ( a > b ) and E₂: x^2 b^2 + y^2 a^2 = 1 intersect at four points. If the distance between the foci of E₁ is 2 3 and the area of the quadrilateral formed by their points of intersection is 144 7 , then the eccentricity of E₁ is :

Step-by-step solution

By symmetry, the points of intersection of E₁ and E₂ in the first quadrant lie on the line y = x . Substituting y = x into the equation of E₁ , we get: x^2 ( 1 a^2 + 1 b^2 ) = 1 x^2 = a^2b^2 a^2+b^2 The quadrilateral formed by the four symmetric intersection points is a rectangle (and a square in this case). Its area is: 4x^2 = 4a^2b^2 a^2+b^2 = 144 7 a^2b^2 a^2+b^2 = 36 7 The distance between the foci of E₁ is 2ae = 2 3 , which gives ae = 3 , so a^2e^2 = 3 . Using the eccentricity relation b^2 = a^2(1 - e^2) = a^2