Let the ellipse ( E ₁: x^2 a ^2 + y^2 ~b ^2 =1, a b ) and ( E ₂: x^2 ~A ^2 + y^2 ~B ^2 =1, ~A B ) have same eccentricity ( 1 3 ). Let the product of their lengths of latus rectums be ( 32 3 ), and the distance between the foci of (E₁ ) be 4. If (E₁ ) and (E₂ ) meet at (A, B, C ) and (D ), then the area of the quadrilateral (A B C D ) equals :

Step-by-step solution

aligned & 2 a e=4 & a=2 3 & 1- b^2 12 = 1 3 b^2=8 & & 2 b^2 a 2 A^2 B = 32 3 & & 2 8 2 3 2 A^2 B = 32 3 & & A^2 B =2 A^2=2 B & & 1- A^2 B = 1 3 & & B=3 A^2=6 aligned E₁: x^2 12 + y^2 8 =1 ....(i) E₁: x^2 6 + y^2 9 =1 ...(ii) On solving (i) & (ii) aligned &(x, y)= ( 6 5 , 6 5 ), ( - 6 5 , 6 5 ), ( 6 5 , -6 5 ), & ( - 6 5 , -6 5 ) aligned Four points are vertices of rectangle area = 24 6 5