Let E : x^2 a ^2 + y^2 ~b ^2 =1, a b and H : x^2 ~A ^2 - y^2 ~B ^2 =1 . Let the distance between the foci of E and the foci of H be 2 3 . If a-A=2 , and the ratio of the eccentricities of E and H is 1 3 , then the sum of the lengths of their latus rectums is equal to:

Step-by-step solution

aligned & x ^2 a ^2 + y ^2 ~b ^2 =1 foci are ( ae , 0) and (- ae , 0) & x ^2 ~A ^2 + y ^2 ~B ^2 =1 foci are ( Ae ^ , 0 ) and (- Ae ^ , 0 ) & 2 ae =2 3 ae = 3 & and 2 Ae ^ =2 3 Ae ^ = 3 & ae = Ae ^ e e ^ = A a & 1 3 = A a a =3 ~A aligned Now a - A =2 a - a 3 -2 a =3 and A =1 Ae = 3 e = 1 3 and e ^ = 3 aligned & b ^2= a ^2 (1- e ^2 ) & b ^2=6 aligned and B ^2= A ^2 ( ( e ^ )^2-1 )=(2) B ^2=2 sum of LR = 2 ~b ^2 a + 2 ~B ^2 ~A =8