F1 (- 3,0) and F2 (3,0) are the two focal points of the ellipse x ^ 2 / A + y ^ 2 / b = 1. When p is on the ellipse, ∠ f1pf2 = 120 °, the triangle f1pf2 has the largest area, a + B I have worked out that s = √ 3B doesn't seem to have a maximum?

F1 (- 3,0) and F2 (3,0) are the two focal points of the ellipse x ^ 2 / A + y ^ 2 / b = 1. When p is on the ellipse, ∠ f1pf2 = 120 °, the triangle f1pf2 has the largest area, a + B I have worked out that s = √ 3B doesn't seem to have a maximum?

Point P must be the upper or lower vertex of the ellipse. So the coordinates of point P are (radical 3,0) or (- radical 3,0) brought into the elliptic equation, B = 3. According to the definition of ellipse: the sum of the distances from the point on the ellipse to F1 and to F2 is constant, here = Pf1 + PF2 = 4, radical 3, so that the rightmost point (x, 0) is (x + radical 3) + (x-radical)