The major axis length of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 4, F1 and F2 are its left and right focus respectively, and the focus of parabola y2 = - 4x is F1 The line L passing through the focus F1 intersects with the ellipse at p.q. the maximum area of the triangle f2pq is calculated

The major axis length of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 4, F1 and F2 are its left and right focus respectively, and the focus of parabola y2 = - 4x is F1 The line L passing through the focus F1 intersects with the ellipse at p.q. the maximum area of the triangle f2pq is calculated

2a=4,a=2
F1 (- 1,0), that is, C = 1
b^2=a^2-c^2=3
C:x^2/4+y^2/3=1
When l is perpendicular to the x-axis, take the maximum area, PQ coordinates x = - 1, y = + - 1.5
PQ=2*1.5=3
S(F2PQ)=3*2/2=3