If the left and right focus of the ellipse X & # 178 / 4 + Y & # 178 / 3 = 1 is F1F2 and P (x, y) is on the ellipse, then the value range of vector Pf1 * vector PF2?

If the left and right focus of the ellipse X & # 178 / 4 + Y & # 178 / 3 = 1 is F1F2 and P (x, y) is on the ellipse, then the value range of vector Pf1 * vector PF2?

F1 (- 1,0); F2 (1,0)
∵P(x,y)
The vector Pf1 = (- 1-x, 0-y) = (- 1-x, - y)
Vector PF2 = (1-x, 0-y) = (1-x, - y)
The vector Pf1 * vector PF2 = (- 1-x) × (1-x) + (- y) × (- y) = x & # 178; - 1 + Y & # 178;
From X & # 178 / 4 + Y & # 178 / 3 = 1, Y & # 178; = 3 - (3 / 4) x & # 178;
The vector Pf1 * vector PF2 = (- 1-x) × (1-x) + (- y) × (- y) = x & # 178; - 1 + Y & # 178;
=x²-1+3-(3/4)x²
=(1/4)x²+2 (-2≤x≤2)
∴0≤x²≤4
So the range of vector Pf1 * vector PF2 is [2,3]

It is known that the left and right focal points of the ellipse x ^ 2 / 4 + y ^ 2 = 1 are F1F2, and the point P is on the ellipse. When the area of the triangle f1pf2 is 1, the vector Pf1 * vector PF2=

If the distance between the 3P point and the y-axis is | y | y 124; y 124; y 124; then s \124; f1pf2 = \\\\\\\\\\\\\| (189; \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ / 3-3 + 1 / 3 = 0, Pf1 ⊥ PF2