Let the right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 be f (C, 0) and a = 2C. Where are the two real roots of the equation AX ^ 2 + bx-c = 0 (x1, x2)

Let the right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 be f (C, 0) and a = 2C. Where are the two real roots of the equation AX ^ 2 + bx-c = 0 (x1, x2)

e=c/a=1/2 => b/a=√3/2
X1 and X2 are the two real roots of the equation AX ^ 2 + bx-c = 0, which satisfy the Weida theorem
x1+x2=-b/a=-√3/2,x1x2=-c/a=-1/2
So X1 & # 178; + x2 & # 178; = (x1 + x2) &# 178; - 2x1x2 = 3 / 4 + 1 = 7 / 4

Let the eccentricity of the ellipse be 1 / 2 and the two real roots of the right focus f (C, 0) equation AX ^ 2 + bx-c = 0 be x1, x2. Then p (x1, X20) must be on the circle x ^ 2 + y ^ 2 = 2?
Let the eccentricity e = 1 / 2 of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), the right focus f (C, 0), and the two real roots of the equation AX ^ 2 + bx-c = 0 be X1 and X2 respectively, then the point P (x1, x2) ()
A. Must be in the circle x ^ 2 + y ^ 2 = 2. B. must be on the circle x ^ 2 + y ^ 2 = 2
C. It must be outside the circle x ^ 2 + y ^ 2 = 2 d. The above three situations are possible
E = root (1-b2 / A2) is this a formula? Why haven't I seen it? Can you prove it for me

Let e = CA = 12, let Ba = 32, let x1, X2 be the two real roots of the equation AX2 + bx-c = 0, let X1 + x2 = - BA = - 32, x1x2 = - Ca = - 12, so X12 + X22 = (x1 + x2) 2-2x1x2 = 34 + 1 = 74 ＜ 3, and let P (x1, x2) be in the circle x2 + y2 = 3