Let the eccentricity of the ellipse x2a2 + y2b2 = 1 (a > b > 0) be e, the right focus f (C, 0), and the two real roots of the equation AX2 + bx-c = 0 be X1 and X2 respectively, then the point P (x1, x2) () A. Must be in the circle x2 + y2 = 1 B. must be on the circle x2 + y2 = 1 C. must be outside the circle x2 + y2 = 1 D. the relation with x2 + y2 = 1 is related to E

The two real roots of the equation AX2 + bx-c = 0 are X1 and X2 respectively. According to Weida's theorem, it is obtained that X1 + x2 = - BA, x1x2 = - Ca, X12 + X22 = (x1 + x2) 2-2x1x2 = b2a2 + 2ca = B2 + 2aca2 = B2 + 2acb2 + C2 > 1, and the point P (x1, x2) is outside the circle x2 + y2 = 1

If the eccentricity of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 1 / 2 and the right focus is f (C, 0), the two real roots of equation AX ^ 2 + bx-c are X1 and X2 respectively,
Then point P (x1, x2)
A must be in x ^ 2 + y ^ 2 = 2
B must be on x ^ 2 + y ^ 2 = 2
C must be outside x ^ 2 + y ^ 2 = 2
D all three situations are possible

E = C / a = 1 / 2, C = A / 2, a ^ 2-B ^ 2 = C ^ 2: A ^ 2-B ^ 2 = a ^ 2 / 4, B ^ 2 = 3A ^ 2 / 4, the two real roots of the equation AX ^ 2 + bx-c are X1 and x2z respectively: X1 + x2 = - B / ax1x2 = - C / a = - 1 / 2x1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2 = B ^ 2 / A ^ 2 + 1 = 3 / 4 + 1 = 7 / 4

Let the eccentricity of the ellipse X & # 178 / A & # 178; + Y & # 178 / / B & # 178; = 1 be e, and the two real roots of the right focus f (C, 0) equation AX & # 178; + bx-c = 0
For x1, X2, point P (x1, x2). The answer is that it must be outside the circle X & # 178; + Y & # 178; = 1. Why, how can it come out of the circle?

x1²+x2²=(x1+x2)²-2x1x2
=(-b/a)²-2(-c/a)
=(b²+2ac)/a²
=(a²+2ac-c²)/a²
=1+c(2a-c)/a²>1
So, point P is outside the circle

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