The quadratic function f (x) = ax ^ 2 + BX + C satisfies the following conditions 1) F (- 1) = f (- 3); 2) f (x) has a minimum value of 3; 3) f (x) image passes through point (0,4), find the analytic expression of function f (x)

The quadratic function f (x) = ax ^ 2 + BX + C satisfies the following conditions 1) F (- 1) = f (- 3); 2) f (x) has a minimum value of 3; 3) f (x) image passes through point (0,4), find the analytic expression of function f (x)

F (x) image through the point (0,4), get: 0 + 0 + C = 4, C = 4, that is, f (x) = ax ^ 2 + BX + 4f (- 1) = f (- 3), get: symmetry axis is: x = (- 1-3) / 2 = - 2-B / 2A = - 2, B = 4A, that is, f (x) = ax ^ 2 + 4ax + 4 = a (x + 2) ^ 2 + 4-4a minimum value is 3, get: 4-4a = 3, a = 1 / 4, so f (x) = (1 / 4) x ^ 2 + X + 4

It is known that F1 and F2 are the focal points of the ellipse, P is any point on the ellipse, and the eccentricity of the ellipse is 1 / 3
It is known that F1 and F2 are the focus of the ellipse, P is any point on the ellipse, the eccentricity of the ellipse is 1 / 3, take P as the center of the circle, the length of PF2 as the radius to make the circle P, when the circle P is tangent to the X axis, the chord length obtained by cutting the Y axis is (12 root sign 55) / 9
Proof: no matter how the point P moves on the ellipse, there must be a definite circle tangent to the circle P. try to solve the equation of definite circle

Because the eccentricity is 1 / 3, let the elliptic equation be: x ^ 2 / (9K ^ 2) + y ^ 2 / (8K ^ 2) = 1. The coordinates of point F2 are (k, 0). Because P is the point on the ellipse, and the circle P is tangent to the X axis, and the circle P takes PF2 as the radius, so PF2 is perpendicular to the X axis

It is known that the left and right focal points of the ellipse are F1 and F2 respectively, and the eccentricity is e, if there is a point P on the ellipse
The left and right focus of the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 are F1 and F2 respectively, and the eccentricity is e. if there is a point P on the ellipse such that Pf1 / PF2 = e, then the value range of eccentricity of the ellipse is?
The point m is the point on X * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0). The circle centered on M is tangent to the X axis at the focal point F
If the triangle PQM is an obtuse triangle, calculate the value range of eccentricity of the ellipse

1. Let Pf1 = x, PF2 = y (x < y)
From the question x + y = 2A... ①
x/y=c/a ...②
y-x＜2c ...③
From (1) and (2), we get y = 2A ^ 2 / (a + C)... And (4)
① (3) y < A + c... 5
A ^ 2-C ^ 2-2ac < 0
Divide by a ^ 2
1-e^2-2e＜0
The solution is - √ 2-1 ＜ e or e ＞ 2-1
∵ 0＜e＜1
∴ √2-1＜e＜1

It is known that the left focus of the ellipse is f, the left and right vertices are AC, the upper vertex is B, and the circle P is made through FBC
It is known that the left focus of the ellipse x ^ 2 + y ^ 2 / b ^ 2 = 1 (b belongs to (0,1)) is f, the left and right vertices are AC, the upper vertex is B, and the circle P is made through three points of FBC, where the coordinates of the center P are (m, n)
(1) When m + n > 0, the range of eccentricity of ellipse is larger
(2) Can the line AB be tangent to the circle p

I can't upload pictures yet, so you can just make do with it
(1) Because e = C / A, a = 1, C = e, then f (- E, 0), B (0, b), C (1,0)
The distance from point P to three points is equal: (M + e) 2 + N2 = M2 + (N-B) 2 = (m-1) 2 + N2
We can get m = (1-e) / 2, n = (b2-e) / (2b)
From M + n > 0, E2 can be obtained