It is known that F1 and F2 are the upper and lower focus of ellipse C1: y ^ / A ^ 2 + x ^ 2 / b ^ 2 = 1 respectively, where F1 is also the focus of parabola x ^ 2 = 4Y, and point m is C1, C2 is at the It is known that F1 and F2 are the upper and lower focus of the ellipse C1: y ^ / A ^ 2 + x ^ 2 / b ^ 2 = 1, where F1 is also the focus of the parabola x ^ 2 = 4Y, and point m is C1 and C2 in the second image And MF2 = 5 / 3 1. Find the equation of ellipse C1

It is known that F1 and F2 are the upper and lower focus of ellipse C1: y ^ / A ^ 2 + x ^ 2 / b ^ 2 = 1 respectively, where F1 is also the focus of parabola x ^ 2 = 4Y, and point m is C1, C2 is at the It is known that F1 and F2 are the upper and lower focus of the ellipse C1: y ^ / A ^ 2 + x ^ 2 / b ^ 2 = 1, where F1 is also the focus of the parabola x ^ 2 = 4Y, and point m is C1 and C2 in the second image And MF2 = 5 / 3 1. Find the equation of ellipse C1

F1 (0,1) ∵ the intersection point of quadrant A is set as a (x1, Y1) x1 ＜ 0, Y1 ＞ 0 parabola definition | AF1 | = Y1 + P / 2 = Y1 + 1 = 5 / 3, Y1 = 2 / 3x1 ^ 2 = 4Y1 = 8 / 3, 8 / (3b ^ 2) + 4 / (9a ^ 2) = 1, a ^ 2-B ^ 2 = 1, B ^ 2 = 3, a ^ 2 = 4Y ^ 2 / 4 + x ^ 2 / 3 = 1
It is known that the parabola C1: x ^ 2 + by = B ^ 2 passes through the two focal points of the ellipse C2: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0)
2. Let Q (3, b), m and n be the intersection points of C1 and C2 which are not on the y-axis. If the center of gravity of the triangle QMN is on the parabola C1, the equations of C1 and C2 are solved
1) The focus coordinate of ellipse is f (positive and negative C, 0), which is substituted into C2 equation to get C ^ 2 = B ^ 2 = a ^ 2-C ^ 2, so the eccentricity = C / a = √ 2 / 22 > from (1) to get a ^ 2 = 2B ^ 2, the combination of the changed ellipse equation and parabola is 2Y ^ 2-by ^ 2-B ^ 2 = O to get y = - B / 2 or B (it should be omitted from the figure), so x = positive and negative (√ 6 / 2) B
(1) Because the parabola C1 passes through the two focuses F1 (-, C, 0), F2 (C, 0) of the ellipse C2,
So C2 + B × 0 = B2, that is, C2 = B2, from A2 = B2 + C2 = 2c2
The eccentricity of ellipse C2 is 22
(2) It can be seen from (1) that A2 = 2B2, and the equation of ellipse C2 is as follows:
x22b2+y2b2=1
The equation x2 + by = B2 of simultaneous parabola C1 is: 2y2-by-b2 = 0,
The solution is y = - B2 or y = B (rounding)... Expansion
(1) Because the parabola C1 passes through the two focuses F1 (-, C, 0), F2 (C, 0) of the ellipse C2,
So C2 + B × 0 = B2, that is, C2 = B2, from A2 = B2 + C2 = 2c2
The eccentricity of ellipse C2 is 22
(2) It can be seen from (1) that A2 = 2B2, and the equation of ellipse C2 is as follows:
x22b2+y2b2=1
The equation x2 + by = B2 of simultaneous parabola C1 is: 2y2-by-b2 = 0,
The solution is y = - B2 or y = B (rounding off), so x = ± 62b,
M (- 62b, - B2), n (62b, - B2), so the barycenter coordinates of △ QMN are (1,0)
Because the center of gravity is on C1, so 12 + B × 0 = B2, B = 1
So A2 = 2
So the equation of parabola C1 is: x2 + y = 1,
The equation of ellipse C2 is: X22 + y2 = 1
It is known that the ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), the right focus F2 coincides with the focus (1,0) of the parabola C2: y ^ 2 = 4x, and C1 and C2 intersect at a point P
X ^ 2 is the square of X. it's urgent. Please help me~
Given that PF2 = 5 / 3, the center T of circle C3 is a moving point on C2, and that C3 intersects Y axis at Mn, / Mn / = 4, it is proved that when t moves, circle C3 always passes a certain point on C1?
First of all, draw a graph, t point is on the parabola, so we can set t point as (T ^ 2 / 4, t), Mn chord length as 4, we use t to make a vertical line to the Y axis, t is the center of the circle, then the vertical line must bisect the chord, let C3 radius be r, then R ^ 2 = 2 ^ 2 + (T ^ 2 / 4) ^ 2.. ①, let C3 equation be (x-t ^ 2 / 4) ^ 2 + (Y-T) ^ 2 = R ^ 2, replace ① with
In order to know the quadratic equation 2x ^ 2 + ax-2a + 1 = 0 about X, the sum of the squares of two real roots is 29 / 4, the value of a is obtained
analysis,
According to Veda's theorem,
x1+x2=-a/2【1】
x1*x2=(1-2a)/2【2】
In addition, X1 & # 178; + x2 & # 178; = (x1 + x2) &# 178; - 2x1 * x2 = 29 / 4
【1】 And [2] Dai Ren Shang Shi,
The solution is a = 3, or a = - 11
When a = - 11, the surrogate equation,
2x²-11x+23=0
Δ = 11 & # 178; - 23 × 8 ＜ 0, not real root, rounding off
∴a=3.
x1²+x2²=29/4
(x1+x2)²-2x1x2=29/4
a²/4-2(1-2a)/2=29/4
a²-8a-33=0
a1=11; a2=-3
When a = 11, Δ
The solution of the inequality f (y) = f (x / y) = f (x / y) is known
∫ f (x / y) = f (x) - f (y), f (2) = 1 take x = 4, y = 2 ∫ f (2) = f (4 / 2) = f (4) - f (2) ∫ f (4) = 2F (2) = 2 take x = 8, y = 2 ∫ f (4) = f (8 / 2) = f (8) - f (2) ∫ f (8) = f (4) + F (2) = 3f (x / y) = f (x) - f (y), that is, f (x / y) + F (y) = f (x) let X / y = m, y = n, x = Mn, so f (m) + F (n) = f (MN) inequality f (x