The focus of ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) is f (1,0), and the eccentricity is 1 / 2 Suppose that the ellipse C passing through point F intersects two points m and N, and the vertical bisector of line Mn intersects Y axis at point P (0, Y0)

The focus of ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) is f (1,0), and the eccentricity is 1 / 2 Suppose that the ellipse C passing through point F intersects two points m and N, and the vertical bisector of line Mn intersects Y axis at point P (0, Y0)

The results show that C = 1, e = C / a = 1 / 2, a = 2, B ^ 2 = 4-1 = 3
So the elliptic equation is x ^ 2 / 4 + y ^ 2 / 3 = 1
Please see the answer to question 18 of Baidu Library

If an ellipse and the ellipse x2 / A2 + Y2 / B2 = 1 (a > 0, b > 0) are in common focus, then its equation can be set as x2 / M + Y2 / [M - (A2-B2)] = 1 (M > A2-B2)
If the focus is on the y-axis, does the confocal elliptic equation just need to change the positions of x2 and Y2?
② How to deduce this conclusion? Please prove it in detail

To change the position, we usually take a as the semi major axis, that is to say, a is greater than B, and the focus is on the Y axis. Then the standard equation of the ellipse is Y2 / A2 + x2 / B2 = 1. Of course, the confocal elliptic equation also needs to exchange the positions of x2 and Y2

It is known that the center is at the origin of the coordinate, the eccentricity of the ellipse C of the foci F 1 and F 2 on the x-axis is (√ 3) / 2, and the focus of the parabola x ^ 2 = 4Y is a vertex of the ellipse
(1) Solving the equation of ellipse C
(2) Given that the two intersections of the line L passing through the focus F2 and the ellipse C are a and B, and | ab | = 3, find | AF1 | + | BF2|

If the focus of parabola X & # 178; = 4Y is (0,1), then the minor axis of ellipse B = 1C / a = √ 3 / 2C & # 178; = 3 / 4A & # 178; C & # 178; + B & # 178; = A & # 178; a & # 178; = 4A = 2, C = √ 3 ellipse: X & # 178; = 4 + Y & # 178; = 1f2 (√ 3,0) e = C / a = √ 3 / 2, let a (x1, Y1) B (X2, Y2) line AB: y = K (X -...)

It is known that the eccentricity of the ellipse with the center at the origin and the focus on the x-axis is e = e = √ 2 / 2, passing through the focus of the parabola x ^ 2 = 4Y
If the line L (slope not equal to zero) passing through point B (2,0) intersects the ellipse at two different points e, f (E is between B and F), try to find the value range of the area ratio of △ OBE and △ oBf

It is known that the eccentricity of the ellipse with the center at the origin and the focus on the x-axis is e = √ 2 / 2, passing through the focus of the parabola x ^ 2 = 4Y
The solution is a = √ 2, B = 1, C = 1,
The equation of the ellipse is X & # 178; / 2 + Y & # 178; = 1,
We know that the slope of L exists and is not zero,
Let l equation be y = K (X-2) (K ≠ 0) = 1
X & # 178; / 2 + Y & # 178; = 1
(2k & # 178; + 1) x2-8k & # 178; &# 8226; X + (8K & # 178; - 2) = 0, from △ 0 to 0 < K & # 178; < 1 / 2
Let e (x1, Y1), f (X2, Y2), X1 + x2 = 8K and 178; / 2k and 178; + 1, x1x2 = 8K and 178; - 2 / 2k and 178; + 1,
Let λ = s △ OBEs △ oBf,
Be = λ &; BF, λ = x1-2 / x2-2, and 0 ＜λ＜ 1
(x1-2)+(x2-2)=-4／1+2k²,
(x1-2)•(x2-2)=x1x2-2(x1+x2)+4=2／1+2k²．
∴ λ／(1+λ)²=2k²+1／8,
k²=4λ(1+λ)²-1／2．
∵ 0＜k²＜1／2,∴ 0＜4λ／(1+λ)²-1／2＜1／2,
3-2√2＜λ＜3+2√2．
And ∵ 0 ＜λ＜ 1, ∵ 3-2 √ 2 ＜λ＜ 1,
The value range of the area ratio of △ OBE to △ oBf is (3-2 √ 2,1)