It is known that the right focus of the ellipse C1: x2 / A2 + Y2 / B2 = 1 (a greater than b greater than 0) is f, and the upper vertices a and P are any points on C1, Mn Is a diameter of the circle C2: x2 + (Y-3) 2 = 1, if the line L parallel to AF and with an intercept of 3-radical 3 on the Y axis is just tangent to the circle C2 1: Find the eccentricity of C1 2: If the maximum value of vector PM multiplied by vector PN is 49, the equation of C1 is obtained 3: If the line L passing through the right focus F of the ellipse C1 is called the point s, t of the ellipse C1, intersecting the Y axis at the R point, the vector rs = V1 multiplied by the vector SF, the vector RT = V2 multiplied by the vector TF, it is proved that V1 + V2 is the fixed value

It is known that the right focus of the ellipse C1: x2 / A2 + Y2 / B2 = 1 (a greater than b greater than 0) is f, and the upper vertices a and P are any points on C1, Mn Is a diameter of the circle C2: x2 + (Y-3) 2 = 1, if the line L parallel to AF and with an intercept of 3-radical 3 on the Y axis is just tangent to the circle C2 1: Find the eccentricity of C1 2: If the maximum value of vector PM multiplied by vector PN is 49, the equation of C1 is obtained 3: If the line L passing through the right focus F of the ellipse C1 is called the point s, t of the ellipse C1, intersecting the Y axis at the R point, the vector rs = V1 multiplied by the vector SF, the vector RT = V2 multiplied by the vector TF, it is proved that V1 + V2 is the fixed value

(1) The equation of line L is BX + cy - (3 -) C = 0
Because the line L is tangent to the circle C2: x2 + (Y-3) 2 = 1, d = = 1
We can get 2c2 = A2, thus e =
(2) Let P (x, y), then · = (+) (+) = - = x2 + (Y-3) 2-1 = - (y + 3) 2 + 2c2 + 17 (- C ≤ y ≤ C),
Or let m (x1, Y1), n (X2, Y2), P (x, y), because X1 + x2 = 0, Y1 + y2 = 6, X12 + y12-6y1 + 8 = 0
So · = (x1-x) (x2-x) + (y1-y) (y2-y) = x2 + Y2 - (x1 + x2) x + (Y1 + Y2) y + x1x2 + y1y2 = x2 + Y2 + 6y-x12 + Y1
(6-y1)=x2+y2+6y+8=-(y+3)2+2c2+17.
① When C ≥ 3, (·) max = 17 + 2c2 = 49, the solution is C = 4, and the ellipse C1 is + = 1
② When 0 < C < 3, (·) max = - (- C + 3) 2 + 17 + 2c2 = 49,
The solution is C = 5-3, but (5-3) - 3 = - 6 > 0,
So 5-3 is more than 3, so C = 5-3 is omitted
To sum up, the equation of ellipse C1 is = 1

It is known that the right focus of the ellipse x2a2 + y2b2 = 1 (a > b > 0) is f, the intersection of the right guide line and the X axis is a, and there is a point P on the ellipse that satisfies the vertical bisector of the line AP passing through point F, then the value range of the eccentricity of the ellipse is______ ．

Because there is a point P on the ellipse satisfying that the vertical bisector of line AP passes through point F, the distance from point F to point P and point a is equal; because | FA | = A2C − C = B2C, | PF | ∈ [a-c, a + C], so B2C ∈ [a-C, a + C], we can get ac-c2 ≤ B2 ≤ AC + C2, that is, ac-c2 ≤ a2-c2 ≤ AC + C2, the solution is ca ≤ 1ca ≤− 1 or

Given that point m is the golden section point (am > MB) of line AB, if AB = 6, then am=

According to the definition of golden section: am = {(root 5-1) / 2} × 6 = 3 root 5-3. & nbsp; & nbsp; & nbsp; or am = 6 × 0.618 = 3.708

If M is the golden section of line AB, MB = 3cm, then AB = ()

There are two answers: one is that MB is a longer segment of AB, the other is that MB is a shorter segment of ab