Given ellipse: two foci F1 and F2 of x ^ 2 / 2 + y ^ 2 = 1, point P (x0, Y0) satisfies 0

Given ellipse: two foci F1 and F2 of x ^ 2 / 2 + y ^ 2 = 1, point P (x0, Y0) satisfies 0

2 & lt; = Pf1 + PF2 & lt; 2 √ 2. Because the circle 0 & lt; x0 ^ 2 + Y0 ^ 2 & lt; 1 is completely contained in the ellipse (it has a junction with the minor axis of the ellipse), so Pf1 + PF2 & lt; 2A = 2 √ 2. When p approaches (0,1) or (0, - 1), the maximum value is taken. On the other hand, due to the triangular trilateral relationship, Pf1 + PF2 & gt; = | F1F2 | = 2. At this time, P is on the line F1F2, for example, P takes (0,0), The blue circle is the boundary of x ^ 2 + y ^ 2 & lt; 1. The red ellipse is x ^ 2 / 2 + y ^ 2 = 1, and the green ellipse is the line from (- 1,0) to (1,0). The green line - & gt; the black dotted line - & gt; the red line is actually equivalent to a family of ellipses with (- 1,0) and (1,0) as the focus (i.e., C = 1 as the fixed value) but the long axis gradually increases. Each ellipse represents the set of points (those black dotted lines) with Pf1 + PF2 as the fixed value, which expand from the inside to the outside, The smallest ellipse is b = 0, which is the line F1F2 (degenerate ellipse), which is the green line, which is the minimum value of Pf1 + PF2. The largest ellipse is b = 1, which is the ellipse x ^ 2 / 2 + y ^ 2 = 1, which is the red line. The red line is tangent to the boundary of the blue circle, so it is the maximum value of Pf1 + PF2

It is known that the two focal points of ellipse C: X * 2 / 2 + y * 2 = 1 are F1 and F2, and the point P (x0, Y0) satisfies 0 < x0 * 2 / 2 + Y0 * 2 < 1
Given that the two focal points of ellipse C: X * 2 / 2 + y * 2 = 1 are F1 and F2, and the point P (x0, Y0) satisfies 0 < x0 * 2 / 2 + Y0 * 2 < 1, then how many common points are the line x0 * 2 / 2 + y0y = 1 and ellipse C?
There's no common point. Why?
I get a saying: "the (x0, Y0) is inside the ellipse x ^ 2 / 2 + y ^ 2 = 1, and the points (x, y) on X * x0 / 2 + y * Y0 = 1 are outside the ellipse, so the line x0 * 2 / 2 + y0y = 1 and ellipse C have no common points."
Why (x0, Y0) is inside the ellipse x ^ 2 / 2 + y ^ 2 = 1? Why are the points (x, y) on X * x0 / 2 + y * Y0 = 1 outside the ellipse?

To judge that there are several common points between ellipse and straight line, just judge the symbol of delt. X ^ 2 / 2 + y ^ 2 = 1 and X * x0 / 2 + y * Y0 = 1 are (x0 ^ 2 + 2y0 ^ 2) x ^ 2-4x0 * x + 4-4y0 ^ 2 = 0delt = (- 4x0) ^ 2-4 (x0 ^ 2 + 2y0 ^ 2) * (4-4y0 ^ 2) = - 32y0 ^ 2 + 16x0 ^ 2y0 ^ 2 + 32y0 ^ 2 = 16y0 ^ 2 (x0 ^ 2 + 2y0 ^ 2-2) (x0, Y0

It is known that the focus of the ellipse is F1 (- 6,0), F2 (6,0), and the ellipse passes through the point P (5,2). (1) the standard equation of the ellipse is solved. (2) if the point m (x0, Y0) on the ellipse satisfies MF1 ⊥ MF2, the value of Y0 is obtained

(1) Let x2a2 + y2b2 = 1 (A & gt; B & gt; 0), and its half focal length C = 6. ∵ point P (5,2) is on the ellipse, ∵ 2A = | Pf1 | + | PF2 | = (5 + 6) 2 + 22 + (5-6) 2 + 22 = 65. ∵ a = 35, so B2 = a2-c2 = 9. ∵ so the standard equation of the ellipse is & nbsp; X245 + Y29 = 1. (2) MF1 ⊥ MF2 = (- 6-x0, - Y0) · (6-x0, - Y0) = x20-36 + Y20 = 0 from MF1 ⊥ MF2, that is, xo2 = 36-y02. Substituting into the elliptic equation, yo2 = 94, so & nbsp; Y0 = ± 32

The ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 (1,1) intersects with the straight line y = KX + 1-k at a and B. the left focus of the ellipse is F1. When the perimeter of the triangle f1ab is the largest, k =?
The ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 intersects with the straight line y = KX + 1-k at a. the left focus of the ellipse is F1. When the perimeter of the triangle f1ab is the largest, k =?
Hope to have a problem-solving process

Straight line y = KX + 1-k = > Y-1 = K (x-1)
The straight line passes through the fixed point (1,1),
In the ellipse The slope of the line passing through a certain point (1,1) in the interior is set as K Ask for the circumference again
It's easy to draw a picture. The maximum perimeter is 2a, which is 4
Then K has no real solution!
Non professional standard, ha ha, it's right not to guarantee

The left and right focal points of the ellipse x ^ 2 / 2 + y ^ 2 = 1 are F1 and F2 respectively. The line L passing through F1 intersects with the ellipse m and n
And │ vector F2m + vector f2n │ = 2 √ 26 / 3, find the linear l equation

C = 1, F1 (- 1,0), F2 (1,0). Let m (x1, Y1), n (X2, Y2). Let the slope of the line l be K, then the L equation: y = K (x + 1). ① substitute into & nbsp; X ^ 2 / 2 + y ^ 2 = 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; and arrange: (1 + 2K ＾ 2) x ^ 2 + 4K ^ 2x + 2K ^ 2-1 = 0. Then X1 + x2 = (4K ^ 2) / (1 + 2K ＾ 2), Y1 +

Given that P is the moving point on the ellipse x ^ 2 / 16 + y ^ 2 / 8 = 1, F1 and F2 are the left and right focal points of the ellipse
O is the origin of the coordinate, if M is a point on the angular bisector of ∠ f1pf2 and the vector f1m * vector MP = 0, then the value range of the vector OM

When point P is at the intersection of ellipse and y-axis, point m coincides with origin o, and | om → | takes the minimum value of 0
When point P is at the intersection of ellipse and x-axis, point m coincides with focus F1, and then | om → | takes the maximum 2 root sign 2
The value range of ∵ XY ≠ 0 and | om →| is (0,2 * radical 2)

The focus F1, F2 and point P of ellipse X & sup2 / 9 + Y & sup2 / 4 = 1 are the moving points. When ∠ f1pf2 is an obtuse angle, the
Point P abscissa value range. Thank you, please write the process

When ∠ f1pf2 is a right angle, x + y = 2A = 6x & sup2; + Y & sup2; = (2C) & sup2; = (2 √ 5) & sup2; = 20 = > X & sup2; - 6x + 8 = 0 (X-2) (x-4) = 0x = 2, x = 4, substitute X & sup2 / 9 + Y & sup2 / 4 = 1C (- √ 5,0) C (√ 5,0) (x0 + √ 5) ^ 2 + Y0 ^ 2 = 2 ^ 2 = > x0 = 3 √ 5 (x0 + √ 5) ^ 2 + Y0

It is known that the left and right focal points of the ellipse x216 + Y29 = 1 are F1 and F2 respectively, and the point P is on the ellipse. If P, F1 and F2 are the three vertices of a right triangle, then the distance from the point P to the X axis is ()
A. 95B. 3C. 977D. 94

Let m be an end point of the minor axis of the ellipse. Since a = 4, B = 3, C = 7 ＜ B ∈ f1mf2 ＜ 90 °, only ∠ pf1f2 = 90 ° or ∠ pf2f1 = 90 ° can be obtained. Let x = ± 7, then y2 = 9 (1 − 716) = 9216,  y | = 94. That is, the distance from P to X axis is 94

Ellipse X & sup 2 / 25 + Y & sup 2 / 9 = 1. F 1, F 2 are two focal points, P is on the ellipse
Pf1f2 forms a triangle, and the trajectory equation of the center of gravity m of △ pf1f2 is obtained
I will calculate the ellipse, just how to prove that the trajectory of M is an ellipse, the rest of the trouble will not be written
Just prove that the trajectory of M is an ellipse

c²=25-9=16
So F1 (- 4,0), F2 (4,0)
P(a,b)
M(x,y)
M is the center of gravity
So x = (- 4 + 4 + a) / 3, y = (0 + 0 + b) / 3
a=3x,b=3y
P is on the ellipse
So (3x) & sup2 / / 25 + (3Y & sup2;) / 9 = 1
x²/(25/9)+y²=1
Pf1f2 not collinear
So B ≠ 0
So y ≠ 0
So x & sup2; / (25 / 9) + Y & sup2; = 1, excluding (- 5 / 3,0) and (5 / 3,0)
So it's an ellipse with two major axis vertices removed

It is known that F1 and F2 are the two focal points of the ellipse X & sup2 / 16 + Y & sup2 / 9 = 1. The straight line passing through point F2 intersects the ellipse at points a and B. If | ab | = 5, then | AF1 | - | BF2 | = ()
A.3 B.8 C.13 D.16

Choose a
Let | af2 | = m, | BF2 | = n, then | AF1 | = 2a-m = 8-m, so | AF1 | - | BF2 | = 8-m-n = 8 - (M + n) = 8-5 = 3