If the line passing through a focus F1 of the ellipse 4x ^ 2 + y ^ 2 = 1 intersects with the ellipse at two points a and B, then the perimeter of △ abf2 formed by a and B and another focus F2 of the ellipse? Seek the truth

If the line passing through a focus F1 of the ellipse 4x ^ 2 + y ^ 2 = 1 intersects with the ellipse at two points a and B, then the perimeter of △ abf2 formed by a and B and another focus F2 of the ellipse? Seek the truth

No matter in which ellipse there are AF1 + af2 = BF1 + BF2 = 2A, the perimeter is AF1 + af2 + BF1 + BF2 = 4A = 4

The left and right focal points of the ellipse X225 + y216 = 1 are F1, F2 respectively, and the chord AB passes F1. If the circumference of the inscribed circle of △ abf2 is π, and the coordinates of a and B are (x1, Y1), (X2, Y2) respectively, then the value of | y1-y2 | is ()
A. 53B. 103C. 203D. 53

Ellipse: X225 + y216 = 1, a = 5, B = 4, ｜ C = 3, left and right focus F1 (- 3, 0), F2 (3, 0), △ abf2 inscribed circle circumference is π, then the inscribed circle radius is r = 12, and △ abf2 area = △ af1f2 area + △ bf1f2 area = 12 ×| Y1 ×| F1F2 | + 12 ×| Y2 ×| F1F2 | = 12 ×|

Let F1 and F2 be the left and right focus of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > 0, b > 0)
(1) If the sum of the distances from a point a (1,3 / 2) on the ellipse C to F1 and F2 is equal to 4, the equation of the ellipse C and the coordinates of the focus can be obtained
(2) The left and right ellipses have the following properties: if M and N are two symmetrical points on the origin of the ellipse C, and P is any point on the ellipse, when the slopes of the lines PM and PN exist and are denoted as KPM and KPN, then the product of KPM and KPN is a fixed value independent of the position of point P. try to write out the similar properties of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) and prove them

(1) from the first definition of ellipse, 2A = 4, a = 2, substituting a point a (1,3 / 2) and a = 2 on ellipse C into elliptic equation, we can get B & # 178; = 3, so the elliptic equation is X & # 178 / 4 + Y & # 178 / 3 = 1, C = √ A & # 178

It is known that F 1 and F 2 are respectively the left and right focal points of the ellipse C: X 2A2 + y 2B2 = 1 (a > b > 0). The sum of the distances between a (1,32) and F 1 and F 2 on the ellipse C is equal to 4. (1) write out the equation and focal coordinates of the ellipse C; (2) set point K as the moving point on the ellipse, and find the trajectory equation of the midpoint of line f1k

(1) ∵ ellipse C: the focus of x2a2 + y2b2 = 1 (a > b > 0) is on the x-axis, and the sum of the distances from point a to focus F1 and F2 is 4, ∵ 2A = 4, that is, a = 2; and ∵ point a (1, 32) is on the ellipse, ∵ 122 + 94b2 = 1, ∵ B2 = 3, ∵ C2 = A2-B2 = 1; the equation of ellipse C is x24 + x23 = 1, focus F1

Let F1 and F2 be the left and right focus of ellipse C: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > b > 0)
(1) If the sum of the distances from the points (1,1.5) on the ellipse C to F1 and F2 is 4, write out the equation and focal coordinates of the ellipse C
(2) Let y be the moving point on the ellipse obtained in (1), and find the trajectory equation of the midpoint of line segment F1y

1、
P(1,1.5)
PF1+PF2=2a=4
a=2
x²/4+y²/b²=1
Over (1,1.5)
1/4+4/9b²=1
b²=16/27
x²/4+27y²/16=1
2、
Midpoint Q (x, y)
Y(m,n)
c²=a²-b²=4-16/27=92/27
F1=(-2√69/9,0)
So x = (m-2 √ 69 / 9) / 2, y = n / 2
m=2x+2√69/9,n=2y
Y is in the ellipse
(2x+2√69/9)²/4+27*4y²/16=1
That is, (2x + 2 √ 69 / 9) & sup2 / 4 + 27y & sup2 / 4 = 1

Let F1 (- C, 0) and F2 (C, 0) be the two focal points of the ellipse x2a2 + y2b2 = 1 (a > b > 0), and p be the intersection of the circle with the diameter of F1F2 and the ellipse. If ∠ pf1f2 = 5 ∠ pf2f1, then the eccentricity of the ellipse is ()
A. 32B. 63C. 22D. 23

P is the intersection point of the circle and ellipse with the diameter of F1F2 as the diameter of F1F2, which is the intersection point of the circle and ellipse with the diameter of F1F2 as the diameter of F1F2 as the diameter of F1F2. The \124; P is the intersection point of the circle and ellipse with the diameter of F1F2 as the diameter of F1F2 as the diameter of F1F2 as the diameter of F1F2, and the \124; f1pf2 = 90 \\ \\| pf1f1f1f1f1f1f2 = 5  pf2f2f2f2f2f2f2f2f2 = 5 pf2f2f2f2f2f2f2f2f2f2f2f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f1f2 = 5 = 5 = 5 = 5 = 5 = 5  5 therefore, B is chosen

It is known that the left and right focal points of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are F1 (- C, 0) and F2 (C, 0), respectively. If there is a point P on the ellipse such that asin ∠ pf1f2 = csin ∠ pf2f1, then the value range of eccentricity of the ellipse is ()
A. （0，2-1）B. （22，1）C. （0，22）D. （2-1，1）

In △ pf1f2, from the sine theorem: pf2sin ∠ pf1f2 = pf1sin ∠ pf2f1, then from the known: aPF2 = cpf1, that is: apf1 = cpf2, set point P (x0, Y0) from the formula of focus radius, then: Pf1 = a + ex0, PF2 = a-ex0, then a (a + ex0) = C (a-ex0) solution: x0 = a (C-A) e (c + a) = a (E-1) e (E + 1) from the geometric properties of ellipse: x0 ＞ - A, then a (E-1) e (E + 1) ＞ - A, then E2 + 2e-1 ＞ 0, the solution: e ＜ - 2 -1 or E > 2-1, and E ∈ (0, 1), so the eccentricity of ellipse is e ∈ (2-1, 1), so D

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) e = √ 2 / 2, the left and right focus are f1.f2, fixed point P (2, √ 3), and | F1F2 | = | PF2|
1. Find the equation of ellipse C
2. Let the line L: y = KX + m and the ellipse C intersect at two points M. n, and the sum of slopes of the lines F2m and f2n is zero, then find the relationship between M and K

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) e = √ 2 / 2, the left and right focus are f1.f2, fixed point P (2, √ 3), and | F1F2 | = | PF2|
1. Find the equation of ellipse C
2. Let the line L: y = KX + m and the ellipse C intersect at two points M. n, and the sum of slopes of the lines F2m and f2n is zero, then find the relationship between M and K
1. Analysis: ∵ ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), the left and right focus are f1.f2, fixed point P (2, √ 3), and | F1F2 | = | PF2|
4C ^ 2 = (2-C) ^ 2 + 3 = = > 3C ^ 2 + 4c-7 = 0 = = > C1 = 1, C2 = - 7 / 3 (rounding)
The centrifugation rate e = √ 2 / 2, a = √ 2, B = 1
The ellipse C: x ^ 2 / 2 + y ^ 2 = 1
2. Analysis: let the line L: y = KX + M = = > y ^ 2 = k ^ 2x ^ 2 + 2kmx + m ^ 2
Substituting ellipse x ^ 2 / 2 + y ^ 2 = 1
We get (1 + 2K ^ 2) x ^ 2 + 4kmx + 2m ^ 2-2 = 0
Let m (x1, Y1), n (X2, Y2)
Then X1 + x2 = - 4km / (1 + 2K ^ 2), x1x2 = (2m ^ 2-2) / (1 + 2K ^ 2)
∵ the sum of slopes of F2m and f2n is zero
Y1/(x1-1)=-y2/(x2-1)==>y1/y2=-(x1-1)/ (x2-1)
(kx1+m)/ (kx2+m)=(1-x1)/ (x2-1)
kx2+m- kx1x2-mx1=kx1x2+mx2-kx1-m
2kx1x2+(m-k)(x1+x2)-2m=0
2k(2m^2-2)-(m-k)4km-2m(1+2k^2)=0
∴m+2k=0==>m=-2k

If point m is the first moving point on the ellipse x ^ 2 / 4 + y ^ 2 / 3, and F1 and F2 are the two focal points of the ellipse, then the minimum value of MF1 and MF2 is ()
If point m is the first moving point on the ellipse x ^ 2 / 4 + y ^ 2 / 3, and F1 and F2 are the two focuses of the ellipse, then the minimum value of │ MF1 │· │ MF2 │ is ()
A.1
B.3
C.4
D. One half

c^2=4-3=1,c=1
Let m (2cosa, √ 3sina)
Then:
|MF1||MF2|=√[(2coaa-1)^2+3sin^2a)*√[(2coaa+1)^2+3sin^a]
=4-cos^2a
So, when cos ^ 2A = 1
The minimum value of MF1 is 4-1 = 3
B.3

It is known that P (x0, Y0) is any point on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), and F1 and F2 are the focus
A circle with diameter of PF2 must be inscribed with a circle with diameter of major axis of ellipse
Can you not use the knowledge related to centrifugal rate? I haven't learned there yet

First of all: for a circle with the diameter of the long axis of the ellipse, the center of the circle is obviously the origin, and the radius is a, so the equation is: x ^ 2 + y ^ 2 = a ^ 2;
Let F2 be the right focus, then F2 (C, 0), P (x0, Y0), and,
Therefore, the center of a circle with PF2 as its diameter is ((c + x0) / 2, Y0 / 2);
According to the focal radius formula, if the diameter PF2 = a-ex0, then the radius = (a-ex0) / 2;
Therefore, the equation of circle with PF2 as diameter is: [x - (c + x0) / 2] ^ 2 + (y-y0 / 2) ^ 2 = (a-ex0) ^ 2 / 4;
Two circles: x ^ 2 + y ^ 2 = a ^ 2 and [x - (c + x0) / 2] ^ 2 + (y-y0 / 2) ^ 2 = (a-ex0) ^ 2 / 4;
R1-R2=a-(a-ex0)/2=(a+ex0)/2；
And the center distance d ^ 2 = (c + x0) ^ 2 / 4 + Y0 ^ 2 / 4 = (C ^ 2 + 2cx0 + x0 ^ 2 + Y0 ^ 2) / 4, ①
Because P is on the ellipse, Y0 ^ 2 = B ^ 2 - (b ^ 2 * x0 ^ 2 / A ^ 2)
Substituting into Formula 1, we get d ^ 2 = [C ^ 2 + B ^ 2 + 2cx0 + x0 ^ 2 - (b ^ 2 * x0 ^ 2 / A ^ 2)] / 4
D ^ 2 = [a ^ 2 + 2cx0 + (ex0) ^ 2] / 4
Note: A ^ 2 + 2cx0 + (ex0) ^ 2 just = (a + ex0) ^ 2, so d ^ 2 = (a + ex0) ^ 2 / 4
That is, the center distance d = (a + ex0) / 2
It is proved that the distance between the center of a circle is equal to the difference in radius, so the two circles are tangent to each other
If you don't understand, please hi me,