The two focal points of ellipse C: X * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0) are F1 and F2. The point P is on ellipse C, and Pf1 ⊥ F1F2, Pf1 = 4 / 3, PF2 = 14 / 3 (1) Solving elliptic c-equation (2) If the line L passes through the center m of the circle x * 2 + y * 2 + 4x-2y = 0, the intersection ellipse C lies at two points a and B, and a and B are symmetric with respect to point m, the equation of the line L is obtained

The two focal points of ellipse C: X * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0) are F1 and F2. The point P is on ellipse C, and Pf1 ⊥ F1F2, Pf1 = 4 / 3, PF2 = 14 / 3 (1) Solving elliptic c-equation (2) If the line L passes through the center m of the circle x * 2 + y * 2 + 4x-2y = 0, the intersection ellipse C lies at two points a and B, and a and B are symmetric with respect to point m, the equation of the line L is obtained

PF1|=4/3,|PF2|=14/3 2a=|PF1|+|PF2|=6 a=3 a^2=9
PF1⊥PF2 |PF1|^2+|PF2|^2==|F1F2|^2=212/9=(4c)^2 c^2=53/9
The equation of B ^ 2 = a ^ 2-C ^ 2 = 28 / 9 ellipse C: x ^ 2 / 9 + 9y ^ 2 / 28 = 1
two
If the line L passes through the center m of the circle x ^ 2 + y ^ 2 + 4x-2y = 0, the intersection ellipse C: x ^ 2 / 9 + y ^ 2 / 4 = 1 lies at two points a and B, and a and B are symmetric with respect to point m, the equation of the line L is obtained
Let the slope of the line be K,
Write the linear equation according to m and K and substitute it into the ellipse C: x ^ 2 / 9 + y ^ 2 / 4 = 1 to get X1 + x2 = Y1 + Y2=
At the same time, X1 + x2 = 2 times the abscissa of M, Y1 + y2 = 2 times the ordinate of M
Find K

It is known that F1 (- 3,0) F2 (3,0) are the left and right focal points of the ellipse respectively, and P is the point on the ellipse, which satisfies the bisector intersection F1F2 of Pf1 ⊥ F1F2, ∠ f1pf2
In M (1,0), find the equation of ellipse

C = 3
Let | Pf1 | = x, then | PF2 | = 2a-x
Because the bisector of ∠ f1pf2 intersects F1F2 at M (1,0)
So | Pf1 | / | PF2 | = | f1m | / | MF2 | = 4 / 2 = 2
So x / (2a-x) = 2, that is, x = 4a-2x, that is, 3x = 4A, so x = 4A / 3
That is, | Pf1 | = 4A / 3, | PF2 | = 2A / 3
At the same time, Pf1 ⊥ F1F2,
So | Pf1 | ^ 2 + | PF2 | ^ 2 = (2C) ^ 2 = 4C ^ 2 = 36
That is, (4a / 3) ^ 2 + (2A / 3) ^ 2 = 36
The solution is a ^ 2 = 81 / 5
So B ^ 2 = a ^ 2-C ^ 2 = 81 / 5-9 = 36 / 5
So the equation of ellipse is x ^ 2 / (81 / 5) + y ^ 2 / (36 / 5) = 1

Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, e = - 1 + root 5 / 2, a is the left vertex, f is the right focus, B is a vertex of the minor axis, calculate the angle ABF
a>b>0

X 2 / a 2 + y 2 / B 2 = 1 (a > b > 0) the left and right focus is F 1, F 2 segment is divided into 5:3 segments by Y 2 = 2bx, and E is calculated

The focal coordinate of y2 = 2bx is f (B / 2,0)
F1(-c,0) F2(c,0)
(c+b/2)/(c-b/2)=5/3
c=2b
a^2=b^2+c^2=5b^2
a=√5b
e=c/a=2b/√5b=(2√5)/5