Let the equation of the ellipse be x square + y square / 4 = 1, the straight line passing through M (0,1) intersects the ellipse at two points AB, O is the coordinate origin, Op vector = 1 / 2 (OA vector + ob vector). When l rotates around point m, the trajectory equation of the moving point P is obtained

Let the equation of the ellipse be x square + y square / 4 = 1, the straight line passing through M (0,1) intersects the ellipse at two points AB, O is the coordinate origin, Op vector = 1 / 2 (OA vector + ob vector). When l rotates around point m, the trajectory equation of the moving point P is obtained

E: x^2+y^2/4 = 1 (1)
M(0,1)
OP = (1/2)(OA+OB)
L: passing through M(0,1)
y = mx +c
1= c
ie
L: y = mx +1 (2)
Sub (2) into (1)
x^2 + (mx+1)^2/4 =1
4x^2 + (mx+1)^2 = 4
(4+m^2)x^2 + 2mx -3 =0
Let P be (x,y)
then
2x = -2m/(4+m^2) (3)
from (2)
y = mx+1
x = (y-1)/m (4)
Sub (4) into (1)
(y-1)^2/m^2 + y^2/4 = 1
4(y-1)^2 + m^2y^2 = 4m^2
(4+m^2)y^2 - 8y + 4(1-m^2) =0
then
2y = 8/(4+m^2)
4+m^2 = 4/y
m = √[4(1-y)/y] (5)
Sub (5) into (3)
2x = -2m/(4+m^2)
x = -√[4(1-y)/y]/ (4/y)
x^2 = [4(1-y)/y] / [4/y]^2
= y(1-y)/4
4x^2 = y(1-y)
The trajectory equation of P is as follows
4x^2 = y(1-y)

The point m (1,1) is located in the ellipse 4 / x square + 2 / 2 square = 1, the line passing through the point m intersects the ellipse at two points a and B, and the point m is the midpoint of the line ab
Find the equation of line AB and the value of absolute value ab

It can be seen that the slope of line AB must exist, and it is set as K
Let a and B coordinates be (x1, Y1) (X2, Y2) respectively
So the linear equation is Y-1 = K (x-1)
The simultaneous elliptic equations are: (2 * k ^ 2 + 1) x ^ 2 - (4 * k ^ 2-4 * k) * x + 2 * k ^ 2-4 * K-2 = 0
So X1 + x2 = (4 * k ^ 2-4 * k) \ (2 * k ^ 2 + 1)
And because X1 + x2 = 2
So k = - 0.5
From the value of K, you can get the linear equation and the value of ab

It is known that the line x + Y-1 = 0 and the ellipse x square / a square + y square / b square = 1 (a > b > 0) intersect at a and B, and the midpoint m of line AB is on the line L: y = 1 / 2x
1) Find the eccentricity of ellipse
2) If the symmetric point of the right focus of the ellipse with respect to the line L is on the unit circle x square + y square = 1, the elliptic equation is solved

Let a (x1, Y1), B (X2, Y2), then (y2-y1) / (x2-x1) = - 1, (Y2 + Y1) / (x2 + x1) = 1 / 2. Then X1 ^ 2 / A ^ 2 + Y1 ^ 2 / b ^ 2 = 1, X2 ^ 2 / A ^ 2 + Y2 ^ 2 / b ^ 2 = 1, subtracting the two expressions to get (x2-x1) (x2 + x1) / A ^ 2 + (y2-y1) (Y2 + Y1) / b ^ 2 = 0, so a ^ 2 / b ^ 2 = 2, that is, a ^ 2 = 2B ^ 2, so C ^ 2 = B ^ 2 = (...)

It is known that the elliptic equation is x square / 25 + y square / 9 = 1, the line L passing through the right focus intersects the ellipse at two points a and B, and the circle with diameter AB passes through the origin
I calculate X1 + x2 = 200K & # 178 / 9 + 25K & # 178;
x1x2=400k²-225/9+25k²

The right focus of the ellipse is (4,0). Let the linear equation be y = K (x-4)
Substituting into the elliptic equation, solve x1, X2, Y1, Y2
(this step is omitted)
The solution is: X1 * x2 = (400k & # 178; - 225) / (9 + 25K & # 178;)
y1*y2=-81K²/(9+25k²)
Because the two intersections and the origin are on the same circle,
X1 & # 178; + Y1 & # 178; + x2 & # 178; + Y2 & # 178; = (y2-y1) & # 178; + (x2-x1) & # 178; (circle angle is 90 degrees, right triangle, Pythagorean theorem)
x1*x2+y1*y2=0
Substitute the values of the above equations X1 * X2, Y1 * Y2 to solve the equation:
K = - 15 / √ 319 or K = 15 / √ 319
So there are two equations for line I
y=15/√319(x-4)
y=-15/√319(x-4)