Let the center of the ellipse C be at the origin, the focus on the Y axis, and the eccentricity be 2 / 2 of the root sign 1. Find the standard equation of ellipse C 2. If the line L with slope 2 passes through the focus of ellipse C on the positive half axis of Y axis and intersects with the ellipse at two points AB, then | AB is obtained|

Let the center of the ellipse C be at the origin, the focus on the Y axis, and the eccentricity be 2 / 2 of the root sign 1. Find the standard equation of ellipse C 2. If the line L with slope 2 passes through the focus of ellipse C on the positive half axis of Y axis and intersects with the ellipse at two points AB, then | AB is obtained|

1. Centrifugation e = C / a = √ 2 / 2 a = √ 2C
One vertex is (1,0), so B = 1
A & # 178; - C & # 178; = B & # 178; a = √ 2c, C = b = 1, a = a = √ 2
The standard equation is Y & # 178 / 2 + X & # 178; = 1
2. The line L passes through the upper intersection (0,1), the slope is 2, and the linear equation is y = 2x + 1
Simultaneous equations: 6x & # 178; + 4x-1 = 0
Let a (x1, Y1) B (X2, Y2)
According to Weida's theorem: X1 + x2 = - 2 / 3, x1x2 = - 1 / 6
|AB|²=(x1-x2)²+(y1-y2)²=(1+2²)(x1-x2)²=5[(x1+x2)²-4x1x2]=5(4/9+2/3)=50/9
Then | ab | = 5 √ 2 / 3