In the plane rectangular coordinate system, the center of the ellipse is at the origin of the coordinate, its focus is on the X axis, and its fixed point is on the straight line x + 2y-2 = 0 In the plane rectangular coordinate system, the center of the ellipse is at the coordinate origin, its focus is on the X axis, and its fixed point is on the straight line x + 2y-2 = 0. (1) find the standard equation of the ellipse. (2) the intersection ellipse of the straight line passing through the coordinate origin o is at two points B and C, f is the left focus of the ellipse, and find the maximum area of the triangle FBC

In the plane rectangular coordinate system, the center of the ellipse is at the origin of the coordinate, its focus is on the X axis, and its fixed point is on the straight line x + 2y-2 = 0 In the plane rectangular coordinate system, the center of the ellipse is at the coordinate origin, its focus is on the X axis, and its fixed point is on the straight line x + 2y-2 = 0. (1) find the standard equation of the ellipse. (2) the intersection ellipse of the straight line passing through the coordinate origin o is at two points B and C, f is the left focus of the ellipse, and find the maximum area of the triangle FBC

(1)
X＋2Y－2＝0
Intersection with X-axis (2,0) and y-axis (0,1)
Ellipse a = 2 b = 1
The solution is x ^ 2 / 4 + y ^ 2 / 1 = 1
(2)
C = sqrt (a ^ 2-B ^ 2) = sqrt (3) (square root)
S = FO * h = sqrt(3) * b = sqrt(3)

Given the ellipse X & # 178; + 2Y & # 178; = 4, find the length of the chord with P (1,1) as the midpoint

Let y = kx-k + 1 be taken into x ^ 2 + 2Y ^ 2 = 4 to get (1 + 2K ^ 2) x ^ 2-4k (k-1) x + 2K ^ 2-4k-2 = 0, so X1 + x2 = (4K (k-1)) / (1 + 2K ^ 2) because (1,1) is the midpoint, so (x1 + x2) / 2 = 1, the solution is k = - 1 / 2 (x1-x2) ^ 2 = (x1 + x2) ^ 2-4 * X1 * x2 = 8 / 3lx1-x2l = 2 √ 6 /

Given the ellipse x square + 2Y square = 4, then the linear equation of the chord with (1,1) as the midpoint is?

The ellipse x ^ 2 + 2Y ^ 2 = 4 is known,
Let the chord circle with (1,1) as the midpoint be
A(x1,y1),B(x2,y2)
Then X1 ^ 2 + 2y1 ^ 2 = 4. (1)
x2^2+2y2^2=4.(2)
(1)-(2)
(x1^2-x2^2)+2[y1^2-y2^2]=0
(x1+x2)(x1-x2)=-2(y1+y2)(y1-y2)
The intersection circle of chords with (1,1) as the midpoint
(x1+x2)/2=1
(y1+y2)/2=1
x1+x2=2.y1+y2=2
Then (y1-y2) / (x1-x2) = - 2
That is, the slope of the straight line is - 2
And over (1,1)
Then Y-1 = - 2 (x-1)
2x+y-3=0

Given the ellipse X & # 178; + 2Y & # 178; = 4, then the length of the chord with (1,1) as the midpoint is

Intersection (x1, Y1) (X2, Y2) X1 + x2 = 2, Y1 + y2 = 2, chord length is √ (x1-x2) ^ 2 + (y1-y2) ^ 2 = √ (1 + K ^ 2) | x1-x2 | K is the slope of straight line X1 & # 178; + 2y1 & # 178; = 4x2 & # 178; + 2Y2 & # 178; = 4, subtracting to get: (y1-y2) / (x1-x2) = - 1 / 2 (x1 + x2) / (Y1 + Y2) = - 1 / 2, the slope of straight line is - 1 / 2