Finding the midpoint trajectory equation of parallel string with slope 2 in ellipse x ^ 2 + 2Y ^ 2 = 1 Let the linear equation l with slope 2 be y = KX + B = 2x + B Simultaneous: x ^ 2 + 2Y ^ 2 = 1, y = 2x + B 9x＾+8bx+2b＾-1=0 L intersects the ellipse at two points a (x1, Y1) and B (X2, Y2) x1+x2=-8b/9=2x x=-4b/9 b=-9x/4 y1+y2=2（x1+x2）+2b=4x+2b=2y y=2x+b=8x/4-9x/4=-x/4 ∴4y+x=0 Why is x = 8x / 4 in the last y = 2x + B = 8x / 4-9x / 4 = - X / 4 not before x = - 4B / 9 How did this step come about?

Finding the midpoint trajectory equation of parallel string with slope 2 in ellipse x ^ 2 + 2Y ^ 2 = 1 Let the linear equation l with slope 2 be y = KX + B = 2x + B Simultaneous: x ^ 2 + 2Y ^ 2 = 1, y = 2x + B 9x＾+8bx+2b＾-1=0 L intersects the ellipse at two points a (x1, Y1) and B (X2, Y2) x1+x2=-8b/9=2x x=-4b/9 b=-9x/4 y1+y2=2（x1+x2）+2b=4x+2b=2y y=2x+b=8x/4-9x/4=-x/4 ∴4y+x=0 Why is x = 8x / 4 in the last y = 2x + B = 8x / 4-9x / 4 = - X / 4 not before x = - 4B / 9 How did this step come about?

Here is 2x = 8x / 4
Don't change to B

The trajectory equation of the moving chord of ellipse x2 + 2Y2 = 2 with slope - 2 is

Let the linear equation be: y = - 2x + m; let the intersection of the straight line and the ellipse be a, B, a (x1, Y1) B (X2, Y2) because X12 + 2y12 = 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) X22 + 2y22 = 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) (1) - (2) get: x12-x22 = 2y22-2y12 (x1 + x2) (x1-x2) = - 2 (Y1 + Y2) (y1-y2) k = - 2 = - X1 + X22 (Y1 + Y2) & nbsp; let the midpoint be p (x, y), so 2 = x2yx-4y = 0

Given the ellipse X & # 178 / 2 + Y & # 178; = 1, find the secant of the ellipse from the left focus F of the ellipse, and find the trajectory equation of the midpoint P of the chord

1： Known ellipse (x ^ 2 / 2) + y ^ 2 = 1
1. The left focus F of the ellipse leads to the secant of the ellipse to find the trajectory equation of the midpoint P of the cut chord
2. Find the trajectory equation of the midpoint Q of parallel string with slope 2
Left focus f (- 1,0)
The left focus f passing through the ellipse leads to the secant of the ellipse y = K (x + 1)
Cut chord AB a (x1, Y1) B (X2, Y2)
The midpoint of the cut chord P ((x1 + x2) / 2, (Y1 + Y2) / 2) is (x, y)
（X1^2/2）+y1^2=1. (1)
（X2^2/2）+y2^2=1. (2)
(1)- (2)
(x1-x2)(x1+x2)/2+(y1-y2)(y1+y2)=0
(y1-y2)/(x1-x2)=-(x1+x2)/[2(y1+y2)]=2x/4y=-x/2y
k=-x/2y
y=-x/2y *(x+1)
2y^2=-x^2-x
x^2+x+2y^2=0
(x+1/2)^2+2y^2=1/4

Find the trajectory equation of the middle point m of the chord where the line L: y = 2x + m is cut by the ellipse X & # 178; + Y & # 178 / / 4 = 1
Detailed steps are required

Find M
x^2+(2x+m)^2/4=1
x^2+x^2+xm+m^2/4=1
2x^2+xm+m^2/4-1=0
x1+x2=-m/2
M((x1+x2)/2,(y1+y2)/2)
So x = - M / 4, y = (2x1 + m + 2x2 + m) / 2 = - M / 2 + M = m / 2
So m = - 4x, M = 2Y
So 2Y = - 4x
y=-2x