"Ab is the moving chord on the parabola y = x ^ 2, and | ab | = a (a is constant and a > = 1), then the nearest distance between the midpoint m of the chord AB and the X axis"

"Ab is the moving chord on the parabola y = x ^ 2, and | ab | = a (a is constant and a > = 1), then the nearest distance between the midpoint m of the chord AB and the X axis"

Let AB: y = KX + B, Y > 0, K ≠ 0 be a real number
y=x^2=kx+b>0
x^2-kx-b=0
xA+xB=k
xM=0.5(xA+xB)=0.5k
yM=k*xM+b=0.5k^2+b>0
b=yM-0.5k^2
4b=4yM-2k^2
xA*xB=-b
(xA-xB)^2=(xA+xB)^2-4xA*xB=k^2+4b=k^2+4yM-2k^2=4yM-k^2
(yA-yB)^2=k^2*(xA-xB)^2
a^2=AB^2=(xA-xB)^2+(yA-yB)^2=(1+k^2)*(xA-xB)^2=(1+k^2)*(4yM-k^2)≥1
a^2=(1+k^2)*(4yM-k^2)
k^4+(1-4yM)k^2+a^2-4yM=0
If the upper equation with unknown number k ^ 2 has real number solution, then its discriminant △≥ 0, i.e
(1-4yM)^2-4(a^2-4yM)≥0
(1+4yM)^2≥(2a)^2
yM>0,a≥1
1+4yM≥2a≥2
The nearest distance between the midpoint m of the chord AB and the X axis = YM, YM ≥ 1 / 4
A: the closest distance between the midpoint m of the chord AB and the X axis is 1 / 4

AB = x, a + B = y how to solve X and y are constants, is there any way to solve it?

b=y-a
ab=ay-a²=x
a²-ay+x=0
a=[y±√(y²-4x)]/2
b=y-a
therefore
a=[y-√(y²-4x)]/2,b=[y+√(y²-4x)]/2
a=[y+√(y²-4x)]/2,b=[y-√(y²-4x)]/2

(1 + X + x ^ 2 +...) +X ^ 2000 = 0, find the value of x ^ 2001)
Known 1 + X + x ^ 2 + +X ^ 2000 = 0, find the value of x ^ 2001

If you multiply x left and right, you get
x+x^2+x^3+… +x^2001=0
If you subtract the two expressions, you get
x^2001-1=0
x^2001=1
There is a similar question in this question
Given: 1 + X + x square = 0, find: X to the power of 2001 + X2000 + x1999 + +The value of X + 1
Answer: x ^ 2001 + x ^ 2000 + x ^ 1999 + +x+1
=(x^2001+x^2000+x^1999)+…… +x^3+x^2+x+1
=x^1999(x^2+x+1)+.+x(x^2+x+1)+1
=x^1999*0+.+x*0+1
=1