Let a (x1, Y1), B (X2, Y2), then x1x2 =? Let a (x1, Y1) and B (X2, Y2) pass through the focus of the parabola (x ^ 2) = 4Y to make the chord AB, then what is X1 times x2? There is a process or a graph.

X = 4Y, focus (0,1)
Let the straight line AB, Y-1 = KX (k is not equal to 0), that is, kx-y + 1 = 0,
Kx-y + 1 = 0
X = 4Y
We get: x-square-4kx-4 = 0,
According to Weida's theorem, x1x2 = C / a = - 4 / 1 = - 4

Through the focus of the parabola y ^ 2 = 8x, make a straight line intersection parabola at two points a (x1, Y1), B (X2, Y2), and X1 + x2 = 6, then the length of | ab | is?

Parabolic focus (2,0)
Let the linear equation be y = K (X-2)
When it is combined with parabola, KY ^ 2-8y-16k = 0
x1+x2=(y1+y2)/k+4=6
The solution is k ^ 2 = 4
And | ab | = ((1 + 1 / K ^ 2) ((Y1 + Y2) ^ 2-4y1y2)) ^ 0.5
Substituting K ^ 2 = 4
The solution is | ab | = 10

If X1 + x2 = 6, then | ab | =?

So the Quasilinear equation of parabola is x = - 1,
∵ the focus of the parabola y2 = 4x intersects the parabola at two points a (x1, Y1) B (X2, Y2)
∴|AB|=x1+x2+2,
And X1 + x2 = 6
∴∴|AB|=x1+x2+2=8
So the answer is 8

If x 1 + x 2 = 8, then | ab|=

2p=8
Then the guide line x = - P / 2 = - 2
x1+x2=8
Then the distance from a to the guide line + the distance from B to the guide line = (x1 + 2) + (x2 + 2) = 12
It is defined by parabola
The distance to focus is equal to the distance to guide line
So AB = AF + BF = 12

Let a (x1, Y1), B (X2, Y2), then x1x2 =? Let a (x1, Y1) and B (X2, Y2) pass through the focus of the parabola (x ^ 2) = 4Y to make the chord AB, then what is X1 times x2? There is a process or a graph.