If the line L passes through the focus F of the parabola y2 = 4x and intersects with the parabola at two points a and B, then the trajectory equation of the midpoint of the chord AB is______ ．

The focus of parabola is (1,0) when the slope of straight line exists, let it be K, then the chord equation of focus is y = K (x-1), so k2x2 - (2k2 + 4) x + K2 = 0 is obtained by substituting it into the parabola equation, and the slope is not equal to 0. The equation is a quadratic equation of one variable, so the abscissa of midpoint: x 1 + x 2 = 2k2 + 4k2 is substituted into the ordinate of midpoint: y = K (x-1) by Weida's theorem: x 1 + x 2 = 2k2 + 4k2 ）=2K. That is, the midpoint is (K2 + 2k2, 2K) elimination parameter K, and the equation is y2 = 2x-2. When the slope of the line does not exist, the midpoint of the line is (1, 0), which is in line with the meaning of the problem, so the answer is: y2 = 2x-2

Make two straight lines perpendicular to each other through the vertex of parabola y = 6x, intersect the parabola at two points AB, and find the trajectory equation of the midpoint of line ab?

Let the vertex of parabola be o
OA：y=kx,OB:y=(-1/k)x
∵y^2=6x
∴A(6/k^2,6/k),B(6k^2,-6k)
Let AB midpoint (x, y)
∴x=(6/k^2+6k^2)/2=3(1/k^2+k^2)
y=(6/k-6k)/2=3(1/k-k)
The elimination parameter k is obtained
The trajectory equation of midpoint AB: y ^ 2 = 3 (X-6)

If a point a (4,0) outside the circle x2 + y2 = 4 is used as the secant of the circle, then the trajectory equation of the midpoint of the chord whose secant is cut by the circle is
How to solve this problem by eliminating parameters? We must solve it by eliminating parameters

Let the tangent equation be y = K (x-4) and the equation of circle be combined to get x ^ 2 + K ^ 2 (x-4) ^ 2 = 4 (k ^ 2 + 1) x ^ 2-8k ^ 2x + 16K ^ 2-4 = 0x1 + x2 = 8K ^ 2 / (k ^ 2 + 1), so Y1 + y2 = K (x1 + x2-8) = - 8K / (k ^ 2 + 1), so the midpoint coordinate is x = 4K ^ 2 / (k ^ 2 + 1) y = - 4K / (k ^ 2 + 1), so x = - KYK = - X / y brings in x = 4K ^ 2 / (k ^ 2)

If the circle x2 + y2 = 4 is known and the secant ABC of the circle is made through a (4,0), then the trajectory equation of the midpoint of the chord BC is ()
A. （x-2）2+y2=4B. （x-2）2+y2=4（0≤x＜1）C. （x-1）2+y2=4D. （x-1）2+y2=4（0≤x＜1）

Let the midpoint (x, y) of the chord BC, the slope of the straight line passing through a be K, and the equation of secant ABC be y = K (x-4); make the secant ABC of the circle, so the line between the midpoint and the center of the circle is perpendicular to the secant ABC, and the equation is x + KY = 0; because the intersection point is the midpoint of the chord and it is on these two straight lines, the trajectory equation of the point in the chord BC is x2 + y2-4x = 0, as shown in the figure, so select B

Through the point P (3.4), make the secant of circle x2 + y2 = 4, intersect the circle at A.B

Let m (x, y) be the midpoint of AB and O (0,0) be the center of the circle
So om ⊥ PM
PM = (x-3, y-4), OM = (x, y) (PM, OM are vectors)
Then om * PM = x (x-3) + (y-4) y = 0
That is, x ^ 2-3x + y ^ 2-4y = 0 (x ^ 2 + y ^ 2)

If the line L passes through the focus F of the parabola y2 = 4x and intersects with the parabola at two points a and B, then the trajectory equation of the midpoint of the chord AB is______ ．