It is known that the parabola y ^ 2 = 2x (P greater than 0), passing through point (1,0) as a straight line with slope k, intersects parabola at two points a and B. the symmetric point of point a about X axis is C It is known that the parabola y ^ 2 = 2x (P is greater than 0), passing through point (1,0) as a straight line with slope k, l intersects the parabola at two points a and B, the symmetric point of point a about X axis is C, and the straight line BC intersects X axis at Q. try to prove that when k changes, q is a fixed point

It is known that the parabola y ^ 2 = 2x (P greater than 0), passing through point (1,0) as a straight line with slope k, intersects parabola at two points a and B. the symmetric point of point a about X axis is C It is known that the parabola y ^ 2 = 2x (P is greater than 0), passing through point (1,0) as a straight line with slope k, l intersects the parabola at two points a and B, the symmetric point of point a about X axis is C, and the straight line BC intersects X axis at Q. try to prove that when k changes, q is a fixed point

Let a (x1, Y1) or a (Y1 ^ 2 / 2, Y1) B (X2, Y2) or B (Y2 ^ 2 / 2, Y2) y = K (x-1) (1)
y^2=2x (2)
We get x = y ^ 2 / 2 and substitute (1) to get KY ^ 2-2y-2k = 0, and we get Y1 + y2 = 1 / K (3) Y1 * y2 = - 2 (4)
You can also find y1-y2 = + - √ (1 / K ^ 2 + 8) (this problem does not need this result)
Let C (Y1 ^ 2 / 2, - Y1), then BC linear equation is y + Y1 = (Y2 + Y1) / [(Y2 ^ 2-y1 ^ 2) / 2] (X-Y1 ^ 2 / 2)
(4) Results the intersection of Y (y1-y2) = 2x + 2 and X axis is y = 0 (no matter what the value of K is, the left side of the equation is 0), so 2x + 2 = 0 and x = - 1
I.e. over fixed point Q (- 1,0)