Let a straight line passing through the focus of the parabola y2 = 2px (P > 0) intersect with the parabola. Let the coordinates of the two intersections be a (x1, Y1) and B (X2, Y2) respectively. Prove: (1) y1y2 = - P2 (2) x1x2 = p24

Let a straight line passing through the focus of the parabola y2 = 2px (P > 0) intersect with the parabola. Let the coordinates of the two intersections be a (x1, Y1) and B (X2, Y2) respectively. Prove: (1) y1y2 = - P2 (2) x1x2 = p24


It is proved that: (1) let the linear equation be x = my + P2, and substitute y2 = 2px, y2-2mpy + P2 = 0, y1y2 = - P2 (2) x1 · x2 = y12p · y22p = p24



The focus of the parabola y2 = 2px is f, and the coordinates of points a, B and C are (1,2) on this parabola. If point F is exactly the center of gravity of △ ABC, then the equation of line BC is ()
A. 2x+y-1=0B. 2x-y-1=0C. x-y=0D. x+y=0


∵ parabola y2 = 2px, point a (1,2) is in this parabola, ∵ parabola equation y2 = 4x, and f (1,0) can be set B (B2, 2b), C (C2, 2C). From the "two-point equation", the equation of straight line BC is (B + C) y-2bc = 2x. From the problem, point F is just the center of gravity of △ ABC, we can get: 3 = 1 + B2 + C2, 0 = 2 + 2



One focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is f (1,0), and O is the coordinate origin. Let the line L passing through point F intersect the ellipse at two points a and B. if the line L rotates around point F arbitrarily, there will always be OA ^ 2 + ob ^ 2