As shown in the figure: the parabola intersects the x-axis at two points a (- 1,0) and B (3,0), and intersects the y-axis at points c (0, - 3). Let the vertex of the parabola be d (1) find the analytical formula of the parabola and the coordinates of vertex D: (2) prove that the center of the circumscribed circle of △ BCD falls at the midpoint of the BD side: (3) if point P is a moving point on the coordinate axis, it can make the triangle with P, a, C as the vertex similar to △ BCD? If it exists, Please write the coordinates of point P: if not, please explain the reason

As shown in the figure: the parabola intersects the x-axis at two points a (- 1,0) and B (3,0), and intersects the y-axis at points c (0, - 3). Let the vertex of the parabola be d (1) find the analytical formula of the parabola and the coordinates of vertex D: (2) prove that the center of the circumscribed circle of △ BCD falls at the midpoint of the BD side: (3) if point P is a moving point on the coordinate axis, it can make the triangle with P, a, C as the vertex similar to △ BCD? If it exists, Please write the coordinates of point P: if not, please explain the reason

(1) Substituting y = ax ^ 2 + BX + C into the coordinates of a, B, C: A: a - B + C = 0b: 9A + 3B + C = 0C: C = 3A = - 1, B = 2, C = 3Y = - x ^ 2 + 2x + 3 (2) y = - x ^ 2 + 2x + 3 = - (x - 1) ^ 2 + 4D (1,4) the parabola symmetry axis is x = 1p, which is the symmetry of C with x = 1 as the symmetry axis

As shown in the figure, the vertex m of the parabola is on the x-axis, the parabola and the y-axis intersect at the point n, and OM = on = 4, the vertices a and B of the rectangular ABCD are on the parabola, and C and D are on the x-axis. (1) find the analytical formula of the parabola; (2) set the abscissa of the point a as t (T > 4), and the perimeter of the rectangular ABCD as l, and find the functional relationship between L and t

(1) ∵ om = on = 4, ∵ m point coordinate is (4,0), N point coordinate is (0,4), let the analytic formula of parabola be y = a (x-4) 2, substitute n (0,4) into 16A = 4, the solution is a = 14, so the analytic formula of parabola is y = 14 (x-4) 2 = 14x2-2x + 4; (2) ∵ the abscissa of point a is t, ∵ DM = T-4, ∵ CD

Given that the parabola y ^ 2 = - X and the straight line y = K (x + 1) intersect at two points AB, calculate OA vertical ob
It is known that the parabola y ^ 2 = - X and the straight line y = K (x + 1) intersect at two points ab,
1. Calculate OA vertical ob
2. When the area of the triangle OAB is equal to the root 10, find K

1. Prove: the equation of parabola and straight line is simultaneous: y ^ 2 = - x, ① y = K (x + 1) ② substitute formula ② into formula ① to simplify: K ^ 2 * x ^ 2 + (2 * k ^ 2 + 1) * x + K ^ 2 = 0. According to Weida's theorem: XA * XB = 1, replace the parabola equation Ya * Yb = - radical (- XA * - XB) (must be negative, because the straight line passes (- 1,0)) so that there is XA * XB

The line y = x + B intersects the parabola y = (1 / 2) x squared at two points ab. if OA is perpendicular to ob, then B=

Let a (x1, Y1), B (X2, Y2), then OA * ob = x1x2 + y1y2 = 0 from OA perpendicular to ob, because y = x + B, then OA * ob = x1x2 + (x1 + b) (x2 + b) = 2x1x2 + B (x1 + x2) + B ^ 2 = 0. If y = x + B, y = (1 / 2) x ^ 2, then x ^ 2-2x-2b = 0, then the discriminant 4 + 8b > 0 is b > - 1 / 2; X1 + x2 = 1; x1x2 = - 2b