It is known that the vertex of parabola y = a (x-t-1) 2 + T2 (a, t are constants, a ≠ 0, t ≠ 0) is a, and the vertex of parabola y = x2-2x + 1 is B. (1) judge whether point a is on parabola y = x2-2x + 1, why? (2) If the parabola y = a (x-t-1) 2 + T2 passes through the point B, then ① find the value of a; ② can the two intersections of the parabola with the X axis and its vertex a form a right triangle? If you can, find out the value of T; if you can't, explain the reason

It is known that the vertex of parabola y = a (x-t-1) 2 + T2 (a, t are constants, a ≠ 0, t ≠ 0) is a, and the vertex of parabola y = x2-2x + 1 is B. (1) judge whether point a is on parabola y = x2-2x + 1, why? (2) If the parabola y = a (x-t-1) 2 + T2 passes through the point B, then ① find the value of a; ② can the two intersections of the parabola with the X axis and its vertex a form a right triangle? If you can, find out the value of T; if you can't, explain the reason

(1) According to the meaning of the question, we can know that the coordinates of point a are (T + 1, T2). Substituting the coordinates of point a into the parabola y = x2-2x + 1, we can get: (T + 1) 2-2 (T + 1) + 1 = T2 + 2T + 1-2t-2 + 1 = T2; therefore, point a is on the parabola y = x2-2x + 1. (2) from the meaning of the question, we can know that the coordinates of point B are (1, 0). Then we can get: 0 = a (1-t-1) 2 + T2, that is at2 + T2 = 0, so a = - 1 =-(x-t-1) 2 + T2; when y = 0, - (x-t-1) 2 + T2 = 0, the solution is x = 1 or x = 2T + 1. Let the intersection of parabola and X axis be m, N, then the coordinates of m point is (1, 0), and the coordinates of N point is (2t + 1, 0). Therefore, AM2 = T2 + T4, an2 = T2 + T4, Mn2 = 4T2. When △ amn is a right triangle, AM2 + an2 = Mn2, that is, (T2 + T4) × 2 = 4T2, the solution is T1 = 1 or T2 = - 1, so the value of T is 1 Or - 1