As shown in the figure, the two right angle sides OA and ob of RT △ ABC are respectively on the positive half axis of X axis and the negative half axis of Y axis, and C is a point on OA As shown in the figure, the two right angle sides OA and ob of RT △ ABC are respectively on the positive half axis of X axis and the negative half axis of Y axis, C is a point on OA, OC = ob, and the parabola y = (X-2) (x-m) - (P-2) (P-M) (where m and P are constants, and M + 2 ≥ 2p > 0) passes through two points a and C (1) It is proved that: (P, 0) is on a parabola; (2) The length of OA and OC is represented by M and P respectively; (3) When m and P satisfy what relation, the area of △ AOB is the largest

As shown in the figure, the two right angle sides OA and ob of RT △ ABC are respectively on the positive half axis of X axis and the negative half axis of Y axis, and C is a point on OA As shown in the figure, the two right angle sides OA and ob of RT △ ABC are respectively on the positive half axis of X axis and the negative half axis of Y axis, C is a point on OA, OC = ob, and the parabola y = (X-2) (x-m) - (P-2) (P-M) (where m and P are constants, and M + 2 ≥ 2p > 0) passes through two points a and C (1) It is proved that: (P, 0) is on a parabola; (2) The length of OA and OC is represented by M and P respectively; (3) When m and P satisfy what relation, the area of △ AOB is the largest

(1) When x = P, y = (P - 2) (P - M) - (P - 2) (P - M) = 0, so (P, 0) is on the parabola
(2) Let a (a, 0), C (C, 0)
When x = a, y = (a - 2) (a - M) - (P - 2) (P - M) = 0 (1)
When x = C, y = (C - 2) (C - M) - (P - 2) (P - M) = 0 (2)
(1)-(2):(a - 2)(a - m) - (c - 2)(c - m) = 0
a² - (m+2)a + 2m = c² - (m+2)c + 2m
a² - c² = (m+2)a - (m+2)c
(a + c)(a - c) = (m + 2)(a - c)
A + C = m + 2
Because there can only be two intersections between the parabola and the x-axis, (P, 0) must be a or C
(a) (P, 0) is a
a = p
c = m + 2 - a = m + 2 - p
According to the meaning of the title, a > C, M + 2 ≥ 2p
a -c = p - (m + 2 - p) = 2p - (m + 2) < 0
Contradiction with a > C
(b) (P, 0) is C
c = p
a = m + 2 - c = m + 2 - p
According to the meaning of the title, a > C, M + 2 ≥ 2p
a - c = (m + 2 - p) - p = (m + 2) - 2p > 0
In line with the theme
|OC| = c = p
|OA| = a = m + 2 - p
(3)|OB| = |OC| = p
The area of △ AOB s = (1 / 2) | ob | * | OA | = P (M + 2-P) / 2
= -[p - (m+2)/2]² + (m+2)²/8
When p = (M + 2) / 2, s is the largest