Let a fixed point F (0, m) (M > 0) on the symmetric axis of the parabola x ^ 2 = 2PY (P > 0) be a straight line AB and the parabola intersect at two points a and B, And a (x1, Y1), B (X2, Y2) (X10), the line corresponding to point F, l: y = - m, is called the quasi quasilinear of parabola (1) If x1x2 = - 4m, solve parabolic equation (2) Make a vertical line of "quasi collimator line" L: y = - m through point a (x1, Y1), and the vertical foot is A1. Prove that three points A1, O and B are collinear (o is the origin of coordinates) (3) If the point m is any point on the quasi collimator L, note that the inclination angles of the straight lines Ma, MB and MF are D, e and F in turn. Try to explore the relationship between the cotangents of D, e and F and give the proof

Let a fixed point F (0, m) (M > 0) on the symmetric axis of the parabola x ^ 2 = 2PY (P > 0) be a straight line AB and the parabola intersect at two points a and B, And a (x1, Y1), B (X2, Y2) (X10), the line corresponding to point F, l: y = - m, is called the quasi quasilinear of parabola (1) If x1x2 = - 4m, solve parabolic equation (2) Make a vertical line of "quasi collimator line" L: y = - m through point a (x1, Y1), and the vertical foot is A1. Prove that three points A1, O and B are collinear (o is the origin of coordinates) (3) If the point m is any point on the quasi collimator L, note that the inclination angles of the straight lines Ma, MB and MF are D, e and F in turn. Try to explore the relationship between the cotangents of D, e and F and give the proof

(1) Let the equation of AB: y = KX + m, and substitute it into the parabolic equation to get x ^ 2-2pkx-2pm = 0
X1x2 = - 2pm = - 4m, P = 2, so the parabolic equation is: x ^ 2 = 4Y
（2）.A1(x1,-m),O(0,0),B(x2,x2^2/4) ,k(OB)=x2/4 ,k(OA1)=-m/x1=(x1x2/4)/x1=x2/4
K (OB) = K (OA1), so: A1, O, B are collinear
(3) Let m (x0, - M), when x0 = 0, Cote = 0, tand = K (MA) = (Y1 + m) / X1 = (x1 ^ 2 + 4m) / 4x1,
tanF=k(MB)=(y2+m)/x2=(x2^2+4m)/4x2
cotD+cotF=...=0 ,cotD+cotF=cotE
When x0 is not equal to 0, the same can be obtained

As shown in the figure, the parabola y = x ^ 2-2x-3 intersects the X axis at two points a and B, intersects the Y axis at point C, and passes through points (2, - 3a). The axis of symmetry is a straight line x = 1, and the vertex is m
(1) Let the intersection of the line y = - x + 3 and the Y axis be d. on the line BD, a person goes to a point E (not coincident with B and D) and passes through the circular intersection line BC and point F of a, B and E. This is to judge the shape of △ AEF and explain the reason. (2) when e is any point on the line y = - x + 3, does the conclusion in (1) stand?

(1) The axis of symmetry is a straight line x = 1, - B / 2A = 1, through the point (2, - 3a) 4A + 2b-3 = - 3a, the solution is: a = 1, B = - 2, y = x ^ 2-2x-3 (2) when y = 0, x ^ 2-2x-3 = 0 (x + 1) (x-3) = 0, X1 = - 1, X2 = 3, a (- 1,0), B (3,0) when x = 0, y = - 3, C (0, - 3) when x = 1, y = - 4, m (1, - 4) cm: y = KX + H - 3 =

It is known that the parabola y = ax ^ 2 + 2x + C, the symmetric axis line x = - 1, the intersection of parabola and Y axis and point C
The parabola and the y-axis intersect at point C and the x-axis intersect at two points a (- 3,0) and B
1. Finding the analytic expression of linear AC
2. If the point D is a moving point on the parabola below the line AC, the maximum area of the quadrilateral ABCD can be obtained
3. P is a point on the parabola. If the circle with the diameter of line Pb is tangent to line BC at point B, the coordinates of point P are obtained

First question:
The axis of symmetry of ∵ y = ax ^ 2 + 2x + C is x = - 1, ∵ - 2 / (2a) = - 1, and a = 1
In this way, the parabolic equation can be expressed as: y = x ^ 2 + 2x + C
∵ point a (- 3,0) is on the parabola y = x ^ 2 + 2x + C, ∵ 0 = (- 3) ^ 2 + 2 × (- 3) + C, ∵ C = - 3
The coordinates of point C are (0, - 3)
The analytic formula of ∧ line AC is (y-0) / (x + 3) = (0 + 3) / (- 3-0) = - 1, ∧ x + y + 3 = 0
That is: the analytical formula of the line AC is x + y + 3 = 0
Second question:
∵ a, B and C are fixed points, and ∵ ABC's area is fixed value
To maximize the area of quadrilateral ABCD, we only need to maximize the area of △ ACD
The maximum value of height on AC in △ ACD is required
Obviously, when a straight line parallel to AC is tangent to a parabola, the distance from the tangent point to AC is the maximum value of the height of AC in △ ACD
From the equation x + y + 3 = 0 of AC, the slope of AC = - 1
The derivative of y = x ^ 2 + 2x - 3 is y ′ = 2x + 2
Let the coordinates of point d be (n, n ^ 2 + 2n-3), then 2n + 2 = - 1, х n = - 3 / 2
∴n^2＋2n－3＝9/4－3－3＝－15/4.
The coordinates of point D are (- 3 / 2, - 15 / 4)
The distance from point d to AC D1 = - 3 / 2-15 / 4 + 3 / √ 2 = 9 √ 2 / 8
Let y = 0 in y = x ^ 2 + 2x-3, we get: (x + 3) (x-1) = 0, X1 = - 3, X2 = 1
The coordinates of point B are (1,0)
The distance between point B and AC D2 = 1 + 0 + 3 / √ 2 = 2 √ 2
AC = √ [(- 3-0) ^ 2 + (0 + 3) ^ 2] = 3 √ 2
The maximum area of quadrilateral ABCD = the area of △ ABC + the maximum area of △ ACD
＝（1/2）｜AC｜d2＋（1/2）｜AC｜d1＝（1/2）×3√2×2√2＋（1/2）×3√2×9√2/8＝75/8.
That is: the maximum area of quadrilateral ABCD is 75 / 8
Third question:
∵ BC cuts the circle with diameter of Pb to B, ∵ Pb ⊥ BC
From B (1,0), C (0, - 3), the slope of BC = (0 + 3) / (1-0) = 3, and the slope of Pb is - 1 / 3
Let the coordinates of point p be (T, T ^ 2 + 2t-3), then: (T ^ 2 + 2t-3) / (t-1) = - 1 / 3,
∴3t^2＋6t－9＝－t＋1,∴3t^2＋7t－10＝0,∴（t－1）（3t＋10）＝0,
∴t1＝1、t2＝－10/3.
From t = 1, T ^ 2 + 2t-3 = 1 + 2-3 = 0, the coordinates of point P are (1,0)
From t = - 10 / 3, T ^ 2 + 2t-3 = 100 / 9-20 / 3-3 = 13 / 9, the coordinates of point P are (- 10 / 3,13 / 9)
That is: the coordinates of the point P satisfying the condition are (1,0) or (- 10 / 3,13 / 9)

It is known that the parabola y = ax ^ 2 + 2x + C, the axis of symmetry is a straight line x = - 1,
The parabola and the y-axis intersect at point C and the x-axis intersect at two points a (- 3,0) and B
1. Finding the analytic expression of linear AC
2. If the point D is a moving point on the parabola below the line AC, the maximum area of the quadrilateral ABCD can be obtained
3. P is a point on the parabola. If the circle with the diameter of line Pb is tangent to line BC at point B, the coordinates of point P are obtained
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(1) X = - B / 2A = - 1
The solution is a = 1
The intersection of parabola and X-axis at point a (- 3,0) is substituted, C = - 3
So the parabola is y = x ^ 2 + 2x-3
So when x = 0, y = - 3,
Point C (0, - 3)
A (- 3,0) and C (0, - 3) are solved by undetermined coefficient method
The linear AC is y = x-3
(2) When y = 0, x ^ 2 + 2x-3 = 0
The solution is x = - 3 or x = 1
So B (1,0)

Given that the vertex of the parabola y = x-2kx + 9 is on the x-axis, then k =?

From the meaning of the title:
4 / 4 [4 × 1 × 9 - (- 2K) ²] = 0
So:
36-4k²=0
4k²=36
k²=9
K = 3 or - 3

It is known that the vertex of parabola C1: y = - x ^ 2 + 2mx + n (m, n are constants, and M is not equal to 0, n > 0) is a,
Intersection with y axis at point C: parabola C2 and parabola C1 are symmetric about y axis, and their vertex is B, connecting AC, BC, ab,
(1) Please write the analytical formula of parabola C2 directly on the horizontal line__________ .
(2) When m = 1, judge the shape of triangle ABC
(3) Is there a point P on the parabola C1 so that the quadrilateral ABCP is a diamond? If so, find the value of M

⑴Y=-x^2－2mx+n
(2) isosceles right triangle
(3) M = 1 or - 1

It is known that the line y = 2mx + 2m (M > 0) intersects the x-axis and y-axis at two points a and C respectively. The coordinate of point B is (3,0). There is a parabola passing through two points a and B, and the vertex P is in the middle
On the line y = 2mx + 2m (M > 0)

What are you looking for?
A(-1,0),B(3,0)
Let y = ax ^ 2 + BX + C
Substituting a, B, we get
0=a-b+c
0=9a+3b+c
So 2A = - B, then - B / 2A = 1
Axis of symmetry x = 1
The vertex is on the line, so the vertex (1,4m)
A = - m, B = 2m, C = 3M are obtained
So y = - MX ^ 2 + 2mx + 3M = - M (x-1) ^ 2 + 4m

Known parabola y = x square - 2mx + 2m square - 4m + 3
）Find the function relation between the vertical coordinate y and the abscissa X of the vertex of the parabola y = xsquare-2mx + 2msquare-4m + 3;
(2) Is there a real number m such that the parabola y = x square - 2mx + 2m square - 4m + 3 intersects with the X axis
The distance between a (x1,0) and B (x2,0) is ab = 4. If it exists, find out the value of M; if it does not exist, explain the reason

(1) X = m, y = m ^ - 4m + 3, so y = x ^ - 4x + 3
(2) It doesn't exist
If M satisfies the problem condition, then
M ^ - 4m + 7 = 0 (Simplified)
16-4 * 7 is less than 0, so there is no solution

Given that the axis of symmetry of the parabola y = x2 + (M + 1) x-m2-1 is a straight line x = - 1, then the value of M is?
Thank you very much for your help

The axis of symmetry of the parabolic equation y = a * x ^ 2 + b * x + C (a ≠ 0) is - B / (2a), which is independent of C
Here, - 1 = - (M + 1) / 2, M = 1

It is known that the parabola y = ax & # 178; + BX + C passes through the point a (- 1,0), and passes through two intersections B and C of the line y = x-3 and the coordinate axis

1. It is easy to know: B (3,0), C (0, - 3). Let the analytic formula of parabola be y = a (x + 1) (x-3), then a (0 + 1) (0-3) = - 3, a = 1, y = x & # 178; - 2x-3.2. From (1), we know: y = x & # 178; - 2x-3 = (x-1) &# 178; - 4, so the vertex coordinates are (1, - 4)