In the plane rectangular coordinate system, it is known that the parabola passes through two points a (- 4.0) and B (0. - 4), and the axis of symmetry is a straight line x = - 1 If the point m is a point on the parabola in the third quadrant, the abscissa of the point m is m, and the area of the triangle mAb is s, the function of s with respect to m is obtained. If the point P is a moving point on the parabola and the point q is a moving point on the straight line y = - x, judge that there are several positions that can make the quadrilateral with the point pqbo as the vertex be a parallelogram, and write the corresponding point Q coordinates

In the plane rectangular coordinate system, it is known that the parabola passes through two points a (- 4.0) and B (0. - 4), and the axis of symmetry is a straight line x = - 1 If the point m is a point on the parabola in the third quadrant, the abscissa of the point m is m, and the area of the triangle mAb is s, the function of s with respect to m is obtained. If the point P is a moving point on the parabola and the point q is a moving point on the straight line y = - x, judge that there are several positions that can make the quadrilateral with the point pqbo as the vertex be a parallelogram, and write the corresponding point Q coordinates

Let the distance between M and ab be h, m (m, M & # 178 / 2 + m - 4), - 4 & lt; M & lt; 0bo = 4, let Q (Q, - Q), P (Q, Q & # 178 / 2 + Q - 4) (1) QP = Q & # 178 / 2 + Q - 4 - (- Q) = Q & # 178 / 2 + 2q - 4 = Bo = 4q & # 178; + 4q - 16 = 0q = - 2 ± √ 5q (- 2 ±)

It is known that the intersection of hyperbola y = x / K and parabola y = ax + BX + C is a (2,3), B (m, 32) C (- 3, n)
Find the analytic expressions of hyperbola and parabola
2. Point a, point B, point C in the plane rectangular coordinate system, and calculate the area of △ ABC

1. Substituting a into hyperbola y = K / X
3=k/2
k=6
The analytic formula of hyperbola is y = 6 / X
By substituting BC two points into hyperbola, we can get the coordinates of two points
2=6/m m=3 B(3,2)
n=6/(-3) n=-3 C(-3,-2)
Substituting ABC into parabola
3=4a+2b+c
2=9a+3b+c
-2=9a-3b+c
a=-1/3
b=2/3
c=3
The analytical formula of parabola is y = - 1 / 3x ^ 2 + 2 / 3x + 3
2. Find the analytic formula of AB: y = KX + B
3=2k+b
2=3k+b
k=-1
b=5
AB:y=-x+5
Let the intersection of line AB and y = - 2 be D, then the coordinate of point D is (7, - 2)
S△ABC=S△ACD-S△BCD
S△ACD=1/2*10*5=25
S△BCD=1/2*10*4=20
S△ABC=25-20=5

It is known that the parabola y = ax ^ 2 + bx-3 intersects the X axis at two points a and B, and intersects the Y axis at point C. the center m (1, m) of the circle passing through three points a, B and C is exactly on the symmetric axis of the parabola, and the radius of the circle m is the root sign 5. Let the circle m intersect the Y axis at D, and the vertex of the parabola be e
(1) Find the value of M and the analytical formula of parabola
(2) Let DBC = Alpha and CBE = Alta, and find the value of sin (alpha ALTA)
(3) Explore whether there is a point P on the coordinate axis, so that the triangle with P, a, C as the vertex is similar to the triangle BCE? If there is, please point out the position of point P and write the coordinates of P directly. If not, please explain the reason
1) Parabola y = ax ^ 2 + bx-3 to get C (0, - 3)
The center of the circle m (1, m), the radius of the circle m is the root sign 5. Get (1-0) ^ 2 + (M + 3) ^ 2 = 5, get m = - 1 or - 5
So m = - 1, m (1, - 1)
Circular equation: (x-1) ^ 2 + (y + 1) ^ 2 = 5, let y = 0 get x = - 1 or x = 3, that is a (- 1,0), B (3,0)
Let x = 0 get y = - 3 or y = 1, that is C (0, - 3) d (0,1)
Take the parabola y = ax ^ 2 + bx-3 and get a = 1, B = - 2
So, parabola y = x ^ 2-2x-3
2) Angle DBC = α, angle CBE = β
C(0,-3),B(3,0),D(0,1)
Because e is the vertex, e (1, - 4)
BC=3√2,CE=√2,BE=2√5
BO=3,OD=1,BD=√10
So BC: Bo = Ce: od = be: BD
So △ CBE ∽ OBD
Angle CBE = angle OBD = β
Sin (α - β) = sin angle CBO = OC: BC = 3:3 √ 2 = √ 2 / 2
3) There is no point P satisfying △ PAC ∽ BCE on the coordinate axis
If △ PAC ∽ BCE exists, let P (x, y)
PA:BC=AC:CE=PC:BE
PA:3√2=√10:√2=PC:2√5=√5
PA=3√10;PC=10
(x+1)^2+y^2=(3√10)^2=90
Let x = 0 get y = positive and negative √ 89; let y = 0 get x = positive and negative 3 √ 10-1
x^2+(y+3)^2=10^2=100
Let x = 0 get y = 7, y = - 13; let y = 0 get x = plus or minus 3 √ 91
So there's no point P on the axis
Seek complete explanation

The task of the third question is to find the point P on the coordinate axis, so that the two triangles are similar. Assuming that they are similar, according to the similarity property, the equation satisfied by point P is obtained. In the equation, X and y are equal to zero respectively, and the ordinates and abscissa of point P on Y axis and X axis are obtained, Therefore, the contradiction shows that point P does not exist