As shown in the figure, it is known that the parabola Y1 = ax & # 178; + BX + C and the parabola y2 = x & # 178; + 6x + 5 are symmetric about the y-axis, and intersect with the y-axis at point m, and intersect with the x-axis at two points a and B If the image of a function y = KX + B passes through a point m and intersects with a parabola Y1 at another point n (m, n), where m ≠ N and satisfies the following conditions: M & # 178; - M + T = 0 and N & # 178; - N + T = 0 (t is a constant) ① Finding the value of K ② Let the line intersect the x-axis at point D, and p be a point in the coordinate plane. If the quadrilateral with vertices o, D, P, and M is a parallelogram, try to find the coordinates of point P (just write the coordinates of point P directly, and do not require the process of solution)

As shown in the figure, it is known that the parabola Y1 = ax & # 178; + BX + C and the parabola y2 = x & # 178; + 6x + 5 are symmetric about the y-axis, and intersect with the y-axis at point m, and intersect with the x-axis at two points a and B If the image of a function y = KX + B passes through a point m and intersects with a parabola Y1 at another point n (m, n), where m ≠ N and satisfies the following conditions: M & # 178; - M + T = 0 and N & # 178; - N + T = 0 (t is a constant) ① Finding the value of K ② Let the line intersect the x-axis at point D, and p be a point in the coordinate plane. If the quadrilateral with vertices o, D, P, and M is a parallelogram, try to find the coordinates of point P (just write the coordinates of point P directly, and do not require the process of solution)

What about the picture? How can we do it without a picture

It is known that the parabola y = ax & # 178; + X + C (a ≠ 0) of X intersects with the X axis at the points a (- 2,0), B (6,0), and the analytical formula of this parabola is obtained

There is no solution for the moment

The image of y = 4sin (2x + π / 2) is? A about the line x = π / 6 symmetry B about the line x = π / 12 symmetry c about the Y axis symmetry d about the origin symmetry

The image of y = 4sin (2x + π / 2) = 4cos2x is C symmetric about y axis

Given that the line L and the line y = - 2x are symmetric about the X axis, there is a point P (m, - 3), a point a (0,4), O on L as the origin of the coordinate, find △ Po
Finding the area of △ Poa

A line L symmetric to the x-axis and y = - 2x is y = 2x
Then point P is (- 3 / 2, - 3), the distance from point P to X axis is | - 3 | = 3, OA = 4
Then the area of △ POA = 3 * 4 / 2 = 6

Let the image of the first-order function y = 2x + 1 be symmetric with respect to the x-axis image (), symmetric with respect to the y-axis image (), symmetric with respect to the origin（

Y = - 2x-1y = - 2x + 1y = 2x-1 solution: 1, y = 2x + 1 through the point (0,1) and point (- 1 / 2,0), set the mapping image as y = ax + B, then calculate the image through the point (0, - 1) and (- 1 / 2,0), generation to find y = - 2x-1, the same as other. 2, about the X axis symmetry, that x value unchanged, y value is just the opposite, then -

The focus f passing through the parabola y = 2px (P > 0) makes a straight line intersection parabola at any two points a and B. It is proved that the sum of the distances from the point a and B to the axis of symmetry of the parabola is a fixed value

1. If the slope of the straight line AB does not exist, then the ordinates of a and B are all P, and the sum of the distances to the x-axis is p. 2. If the slope of the straight line AB exists, let its slope be K, then AB: y = K (X-P / 2), and the parabola y & sup2; = 2px are connected, and X is eliminated, then y & sup2; - (2P / k) y + P & sup2; = 0, then the sum of the distances from a and B to the x-axis =

If | FA | = 2 | FB |, then k = ()
A. 13B. 23C. 23D. 223

Let the Quasilinear of parabola C: y2 = 8x be l: x = - 2, and the straight line y = K (x + 2) (k > 0) pass through the fixed point P (- 2, 0) as shown in Fig. A and B respectively make am ⊥ L in M, BN ⊥ L in N, from | FA | = 2 | FB |, then | am | = 2 | BN |, point B is the midpoint of AP and connecting ob, then | ob | = 12 | AF |, | ob | = | BF |, and the abscissa of point B is

Let f be the focal point of the parabola y2 = 4x. Let A.B.C be three points on the parabola. If FA + FB + FC = O, then ∣ FA ∣ + ∣ FB ∣ + ∣ FC ∣ =?
Let a (x1, Y1), B (X2, Y2), C (X3, Y3)
Parabolic focal coordinate f (1,0), quasilinear equation: x = - 1
∵FA+FB+FC=O
The point F is the center of gravity of △ ABC
Then X1 + x2 + X3 = 3
y1+y2+y3=0
And | FA | = X1 - (- 1) = X1 + 1
|FB|=x2-(-1)=x2+1
|FC|=x3-(-1)=x3+1
∴|FA|+|FB|+|FC|=x1+1+x2+1+x3+1=(x1+x2+x3)+3=3+3=6
The answer is the above, but I don't understand one thing
Why X1 + x2 + X3 = 3

Upstairs is the solution, also can
FA (vector) = (x1-1, Y1)
FA + FB + FC = (x1-1 + x2-1 + x3-1, Y1 + Y2 + Y3) = zero vector
Then x1-1 + x2-1 + x3-1 = 0
x1+x2+x3=3
Y1+Y2+Y3=0

Let f be the focus of the parabola y2 = 4x, and a, B, C be the three points on the parabola. If FA + FB + FC = 0, then the value of | FA | + | FB | + | FC | is ()
A. 3B. 4C. 6D. 9

Let a (x1, Y1), B (X2, Y2), C (X3, Y3) parabolic focal coordinates f (1, 0), the Quasilinear equation: x = - 1