It is known that a (2,3) is on the image with positive scale function y = KX (1) Find the analytic expression of the function (2) find the coordinates of the symmetric point B of point a about the origin, and judge whether B is on the function y = KX. (3) find the distance between two points a and B

It is known that a (2,3) is on the image with positive scale function y = KX (1) Find the analytic expression of the function (2) find the coordinates of the symmetric point B of point a about the origin, and judge whether B is on the function y = KX. (3) find the distance between two points a and B

1. Substitute x = 2, y = 3 into y = KX
So k = 3 / 2
So the equation is y = 3x / 2
2.B(-2,-3)
Substituting x = - 2, y = - 3 into y = 3x / 2
establish
So B is in the image
3. AB = 2 radical (2 ^ 2 + 3 ^ 2)
=2 root number 13

A straight line L: y = KX + 1. A parabola C: Y & sup2; = 4x. When what is the value of K, do l and C have (1) one common point (2) two common points (3) no common point?

y²=(kx+1)²=4x
k²x²+(2k-4)x+1=0
Discriminant = (2k-4) & sup2; - 4K & sup2; = - 16K + 16
(1) A common point
That is, the equation has a solution, and the discriminant is equal to 0
-16k+16=0
k=1
(2) Two common points
The equation has two solutions and the discriminant is greater than 0
-16k+16>0
k

If the line ly = KX + K + 2 and the parabola C: Y & # 178; = 4x have a unique common point, the equation L is solved

Straight line y = KX + 1 and parabola y = 4x have two common points. Equation x divided by 2 + K minus y divided by K + 1 equals 1, which means hyperbola has a large range of values for real numbers. The steps are very important

The line y = KX + 1 and the parabola y = 4x have two common points, and the elimination result is KY ^ 2-4y + 4 = 0
There are two common point discriminant, 16-16k > 0 K0
k-1
So K

Let the straight line y = 2x + B and the parabola y ^ 2 = 4x intersect at two points a and B, and let | ab | = 3 root sign 5, find the value of B

There are always two symmetries on the parabola y 2 = 4x with respect to the straight line y = k x + 3

Let B and C be symmetric with respect to the line y = KX + 3, so let the BC equation of the line be x = - KY + m, and substitute y2 = 4x to get Y2 + 4ky-4m = 0. Let B (x1, Y1), C (X2, Y2), then If the midpoint of BC is m (x0, Y0), then Y0 = Y1 + Y22 = - 2K, x0 = 2k2 + M. ∵ the point m (x0, Y0) is on the straight line L, ∵ 2K = K (2k2 + m) + 3, ∵ M = - 2K3 + 2K + 3K. The intersection of BC and parabola is at two different points, ∵ Δ= 16k2 + 16m > 0

If there are two symmetric points on the parabola y = - x ^ 2 + 4 about the straight line y = KX + 3, then the value range of K is
No one can answer the question for several days=
Does anyone know how to do it

Let a (x1, Y1) and B (X2, Y2) be two points on the parabola y = - X & # 178; + 4, and the coordinates of the midpoint of a and B ((x1 + x2) / 2, (Y1 + Y2) / 2) are on the straight line y = KX + 3, ■ (Y1 + Y2) / 2 = k * (x1 + x2) / 2 + 3 ①,∵y1=-x1^2+4,y2=-X2^2+4,∴y1-y2=(x2-x1)(x1+x2),x1+x2=-(y1-y2)/(x1...

There are always two symmetries on the parabola y 2 = 4x with respect to the straight line y = k x + 3

Let B and C be symmetric with respect to the straight line y = KX + 3, so let the BC equation of the straight line be x = - KY + m, and substitute y2 = 4x to get Y2 + 4ky-4m = 0. Let B (x1, Y1), C (X2, Y2), then the midpoint m (x0, Y0) of BC, then Y0 = Y1 + Y22 = - 2K, x0 = 2k2 + M. ∵ point m (x0, Y0) is on the straight line L, ∵ - 2K = K (2k2 + m) +

It is known that there are two different points on the parabola C Y & # 178; = 4x, which satisfy the condition of finding the real number k symmetrically with respect to the straight line y = KX + 3
(Y1 + Y2) (y1-y2) = 4 (x1-x2) → 2y0 · (- 1 / k) = 4, Y0 is the midpoint = (Y1 + Y2) / 2

Let two points be a (x1, Y1), B (X2, Y2)
Then the slope of the line AB K '= (y2-y1) / (x2-x1)
Since the line AB is perpendicular to the line y = KX + 3, there is a slope k '= - 1 / K of the line ab
So there is (Y1 + Y2) (y1-y2) = 4 (x1-x2) → 2y0 · (- 1 / k) = 4

In the parabola y ^ 2 = 4x, there are always two symmetries about y = KX + 3. It is urgent to find the range of K. I don't know how to understand it in steps
On the parabola y ^ 2 = 4x, there are always two symmetries about y = KX + 3. It's urgent to find the range of K. I don't understand the steps. Can I write well on the paper and upload photos,

Let a (x1, Y1), B (X2, Y2), AB be symmetric with respect to the line y = KX + 3, Y1 ^ 2 = 4x1 (1), Y2 ^ 2 = 4x2, (2) (y1-y2) / (x1-x2) * (Y1 + Y2) / 2 = 2, K1 * Y0 = 2, Y0 = - 2K, x0 = (x1 + x2) / 2 = (Y1 ^ 2 + Y2 ^ 2) / 8 > 0, - 2K = kX0 + 3, x0 = (- 2k-3) / k > 0, (2k + 3) / k > 0