For the quadratic equation of one variable x 2 + BX + C = 0 of x 2 = 1, x 2 = 2, then the factorization result of x 2 + BX + C is______ ．

For the quadratic equation of one variable x 2 + BX + C = 0 of x 2 = 1, x 2 = 2, then the factorization result of x 2 + BX + C is______ ．

The two known equations are: x 1 = 1, x 2 = 2, we can get: (x-1) (X-2) = 0, ∩ x 2 + B x + C = (x-1) (X-2)

The standard equation of parabola in focus coordinate (- 2.0)

If the focal coordinate of the parabola is (2,0), then the standard equation of the parabola is?

The focal point (2,0) is on the x-axis, and the standard form of the equation is y ^ 2 = 2px
Intersection coordinates x = P / 2 = 2
∴p=2*2=4
The standard equation is y ^ 2 = 8x

The focal coordinate of parabola is (0,1 / 4A) (a)

Because a

The focal coordinate of the parabola y2 = 20x is? The Quasilinear equation is?

The focal coordinate of the parabola y2 = 20x is (5,0) and the Quasilinear equation is x = - 5
reason:
∵ 2p=20,p=10,p/2=5
The focal point is (P / 2,0) and the collimator is x = - P / 2

F is the focal point of the parabola y2 = 4x, the focal point coordinates of the parabola and the Quasilinear equation

The parabolic equation shows that P = 2, the focus is (P / 2, 0), and the Quasilinear is x = - P / 2
Then put P in
(1,0), the Quasilinear equation is x = - 1,

The focal coordinates of the parabola y = 4 (x-1) ^ 2 are_____ The Quasilinear equation is______ (to analyze the process)

Y = 4 (x-1) ^ 2 can be reduced to (x-1) ^ 2 = 1 / 4 * y
The focal coordinate of x ^ 2 = 1 / 4 * y is (0,1 / 16), and the Quasilinear equation is y = - 1 / 16
(x-1) ^ 2 = 1 / 4 * y is obtained by shifting x ^ 2 = 1 / 4 * y one unit to the right,
So the focal coordinate of (x-1) ^ 2 = 1 / 4 * y is (1,1 / 16), and the Quasilinear equation is y = - 1 / 16

The focal coordinate of parabola y = 4x2 is ()
A. （0，1）B. （0，116）C. （1，0）D. （116，0）

The standard equation of parabola y = 4x2 is & nbsp; x2 = 14y, P = 18, the opening is upward, the focus is on the positive half axis of Y axis, so the focus coordinate is (0116), so B

How to solve the equation after the parabola y square = 4x is rotated 90 ° counterclockwise around the focus?

After 90 ° counter clockwise rotation, the vertex is (1, - 1), and the point of intersection with X axis (- 1,0) (3,0). Let the equation be y = ax & sup2; + BX + C substitute three points into a + B + C = - 1 A-B + C = 0 9A + 3B + C = 0, and a = 1 / 4 B = - 1 / 2 C = - 3 / 4 y = x & sup2. / 4-x / 2

Finding the parabolic focus of the square of y = 4x

x²=y/4
2p=1/4
p/2=1/16
So the focus is (0,1 / 16)