It is known that the quadratic equation 2x2 + 4x + k-1 = 0 has real roots 1) Find the value of K; (2) When the equation has two non-zero integer roots, the image of the quadratic function y = 2x ^ 2 + 4x + k-1 about X is translated down 8 units, and the analytic expression of the translated function is obtained; (3) Under the condition of (2), the image of the translated quadratic function below the x-axis is folded along the x-axis, and the rest of the image remains unchanged to get a new image. Try to explore whether the new image intersects with the line y = 1 / 2x + B? If there are several intersections?

(1) It is shown that Δ = 16-8 (k-1) ≥ 0. K ≤ 3. ∵ K is a positive integer, ∵ k = 1,2,3; (2) when k = 1, the equation 2x2 + 4x + k-1 = 0 has a root of zero; when k = 2, the equation 2x2 + 4x + K-1 = 0 has no integer root; when k = 3, the equation 2x2 + 4x + k-1 = 0 has two non-zero integer roots

It is known that the equation AX ^ 2 + 2x-1 = 0 about X has real roots, and the value range of a is obtained

When a = 0, it is consistent with
When a ≠ 0, △ = 4 + 4A ≥ 0, the solution is a ≥ - 1
In conclusion, a ≥ - 1

The four roots of x2-5x + M = 0 and x2-10x + n = 0 are arranged to form an equal ratio sequence with the first term of 1 to calculate M: n

Let 1 be the root of the first equation, then obviously the other root is 4, M = 4
Then, since the final four numbers are arranged into an equal ratio sequence with the first term of 1, if they are 1, 4, 16 and 64, it does not conform to the case that the sum of two numbers in the second equation is 10
So after testing, the four roots should be 1,2,4,8
Then M = 4, n = 16, M / N = 1 / 4
If 1 is the root of the second equation, then n = 9, the two roots are 1 and 9, respectively
If the first two terms of the equal ratio sequence are 1 and 9, it is not in line with the meaning of the problem at all. Then it is only possible that the first and fourth terms are 1 and 9 respectively. In the first equation, the product of the two is equal to 36, but there is no real number root in the equation
Therefore, there is only one solution to this problem
m:n=1/4

What is the parametric equation of a line passing through (12,7) and the origin?

y=12/7x

After importing a solid model into ANSYS, how to move its geometric center to the origin of system coordinates and rotate the solid
After importing a solid model into ANSYS, how to move its geometric center to the origin of the system coordinate? And how to rotate the model, because the calculation results of the model need to be transferred to other software, and the direction of the system coordinate system of the software is different from that of ANSYS

Brother, you're asking the right person. I met this problem in CFX at the beginning,
Move the coordinate system, including global coordinate system and local coordinate system

The equation of the circle passing through the intersection of the line 2x-y + 1 = 0 and the circle x2 + y2-2x-15 = 0 and the origin is______ ．

Let the equation of the circle be x2 + y2-2x-15 + λ (2x-y + 1) = 0, because the circle passing through the intersection of the straight line 2x-y + 1 = 0 and the circle x2 + y2-2x-15 = 0 passes through the origin, so we can get - 15 + λ = 0, and the solution is λ = 15. By substituting λ = 15 into the equation and simplifying, we can get the equation of the circle: x2 + Y2 + 28x-15y = 0. So the answer is: x2 + Y2 + 28x-15y = 0

1. Through P (- 2,0), make a straight line L intersection circle x2 + y2 = 1 at a and B. then PA Pb =?
2. The maximum common chord length of two circles x2 + Y2 + 2aX + 2ay + 2a2-1 = 0 and X2 + Y2 + 2bx + 2by + 2b2-2 = 0?
3. The curve represented by equation x2 + 4xy + 4y2-x-2y-2 = 0 is?

1. If you make a tangent of a circle through P, it is tangent to the circle at point C. according to the cutting line theorem, there is | PC | ^ 2 = | PA | Pb|
It is easy to get | PC | ^ 2 = | op | ^ 2 - | OC | ^ 2 = 3 in triangle POC
∴｜PA｜·｜PB｜=3
2. By sorting out the equations of two circles, we can get (x + a) ^ 2 + (y + a) ^ 2 = 1
(x+b)^2+(y+b)^2=2
There is a big circle and a small circle, so the maximum common chord length is the diameter of the small circle, which is 2
3. By sorting out the equation, we get (x + 2Y) ^ 2 - (x + 2Y) - 2 = 0
Factorization, we get (x + 2y-2) (x + 2Y + 1) = 0
So there is x + 2y-2 = 0 or x + 2Y + 1 = 0, so the curve is two parallel straight lines

Given the equation of circle M: x2 + (Y-2) 2 = 1, line L
It is known that the circle M: x2 + (Y-2) 2 = 1, the straight line L: x-2y = 0, the point P is on the straight line, the tangent PA and Pb of the circle m are made through the point P, and the tangent points are a and B
(1) If ∠ APB = 60 °, try to find the coordinates of point P
(2) If the coordinates of point P are (2,1), a straight line passing through point P intersects circle m at two points c and D, and when CD = root 2, the CD equation of the straight line is obtained
(3) Verification: the circle passing through a, P and M must pass through the fixed point, and the coordinates of all the fixed points must be obtained
(4) Verification: when point P moves on line L, the line passing through a and B always passes through the fixed point

(1) Then | MP | = {[2 (Y0) - 0] ^ 2 + [(Y0) - 2] ^ 2} = 2,5 (Y0) ^ 2-4 (Y0) = 0, Y0 = 0 or Y0 = 4 / 5p

Given the inclination angle a = 30 degrees of line L through P (1,1), the parameter equation of line L is obtained

Parameter equation
♠ it passes through a fixed point (x0, Y0) and its inclination angle α is known
x=x0+tcosα
y=y0+tsinα
If the line L passes through the P (1,1) inclination angle a = 30 & #, then
sina=1/2,cosa=√3/2
So the parameter equation of line L is
x=1+1/2·t
Y = 1 + √ 3 / 2 · t (t is the parameter)

Given that the straight line 1 passes through P (1,1) and the inclination angle a = Pie / 6, write out the parameter equation of 1

When the straight line passes through the point P (1,1), the inclination angle is π / 6, ｜ the linear equation is Y-1 = Tan (π / 6) (x-1) (Y-1) / sin (π / 6) = (x-1) / cos π / 6, let t = (Y-1) / sin (π / 6) = (x-1) / cos π / 6, then x = 1 + T × cos π / 6, y = 1 + T × sin (π / 6) ｜ the parameter equation is x = 1 + (√ 3T) / 2, y = 1 + T / 2, (t is the parameter