The four real roots of the equation x2 + MX + 16 / 3 x2 + NX + 16 / 3 = 0 form an equal ratio sequence of A1 = 3 / 2 If the four real roots of the equation (X & # 178; + MX + 16 / 3) · (X & # 178; + NX + 16 / 3) = 0 form an equal ratio sequence with the first term of 3 / 2, then | M-N|=_______

The four real roots of the equation x2 + MX + 16 / 3 x2 + NX + 16 / 3 = 0 form an equal ratio sequence of A1 = 3 / 2 If the four real roots of the equation (X & # 178; + MX + 16 / 3) · (X & # 178; + NX + 16 / 3) = 0 form an equal ratio sequence with the first term of 3 / 2, then | M-N|=_______

One root is 3 / 2. According to X1 * x2 = C, the corresponding other root is 32 / 9. The latter is divided by the former to get 64 / 27, which is the third power of 4 / 3. It means that the common ratio is 4 / 3, so the other two are 2 and 8 / 3
So | M-N | = | 3 / 2 + 32 / 9-2-8 / 3 | = 31 / 18

If the four roots of the equation x2-5x + M = 0 and x2-10x + n = 0 are properly arranged to form an equal ratio sequence of the first term 1, then the value of M: n is ()
A. 14B. 12C. 2D. 4

Let 1 be the root of the first equation, then it is obvious that the other root is 4, M = 4; since the final four numbers are arranged into an equal ratio sequence with the first term of 1, if it is 1, 4, 16, 64, it does not conform to the situation that the sum of the two in the second equation is 10, so after testing, the four roots should be 1, 2, 4, 8, then M = 4, n = 16. At this time, M: n = 14; if 1 is the root of the second equation, then M: n = 14 If n = 9, the two roots are 1 and 9 respectively. If the first two terms of the proportional sequence are 1 and 9, then the third and fourth terms will not be the roots of the first equation, which is not in line with the meaning of the problem. Then it is only possible that the first term and the fourth term are 1 and 9 respectively. In the first equation, the product of the two is equal to 9, but there is no real root of the equation. Therefore, this problem has only the unique solution M: n = 14

Let the four roots of two equations X & # 178; - ax + 1 = 0, X & # 178; - BX + 1 = 0 form an equal ratio sequence with 2 as the common ratio, then ab=

Let the two roots of the equation x & # 178; - ax + 1 = 0 be x1, and the two roots of the equation x & # 178; - BX + 1 = 0 be X3, X4. By using Veda's theorem, then X1 * x2 = 1, X3 * X4 = 1. Let the proportional sequence be x1, X3, x4, X2, then x2 / X1 = 8, and then X1 * x2 = 1  x2 = 2 √ 2, X1 = √ 2 / 4 ‖ X3 = x2 * 2 = √ 2 / 2, X4 = x2 * 4 = √ 2. By using Veda's Theorem

Given that the line L passes through the point P (2.3) and the inclination angle α = π / 6, the parameter equation of the line L is written out
(1) Write the parametric equation of the line
(2.) let L and circle x ^ 2 + y ^ 2 = 4 intersect at two points a and B, and find the distance between point P and a and B

(1)α=π/6 k=tanα=1/2
∴y-3=1/2(x-2)
2y-x-4=0
(2)x^2+y^2=4
x=2y-4
Substituting can get
5y²-16y+12=0
(5y-6)(y-2)=0
y1=6/5,y2=2
A(-8/5,6/5)B(0,2)
The formula of distance between two points
PA=3√5/5
PB=√5

Given that the straight line L passes through the point P (1,1) and the inclination angle is π / 6, it intersects with the circle x ^ 2 + y ^ 2 = 4 and two points a and B, find the midpoint coordinates of AB?

1. Using the point P and the inclination angle π / 6, the linear equation is obtained;
2. Write the linear equation in the form of y = f (x) and substitute it into the circular equation. In this way, we can get a quadratic equation with X. the two abscissa of this equation are abscissa A and abscissa B. in fact, what you need is (x1 + x2) / 2. Just use the Weida theorem. In this way, we can get the abscissa of the point in AB, and then substitute this coordinate into the linear equation to get the midpoint coordinate

Given that the straight line L passes through the point P (1,1), and the inclination angle a = 30 °, find the parameter equation of the straight line L

1. (Y-1) / (x-1) = tan30 degree = sin30 degree / cos30 degree, so Y-1 = ksin30 degree; X-1 = kcos30 degree, that is, x = 1 + kcos30 degree, y = 1 + ksin30 degree, K is a parameter of R

Let the inclination angle of the line L passing through the point m (1,5) be π / 3, and find the sum and product of the distances from the two intersections of the line L and the circle: x ^ 2 + y ^ 2 = 16 to the point M0
Solving with parametric equation

(1) firstly, the parameter equation of the straight line L is obtained as follows: x = 1 + 1 / 2T ①, y = 5 + (√ 3) / 2 * t ② (t is the parameter)
Then we substitute the deformed t = 2x-2 of ① into ② to get the straight line L: y = 5 + √ 3x - √ 3
(2) let two points be a and B
Substituting x = 1 + 1 / 2T, y = 5 + (√ 3) / 2 * t (t is a parameter) into the circle: x ^ 2 + y ^ 2 = 16, we get: 10 + T ^ 2 + T * (1 + 5 √ 3) = 0
Because | ab | = | T1-T2|
|AB|^2=（t1-t2）^2
So | ab | ^ 2 = (T1 + T2) ^ 2 - 4T1 * T2
From Veda's theorem, T1 + T2 = - 1-5 √ 3
t1*t2=10
So | ab | ^ 2 = (1-5 √ 3) ^ 2-4 * 10
|AB|^2 =36
|AB|=6
|MA|* |MB|
= |t1|* |t2|
= |t1*t2|
=10
So the sum of distances is 6; the product is 10

1. Parametric equation of straight line 2. Intersection of straight line and curve P = 2 at two points, find the product of distance from point to intersection P

The slope 30 ° tangent can be obtained from the inclination angle of 30 ° which is not convenient for me to input. If the slope of the line and a point on the line are known, the equation of the line can be obtained. Curve P = 2? What do you mean? I don't understand

Find the product of the distance between the two intersections of L and circle x ^ 2 + y ^ 2 = 16 and point P
It is known that the line L passes through point P (1, - 5) and the inclination angle is д / 3

According to the cutting line theorem, if l intersects a circle at M and N, then PM * PN = the square of the tangent length from P to the circle,
So, the value is 1 ^ 2 + (- 5) ^ 2 - 16 = 10

Given that the straight line passes through the point P (1,1) and the inclination angle is 30 degrees, let l intersect the circle x * x + y * y = 4 and find the point a and B, and find the product of the distance from P to a and B

The product of the distances from P to a and B is 2
The equation of line L is Y-1 = (x-1) / sqrt (3), that is, sqrt (3) * Y-X + 1-sqrt (3) = 0. Let the origin of the coordinate be o, and the perpendicular line of line L is made through o point. The perpendicular foot is C, that is, the length of OC is the distance from O to line L,
Then | OC | = | 1-sqrt (3) | / sqrt (sqrt (3) ^ 2 + (- 1) ^ 2)
=(sqrt(3)-1)/2
Let the length of chord length AB be D, the distance from P to point a be D1, and the distance from P to point B be d-d1,
And D / 2 = sqrt (2 ^ 2 - | OC | ^ 2)
=sqrt(3+sqrt(3)/2)
The length of OP is | op | = sqrt (1 ^ 2 + 1 ^ 2) = sqrt (2),
So the length of CP | CP | = sqrt (| op | ^ 2 - | OC | ^ 2) = sqrt (1 + sqrt (3) / 2)
So the product of the distances from P to a and B
|PA|*|PB|
=(d/2-|CP|)*(d/2+|CP|)
=(d/2)^2-|CP|^2
=3+sqrt(3)/2-(1+sqrt(3)/2)
=2