Given that m and N are two equations x ^ 2 + 5x + 2 = 0, find the median term of M and n

Given that m and N are two equations x ^ 2 + 5x + 2 = 0, find the median term of M and n

Using the Veda theorem
X1×X2=c/a
So m × n = 2
So the equal ratio median of M and N is the root 2

Given that the real numbers m, 6 and - 9 form an equal ratio sequence, then the eccentricity of conic curve x2m + y2 = 1 is 0______ ．

Given that the real numbers m, 6, - 9 constitute an equal ratio sequence, then: 36 = - 9m solution: M = - 4, conic equation is: Y2 − x24 = 1, then: a = 1, B = 2, according to C2 = A2 + B2 solution: C = 5, according to e = CA solution: e = 5, so the answer is: 5

If M is the median of 2 and 8, then the eccentricity of conic x ^ 2 + y ^ 2 / M = 1 is 0

m²=2x8
m=±4
When m = - 4
x^2-y^2/4=1
e = c/a=√5
When m = 4
x^2+y^2/4=1
a=2,b=1,c=√3
e=c/a=√3/2.
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For a straight line passing through point m (1,5) with an inclination angle of 60, the equation with the displacement t from point m to point P on the straight line as a parameter is

{x=1+tcos60
{y=5+tsin60
.
{x=1+(1/2)t
{y=5+(√3/2)t

Let the line L pass through the point M0 (1,5) and the inclination angle π / 3
(1) Finding the parameter equation of line L
(2) Find the distance from the intersection of line L and line x-y-2 √ 3 = 0 to point M0
(3) Find the sum and product of the distance between the two intersections of the line L and the circle: x ^ 2 + y ^ 2 = 16 and the point M0
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If the answer is good, there will be a reward

(1) According to the meaning of the question, then the parameter equation should be x = 1 + tcos60 ° y = 5 + tsin60 ° that is, x = 1 + 0.5T, y = 5 + T * (root sign 3) / 2T is the parameter (2). The equation of the straight line l can be obtained as y = √ 3x + 5 - √ 3, then x-y-2 √ 3 = 0, and the distance between the intersection (- 4-3 √ 3, - 4-5 √ 3) and M can be obtained

Given that the straight line L passes through the fixed point P (- 3, - 3 / 2), intersects the circle C x = 5cos θ, y = 5sin θ (θ is a parameter), and finds the equation of the straight line L if | ab | = 8
(2) If the point P (- 3, - 3 / 2) is the midpoint of the string AB, the equation of the string AB is obtained

The parametric equation becomes the ordinary equation (1) x = 3-2t y = - 1-4t (2) x = 5cos θ + 1 y = 5sin θ - 1 (T and θ are parameters)

(1) Because x = 3-2t, y = - 1-4t, then 2x-y = 2 (3-2t) - (- 1-4t) = 7. That is, the ordinary equation is 2x-y-7 = 0. (2) because x = 5cos θ + 1, y = 5sin θ - 1, then (x-1) ^ 2 + (y + 1) ^ 2 = (5cos θ) ^ 2 + (5sin θ) ^ 2 = 25, and X belongs to (- 4,6), y belongs to (- 6,4)

Given that the circle C is tangent to the XY axis and the root of the distance from the center of the circle to the straight line y = - x is 2, the equation of circle C can be solved if the straight line X / M + Y / N = 1
(m, n greater than 2) is tangent to circle C, and the value range of Mn is obtained

It's easy to know that the circle C: (x-1) ;; + (Y-1) \\\\\\\\ [[1]]]] [[[[[1]]]]]]]]]] [[[2]]]]]]]] [[[[[2]]]]]]]]]]]]]]], at this time, the circle C: (x-1) (x-1) \ [(x-1) [(x-1) \\[[[x-1) \\ [[[[[[X-1]]]]]]]]]]]]]]]; \\\\\\\+ n) = (m-1) (n-1)

The center of the circle C tangent to the y-axis is on the straight line y = 1 / 2x, and the distance from the center of the circle to the straight line y = x is the root sign 2 / 2. The equation of the circle C is obtained

y=x+b b=±(√2/2)/(cos45)
y=x±(√2/2)/(√2/2)=x±1
Y = x + 1 and y = x / 2 intersect (- 2, - 1), r = | - 2 | - corresponding circle (x + 2) + (y + 1) ^ 2 = 4
Y = X-1 and y = x / 2 intersect (2,1), r = 2, corresponding to circle (X-2) ^ 2 + (Y-1) ^ 2 = 4

It is known that the line and the circle x ^ 2 + y ^ 2 = 1 are tangent in the first image line, and the length of the line sandwiched between the two coordinate axes is equal to 4 root sign 3 / 3

Let the linear equation be x / A + Y / b = 1. From the meaning of the problem, we can get a, b > 0, a ^ 2 + B ^ 2 = (4 root sign 3 / 3) ^ 2 = 16 / 3 straight line, that is: BX + ay AB = 0, the distance from the center of the circle to the straight line = radius = 1, that is, | ab | / root sign (a ^ 2 + B ^ 2) = 1 | ab | ^ 2 = 16 / 3A ^ 2B ^ 2 = 16 / 3 = a ^ 2 + B ^ 2. The solution is: A ^ 2 = 4 / 3 or a ^ 2 = 4, B ^ 2 = 4 / 3 straight line is: X / (2 /