Given that the line L: x + 2Y = 0, P is any point on the ellipse x ^ 2 / 4 + y ^ 2 = 1, find the maximum distance between the point P and the line l?

Given that the line L: x + 2Y = 0, P is any point on the ellipse x ^ 2 / 4 + y ^ 2 = 1, find the maximum distance between the point P and the line l?

Using the parametric equation, the point on the ellipse is (2cos θ, sin θ), and the distance from the point to the straight line is | 2cos θ + 2Sin θ| / √ (1 + 2 ^ 2)
=2|cosθ+sinθ| /√5 =2√2 |sin(θ+π/4)| /√5
When the maximum value is sin (θ + π / 4) = 1 or - 1,
The maximum value is 2 √ 2 / √ 5

Let X & # 178 / M & # 178; + Y & # 178; / (M & # 178; - 1) = 1 (M > 1),
The distance from P to the right guide line is 1

That is 1 + 3 = 2m
m=2
Then a & # 178; = M & # 178; = 4
b²=m²-1=3
c²=1
So e = C / a = 1 / 2
That is, the distance from P to the right focus △ p to the right guide line = e = 1 / 2
So the distance from P to the right guide line is 2

The distance from the right focus of ellipse X & # 178 / 4 + Y & # 178 / 3 = 1 to the straight line y = √ 3x is

According to the elliptic equation, the focus is on the x-axis. The right focus coordinate is (C, 0) and C = √ (A & # 178; - B & # 178;) = 1. Change y = √ 3x into √ 3x-y + 0 = 0, and then get the distance L = √ 3 / 2 according to the formula from point to line

{x+2y+z=8,2x-y-z=-3,3x+y-2z=-1

{x+2y+z=8 （1）
{2x-y-z=-3 （2）
{3x+y-2z=-1 （3）
（1）+（2）
3x+y=5 （4）
（1）*2+（3）
5x+5y=15
x+y=3 （5）
（4）-（5）
2x=2
x=1
Substituting (5) to get the solution
y=2
Substituting x = 1 and y = 2 into (1) yields
1+4+z=8
z=3
The solution of the equations is
{x=1
y=2
z=3

x-y+2z=-1,2x+y-z=-1,3x-2y-2z=-9

X-Y + 2Z = - 1 (1) 2x + Y-Z = - 1 (2) 3x-2y-2z = - 9 (3) get y = x + 2Z + 1 from (1) and substitute it into (2) and (3) respectively, get 3x + Z = - 2 (4) x-6z = - 7 (5) (4) × 6 + (5) 19x = - 19x = - 1 substitute it into (4) z = - 3x-2 = - 3 (- 1) - 2 = 1y = x + 2Z + 1 = - 1 + 2 + 1 = 2x = - 1, y = 2Z = 1

Specific (3x-2y) ^ 2 - (3x + 2Y) ^ 2

(3x-2y)^2-(3x+2y)^2
=(3x-2y+3x+2y)(3x-2y-3x-2y)
=6x*(-4y)
=-24xy
Do not understand can ask, help please adopt, thank you!

If the variables X and y satisfy the constraint condition x + y ≤ 82y − x ≤ 4x ≥ 0y ≥ 0, then the maximum value of Z = 5y-x is 0______ ．

The plane region (shadow part) corresponding to the inequality is given. From z = 5y-x, y = 15x + Z5 is obtained, and the translation line y = 15x + Z5 is obtained. From the image, when the line y = 15x + Z5 passes through point B, the intercept of the line y = - 2X + Z is the largest, and then Z is the largest. From x + y = 82y − x = 4, the solution is x = 4Y = 4, that is, B (4,4)

Given that z = 2x-y, variables X, y satisfy the constraint condition y ≤ XX + y ≥ 1 x ≤ 2, then the maximum value of Z is ()
A. 0B. 5C. 6D. 10

Draw the feasible region, deform the objective function to y = 2x-z, and make its corresponding straight line. When it is translated to (2, - 1), the longitudinal intercept of the straight line is the minimum. At this time, Z is the maximum, and the maximum value of Z is 5, so B

Find the maximum and minimum values of Z = 2x + y, so that X and Y in the formula satisfy the constraint condition {x + y is greater than or equal to 1, X is less than or equal to 1, y is less than or equal to 1

Draw the coordinate diagram under the condition of equality, draw the range according to the symbol, and take the focus
When y = 1 and x = 0, the minimum value is 1
When y = 1 and x = 1, the maximum value is 3

Given: | x | ≤ 1, | y | ≤ 1, let m = | x + 1 | + | y + 1 | + | 2y-x-4 |, find the maximum and minimum of M

∵|x|≤1,|y|≤1
∴-1≤x≤1,-1≤y≤1
∴x+1≥0,y+1≥0
2y-x-4≤0
∴M=|x+1|+|y+1|+|2y-x-4|
=x+1+y+1-(2y-x-4)
=2x-y+6
And - 1 ≤ x ≤ 1, - 1 ≤ y ≤ 1
Then - 2 ≤ 2x ≤ 2, - 1 ≤ - y ≤ 1
So - 3 ≤ 2x-y ≤ 3
So 3 ≤ 2x-y + 6 ≤ 9
That is, 3 ≤ m ≤ 9, the maximum value of M is 9, and the minimum value is 3