Find the minimum value of 2x ^ 2 + 2Y ^ 2 + 4Y + 2x + 2XY + 5

Find the minimum value of 2x ^ 2 + 2Y ^ 2 + 4Y + 2x + 2XY + 5

three

If x and y are real numbers, and 1 ≤ x + 4Y ≤ 2, find the maximum and minimum of x-2xy + 4Y. The gods help

Due to 1

On the ellipse x ^ 2 / 2 + y ^ = 1, find the minimum value of X + y and the maximum value of Y + 2 / x + 2

Let x + y = t, and x ^ 2 / 2 + y ^ = 1, eliminate y to get: x ^ 2 + 2 (t-x) ^ 2-2 = 0, that is, 3x ^ 2-4tx + 2T ^ 2-2 = 0 equation has solution, so Δ = 16t ^ 2-12 (2t ^ 2-2) ≥ 0 ∧ 8t ^ 2 ≤ 24 ∧ - √ 3 ≤ t ∧ 3 ∧ the minimum value of X + y is - √ 32, let (y + 2) / (x + 2) = t ∧ y = TX + 2t-2 and x ^ 2 / 2 + y ^ 2 = 1, eliminate y

The equation of the ellipse is x ^ 2 + y ^ 2 / 2 = 1, and the straight line L: x + 2y-4 = 0. Is there a point on the ellipse so that the distance from it to the straight line L has a maximum? If so, find the maximum

If L1: x + 2y-t = 0 is a line parallel to L, then: x = t-2y Generation X ^ 2 + y ^ 2 / 2 = 1, then: (t-2y) ^ 2 + y ^ 2 / 2 = 19y ^ 2-8ty + 2T ^ 2-2 = 0 discriminant △ = 64t ^ 2-36 (2T ^ 2-2) = - 8t ^ 2 + 72 = 0t ^ 2 = 9t = ± 3. Therefore, when t = 3, y = 4 / 3, x = 1 / 3, point (1 / 3,4 / 3) to the line L has the minimum value = √ 5 / 5t = - 3, y

Finding the shortest distance from the point on the ellipse 16x ＾ 2 + 9y ＾ 2 = 144 to the straight line x + y = 7

Let x = 3cos θ, y = 4sin θ be the point on the ellipse, then the distance from the point (3cos θ, 4sin θ) to the straight line x + y = 7 is d = | 3cos θ + 4sin θ - 7 | / √ (1 & sup2; + 1 & sup2;) = | 5sin (θ + arctan (3 / 4)) - 7 | / √ 2, when θ = π / 2-arctan (3 / 4

Find the maximum and minimum of the distance between the point with x square and Y square = 1 on the ellipse and the root 3 of the line y = x + 2

Let the coordinate of this point be (√ 2cosa, Sina), then:
The distance between the point and the line is
D= │√2COSA-SINA+2√3│/√2
=│√3SIN（A-B）+2√3│/√2,SINB=√6/3,COSB=√3/3
Then Dmin = √ 6, Dmax = 3 √ 6 / 2

The minimum distance between the point on the ellipse X & # 178 / 16 + Y & # 178 / 9 = 1 and the line L: x + Y-9 = 0 is

Set the point as (4cosa, 3sina)
Then the distance is | 4cosa + 3sina-9 | / √ (1 & # 178; + 1 & # 178;)
=|5sin(a+b)-9|/√2
\Where tanb = 4 / 3
-1

The minimum distance between ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1 (x ≥ 0, y ≥ 0) and straight line x-y-5 = 0 is
RT

2

Find the area of the shadow part. There are four petal ellipses in a square. The side length of the square is 8 decimeters, and the shadow part is 4

The area of the shaded area
=Sum of the areas of four semicircles - square area
=The area of two circles with a diameter of 8 decimeters square area
=[3.14×(8÷2)²]×2-8×8
=100.48-64
=36.48 (square decimeter)

Given a certain point m (a, 0) on the major axis of the ellipse x ^ 2 / 36 + y ^ 2 / 20 = 1 and constant a > 0, the minimum distance d from point to point m on the ellipse is obtained

Let's find d ^ 2 First: D ^ 2 = (x-a) ^ 2 + y ^ 2 = x ^ 2-2ax + A ^ 2 + 20 (1-x ^ 2 / 36)
=4X ^ 2 / 9-2ax + A ^ 2 + 20 is a quadratic function whose domain of definition is [- 6,6],
The opening of quadratic function is upward, and the axis of symmetry is x = 9A / 4
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