On the hyperbola y = 2 2 / x, the coordinates of the point with the distance of 2 from the Y axis are_______ .

（2,2）

As shown in the figure, a (- 1,6) is a point on the hyperbola y = KX (x < 0), and P is a point on the positive half axis of the Y axis. Rotate a 90 ° counterclockwise around point P, and just fall on another point B of the hyperbola. Try to find the coordinates of point P

∵ a (- 1,6) is a point on hyperbola y = KX (x ＜ 0), and the analytic expression of the inverse proportion function is y = - 6x. Let P (0, m), ∵ AP ⊥ BP, and AP = BP, ∵ B (M-6, m-1), ∵ point B is on the parabola, ∵ M-1 = - − 6m − 6. The solution is m = 3 or M = 4, ∵ P (0,3) or (0,4)

If the imaginary axis length of the hyperbola MX ^ 2 + y ^ 2 = 1 is twice the real axis length, then M =? 1

The imaginary axis length of the hyperbola MX & sup2; + Y & sup2; = 1 is twice the real axis length
The hyperbolic equation is (- X & sup2 / 4) + Y & sup2; = 1
∴m＝－1/4

How to find the projection domain of a surface on the xoz coordinate plane?
The curve equation is as follows:

Cast it on that surface, so that the other coordinate value except the surface is 0, for example
If we cast it on xoy, then z = 0, if we steal xoz, then y = 0
But if you look at the equation carefully, there is no solution

Given the coordinates of a point and the plane equation, the coordinates of the projection point of the point on the plane are obtained
Given a point coordinate m (x1, Y1, z1) and a plane equation AX + by + CZ + D = 0
Find the coordinates (X2, Y2, Z2) of the projection M 'of point m on the plane
I need a formula to express the coordinates of M 'with x 1, Y 1, Z 1, a, B, C, D
Only the results are required
The process can be roughly described, let me know the result is correct

M 'satisfied
x2=x1+kA
y2=y1+kB
z2=z1+kC
Ax2+By2+Cz2+D=0
The solution is k = - (ax1 + by1 + CZ1 + D) / (a + B + C)
Dai Hui

How to understand the projection of space surface area DS onto xoy plane?
It is said in the book that if the area of the space surface is DS, then the area projected to the xoy plane is DS * cos γ (cosine of the direction of Z axis). I don't understand very well here. If cos γ is not projected to the xoz plane, then DS * cos (pi / 2 - γ) should be projected to the xoy plane, right, A is parallel to xoy, B is projected onto xoy, and the length should be b * sin α. What's the matter, multiplied by cosine

The lower right is the enlarged picture, DS = a * B simple case, a is parallel to xoy, B is projected onto xoy, the length should be b * sin α, what's the matter multiplied by cosine

How to find the projection of space surface on plane (including coordinate plane)

Cast it on that surface, so that the other coordinate value except the surface is 0, for example
On xoy surface, then z = 0

What is the maximum area of projection when a cube with edge length of 1 is projected onto a plane

For convenience, let the cube be abcd-a'b'c'd '
When the projection is maximum, the plane should be parallel to ab'c,
The projection of three planes is three congruent rhombus
The long diagonal of diamond is √ 2
The three diagonals on the projection form an equilateral triangle with a side length of √ 2
And the area of projection = the area of equilateral triangle x 2
S = 1/2 x √2 x √(6)/2 = √3

How to find the projection of an individual on each plane?

"What are the dimensions of our world?"
"Three dimensional!"
"Can you imagine a four-dimensional world?"
A child can't answer by scratching his head
Yes, for human beings living in three-dimensional space, the four-dimensional world is a very mysterious concept. Just as it is difficult for small people living in two-dimensional world (if they exist) to imagine the three-dimensional world, it is also difficult for us to imagine the four-dimensional world. But just as we can study and imagine three-dimensional objects by studying the projection of three-dimensional objects on two-dimensional objects, We can also study the four-dimensional world by the projection of the four-dimensional objects in the three-dimensional world
Figure 1 shows the projection of a cube in the two-dimensional world. The two-dimensional villains can more or less imagine the mysterious figure of the "three-dimensional cube" through these projections. They can count out that the cube has 8 vertices, 12 sides and 6 faces. Don't be frightened by the weird toys in Figure 2, It's just the projection of a four-dimensional cube in the three-dimensional world. We call it a four-dimensional cube. We can imagine that a cube has 16 vertices, 32 edges, 24 faces and 8 individuals
If it's hard to imagine a four-dimensional cube, let's try a sphere instead. For a three-dimensional sphere, no matter from which direction it is projected on the two-dimensional plane, it's just a semicircular equivalent circle, so it's easy to think that the projection of a four-dimensional sphere in the three-dimensional world is just a sphere with the same radius, For example, if a ball passes through a two-dimensional plane, the two-dimensional villain will find a slowly growing point on the plane, and then it will expand into a circle, slowly shrink into a point, and finally disappear suddenly, Then you will be surprised by the following situation: a point appears in front of you out of nothing, expands into a ball, retracts, and then suddenly disappears. How amazing! In fact, this is just the case of a four-dimensional ball crossing the three-dimensional world. With a super sphere, curious people want to study its volume and surface area. Due to the limitation of mathematical knowledge, we can't find it, but we went to college and studied calculus, You can generalize the surface area of a sphere to N5?
Is it just out of human curiosity to study the four-dimensional world? Is there any practical significance? In the physical world, we have found a suitable fourth axis for the four-dimensional world. However, the time axis is very different from the other three axes. Time is measured by a stopwatch, and space is measured by a ruler. In space, you can come and go freely, But time passes like this, never to return, can only advance, cannot retreat
So how to make time equal to space as much as possible? We need a "standard speed" to transform. Last time in the theory of relativity, we knew that the speed of light is constant in any reference frame, so this is a better standard, so time finally has a scale matching with other axes
People are always in pursuit of deeper knowledge. I don't know whether it is due to curiosity or something else. In a word, with the real four-dimensional world, people want to seek the four-dimensional distance of two points. After a lot of lively calculations based on Einstein's theory of relativity, we come to the conclusion that the four-dimensional distance is. Maybe you don't like to use ICT to scale, but you can make do with it
Actually, strictly speaking, I'm a bit lazy. I didn't tell the story of the four-dimensional world strictly according to the development of the laws of physics. In fact, Einstein first got the space-time invariant x2 + Y2 + z2-c2t2 in the theory of relativity, and then put forward that ICT can be used as a four-dimensional axis to get the four-dimensional world in real life
However, science is pluralistic. No matter where you start, the end will always tend to be the same. Human beings have taken many detours. For example, Einstein can directly get the theory of the dynamic expanding universe from his equation, but he foolishly adds a balance term in order to keep the theory of the static universe, which makes the readily available big bang theory disappear, However, it is the bold conjectures and hypotheses of human beings, and human beings are leaping over the ravines on the road of exploration. The confirmation of Planck's quantum hypothesis and Rutherford's atomic model, one by one, makes human science continue to advance and develop

Given the coordinates of three points, solve the plane equation
Three known points a (x1, Y1, z1), B (X2, Y2, Z2), C (X3, Y3, Z3)
I want to know that in the plane formula ax + by + CZ + D = 0
How to use x1, Y1, Z1, X2, Y2, Z2, X3, Y3, Z3 to express a, B, C, D of
That is to help me solve the equation of degree 4 without specific number. It's not clear if I'm not good at mathematics
It doesn't matter. That is to say, let a, B, C and d be represented by the coordinates x1, Y1, Z1, X2, Y2, Z2, X3, Y3 and Z3.
I want results, not methods
The problem itself is either difficult or troublesome

You want the results, right, OK
Vector AB = vector ob - vector OA = (x2-x1, y2-y1, z2-z1)
Vector AC = vector OC vector OA = (x3-x1, y3-y1, z3-z1)
Vector ab × vector AC = ([y1z2-y1z3-y2z1 + y2z3 + y3z1-y3z2], [- x1z2 + x1z3 + x2z1-x2z3-x3z1 + x3z2], [x1y2-x1y3-x2y1 + x2y3 + x3y1-x3y2])
That is, a = y1z2-y1z3-y2z1 + y2z3 + y3z1-y3z2, B = - x1z2 + x1z3 + x2z1-x2z3-x3z1 + x3z2, C = x1y2-x1y3-x2y1 + x2y3 + x3y1-x3y2
(x1, Y1, Y1, Z1, bring (x1, Y1, Y1, Y1, Z1, z1), get (y1z2-y1z2-y1z2-y1z2-y1z2-1z2-y1z2-y1z3-1z3-yz3-yz3-y2z3-y2z1 + y2z3 + y3z3 + y3z3 + y3z3 + y3z3 + y3z1-h3z1-y3z1-y3z1-y3z1-y3z1-y3z1-y3z1-z1-1z3-1z1-1z3-1z3-1yz2-1yz2-1y11y11z2-x1y11z2-x11z2-x11z2-x11z2-x1y11z2-x11z2-x11z2-x1-1z2-x1-1z2-x1-1yz2-11-x2y1z1 + x2y3z1 + x3y1z1-x3y2z1 + D = 0, that is, x1y2z3-x1y3z2-x2y1z3 + x2y3z1 + x3y1z2-x3y2z1 + D = 0, That is, d = - x1y2z3 + x1y3z2 + x2y1z3-x2y3z1-x3y1z2 + x3y2z1
So a = y1z2-y1z3-y2z1 + y2z3 + y3z1-y3z2, B = - x1z2 + x1z3 + x2z1-x2z3-x3z1 + x3z2, C = x1y2-x1y3-x2y1 + x2y3 + x3y1-x3y2, d = - x1y2z3 + x1y3z2 + x2y1z3-x2y3z1-x3y1z2 + x3y2z1

On the hyperbola y = 2 2 / x, the coordinates of the point with the distance of 2 from the Y axis are_______ .