A set of points on a plane that are all at equal distances to a curve

You're talking about least squares, curve fitting to the lattice.

Simple curve fitting can be used to make trend lines of curves in Excel. There are linear, polynomial, power, exponential, logarithmic, etc.

Guo Dunwen: As a simple example, it can be regarded as your review of middle school mathematics knowledge

Give the curve a: function sin x, 0 x 2 (the irregular curve is not easy to describe, and the regular one is given first); Straight line b:y=. Find the set p of points where the vertical distance of curve A from straight line b is.

solution, p=.

The coordinates of A1, A2, A3 and A4 are:

a1（x1，y1），a2（x2，y2），a3（x3，y3），a4（x4，y4），y1= y2=，x1= arc ，x2=π－π/6=（5/6）π；

y3=y4= (,x3= arc sin( 4576°=,x4=2 (The coordinates of each point are:a1( 6,,a2(5 6,,a3(, a4(,

If the curve a is an irregular curve, and the set p of points where the vertical distance of curve a from line b is d, then the curve a needs to be drawn first, and then the line b1 line b and the line b2 line b, the line b1 and the line b2 are located on both sides of the line b, and the distance between the line b1 and the line b and the line b2 and the line b are d, then the set of the intersection points of the line b1 and the line b2 and the curve a is p.

The positional relationship between circles.

1.Intersect. The sum of the distances between the centers of the two circles is less than the sum of the radii of the two circles.

2.Tangent. Inscribed: The sum of the distances between the centers of the two circles is equal to the sum of the radii of the two circles.

Incision: The sum of the distances between the centers of the two circles is equal to the difference between the radii of the two circles.

3.distancing. Alienation: The sum of the distances between the centers of the two circles is greater than the sum of the radii of the two circles.

Contains: The sum of the distances between the centers of the two circles is less than the difference between the radii of the two circles.

It's two straight lines.

In space, the set of points whose distance to a fixed line is equal to the fixed length is a cylinder (excluding the upper and lower bottom surfaces), and that fixed line is actually the axis of the cylinder, you imagine, it is easy to understand. Therefore, if we do not consider the special case of tangent, the intersection of a plane and a cylinder will produce two straight lines, which is the set of the points you want, so the set of all the points on the plane whose distance to the fixed length is equal to the fixed length is two straight lines.

distance determination, 2 pcs.; Distances are variable, innumerable.

Without losing generality, let the hyperbolic equation be: x 2 a 2 y 2 a 2 1, and get:

The focal coordinates of the hyperbola are f1 (2a,0), f2 (2a,0), and the center coordinates of the hyperbola are o(0,0).

Let a(m,n) be a point on the hyperbola. Then:

af1｜＝√m＋√2a）^2＋n^2］，｜af2｜＝√m－√2a）^2＋n^2］。

af1｜｜af2｜＝√m＋√2a）^2＋n^2］［（m－√2a）^2＋n^2］｝

（m^2－2a^2）^2＋（m^2＋2a^2＋2√2am＋m^2＋2a^2－√2am）n^2＋n^4］

m^4－4a^2m^2＋4a^4＋（2m^2＋4a^2）n^2＋n^4］

m^4＋2m^2n^2＋n^4＋4a^4－4a^2m^2＋4a^2n^2）

（m^2＋n^2）^2＋4a^4－4a^2（m^2－n^2）］

Obviously, m, n satisfy the hyperbolic equations, m 2 a 2 n 2 a 2 1, m 2 n 2 a 2.

af1｜｜af2｜＝√m^2＋n^2）^2＋4a^4－4a^2（a^2）］＝m^2＋n^2。

and ao 2 (m 0) 2 (n 0) 2 m 2 n 2.

af1｜｜af2｜＝｜ao｜^2。Proof is complete.

Alas. Only those who have been hit with bricks have a memory. (just kidding).

I drew a brick.

Why? Three-dimensional geometry, from life. So, learn solid geometry:

1: Learn to read a three-dimensional picture. 2：

Learn to draw simple three-dimensional diagrams. 3: Master some reasoning methods and computing skills.

4. Prepare for further study or going out into the society in the future.

In this case, when considering the problem, what are some common things that can be referenced? - Slab brick - cuboid.

Plane 1234 is the "median" parallel plane of A and B. 14//a,12//b.

Plane 5678 is the "median" parallel plane of A and C. 85//a,87//c。

The intersection of plane 1234 and plane 5678 is a straight line mn.

In the same way, there is also a plane, which I did not mark, so that there are three straight lines that have the same point, which is the point p.

If, the distance between these three two perpendicular lines is equal, in other words, they are the three edges of a cube, then there is and only the above point p satisfies the condition of the problem. The distance from this point to the third line is equal to the root number two of the length of the edge of the cube.

If the distance between these three two perpendicular straight lines is not all equal (including that they are not equal), then they are only "cuboids", then let's first analyze: how many points are there to two straight lines perpendicular to each other with equal distances? In which places?

At least 5 points meet the requirements.

One is the midpoint of the "common perpendicular segment". In addition, the two perpendicular feet of the common perpendicular line segment lead to the parallel lines of the straight lines of different surfaces. The "length of the common perpendicular segment" is intercepted from the vertical foot to the four rays, and four dots appear. So a total of 5 points match.

In other cases, it should be: pass a perpendicular foot, lead another parallel line of a straight line, and then make a bisector of the newly formed angle, and the plane formed by this bisector and the "common perpendicular line" (we call it the "angular plane") has four "half planes". The points that meet the needs should all be within these four "half planes".

Below, I haven't had time to analyze it in detail. I don't know, does the above analysis inspire you?

There are the following 4 straight lines, and the equations of the straight lines are given as follows in the form of parameters.

l1：x=t,y=t,z=t

l2：x=t,y=t,z=-t

l3：x=t,y=-t,z=t

l4：x=-t,y=t,z=t

All four straight lines pass through the origin of space, and each line passes through two hexagram limits relative to each other (eight lines of dry hexagrams, each line passes through a pair of hexagram limits).