What is a parabola. Parabola and its properties What equation defines a parabola

ODA 1.parabola is the locus of points in the plane, the distances from which to some point, called the focus, and to some straight line, called the directrix, are equal.

To derive the parabola equation, we introduce a rectangular coordinate system on the plane so that the abscissa axis passes through the focus perpendicular to the directrix, and we will consider its positive direction the direction from the directrix to the focus. The origin of coordinates is located in the middle between the focus and the directrix. Let's derive the parabola equation in the chosen coordinate system.

Let M ( X; at) is an arbitrary point of the plane.

Denote by r distance from point M to focus F, let r= FM,

through d is the distance from the point to the directrix, and through R distance from focus to directrix.

the value R is called the parabola parameter, its geometric meaning is discussed below.

The point M will lie on the given parabola if and only if r=d.

In this case we have

The equation

y 2 = 2 px

called the canonical equation of the parabola .

Parabola Properties

1. The parabola passes through the origin, because the origin coordinates satisfy the parabola equation.

2. The parabola is symmetrical about the OX axis, because points with coordinates ( x, y) And ( x, − y) satisfy the parabola equation.

3. If R> 0, then the branches of the parabola are directed to the right and the parabola is in the right half-plane.

4. Point O is called the vertex of the parabola, the axis of symmetry (axis Oh) is the axis of the parabola.

A parabola is the locus of points in a plane equidistant from a given point F

and a given line dd not passing through a given point. This geometric definition expresses parabola directory property.

Directory property of parabolas

The point F is called the focus of the parabola, the line d is called the directrix of the parabola, the midpoint O of the perpendicular dropped from the focus to the directrix is ​​the vertex of the parabola, the distance p from the focus to the directrix is ​​the parameter of the parabola, and the distance p2 from the vertex of the parabola to its focus is the focal length. The straight line perpendicular to the directrix and passing through the focus is called the axis of the parabola (the focal axis of the parabola). The segment FM connecting an arbitrary point M of the parabola with its focus is called the focal radius of the point

M. The segment connecting two points of the parabola is called the chord of the parabola.

For an arbitrary point of the parabola, the ratio of the distance to the focus to the distance to the directrix is ​​equal to one. Comparing the directory properties of the ellipse, hyperbola and parabola, we conclude that parabola eccentricity is by definition equal to one

Geometric definition of a parabola, expressing its directory property, is equivalent to its analytical definition - the line given by the canonical equation of the parabola:

Properties

• It has an axis of symmetry called parabola axis. The axis passes through the focus and the vertex perpendicular to the directrix.
• optical property. A beam of rays parallel to the axis of the parabola, reflected in the parabola, is collected at its focus. Conversely, light from a source that is in focus is reflected by a parabola into a beam of rays parallel to its axis.
• If the focus of the parabola is reflected with respect to the tangent, then its image will lie on the directrix.
• The segment connecting the midpoint of an arbitrary chord of the parabola and the intersection point of the tangents to it at the ends of this chord is perpendicular to the directrix, and its midpoint lies on the parabola.
• The parabola is the antipodera of the line.
• All parabolas are similar. The distance between the focus and the directrix determines the scale.

Function of one real variable: basic concepts, examples.

Definition: If each value x of a numerical set X according to the rule f corresponds to a single number of the set Y, then they say that the function y \u003d f (x) is given on the numerical set X, the values ​​of x are determined by the set of values ​​included in the domain of the function (X).
In this case, x is called the argument and y is called the value of the function. The set X is called the domain of definition of the function, Y is called the set of values ​​of the function.
Often this rule is given by a formula; for example, y \u003d 2x + 5. The specified method of specifying a function using a formula is called analytical.
The function can also be set by a graph - The graph of the function y - f (x) is the set of points in the plane, the coordinates x, which satisfy the relation y \u003d f (x).

Function of the form , where is called quadratic function.

Graph of quadratic function − parabola.

Consider the cases:

CASE I, CLASSICAL PARABOLA

That is , ,

To build, fill in the table by substituting x values ​​into the formula:

Mark points (0;0); (1;1); (-1;1) etc. on the coordinate plane (the smaller the step we take x values ​​(in this case, step 1), and the more x values ​​we take, the smoother the curve), we get a parabola:

It is easy to see that if we take the case , , , that is, then we get a parabola symmetric about the axis (ox). It is easy to verify this by filling out a similar table:

II CASE, "a" DIFFERENT FROM ONE

What will happen if we take , , ? How will the behavior of the parabola change? With title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;"> парабола изменит форму, она “похудеет” по сравнению с параболой (не верите – заполните соответствующую таблицу – и убедитесь сами):!}

The first picture (see above) clearly shows that the points from the table for the parabola (1;1), (-1;1) were transformed into points (1;4), (1;-4), that is, with the same values, the ordinate of each point is multiplied by 4. This will happen to all key points of the original table. We argue similarly in the cases of pictures 2 and 3.

And when the parabola "becomes wider" parabola:

Let's recap:

1)The sign of the coefficient is responsible for the direction of the branches. With title="Rendered by QuickLaTeX.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}

2) Absolute value coefficient (modulus) is responsible for the “expansion”, “compression” of the parabola. The larger , the narrower the parabola, the smaller |a|, the wider the parabola.

CASE III, "C" APPEARS

Now let's put into play (that is, we consider the case when ), we will consider parabolas of the form . It is easy to guess (you can always refer to the table) that the parabola will move up or down along the axis, depending on the sign:

IV CASE, "b" APPEARS

When will the parabola “tear off” from the axis and will finally “walk” along the entire coordinate plane? When it ceases to be equal.

Here, to construct a parabola, we need formula for calculating the vertex: , .

So at this point (as at the point (0; 0) of the new coordinate system) we will build a parabola, which is already within our power. If we are dealing with the case , then from the top we set aside one unit segment to the right, one up, - the resulting point is ours (similarly, a step to the left, a step up is our point); if we are dealing with, for example, then from the top we set aside one single segment to the right, two - up, etc.

For example, the vertex of a parabola:

Now the main thing to understand is that at this vertex we will build a parabola according to the parabola template, because in our case.

When constructing a parabola after finding the coordinates of the vertex is veryIt is convenient to consider the following points:

1) parabola must pass through the point . Indeed, substituting x=0 into the formula, we get that . That is, the ordinate of the point of intersection of the parabola with the axis (oy), this is. In our example (above), the parabola intersects the y-axis at , since .

2) axis of symmetry parabolas is a straight line, so all points of the parabola will be symmetrical about it. In our example, we immediately take the point (0; -2) and build a parabola symmetrical about the axis of symmetry, we get the point (4; -2), through which the parabola will pass.

3) Equating to , we find out the points of intersection of the parabola with the axis (ox). To do this, we solve the equation. Depending on the discriminant, we will get one (, ), two ( title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, we have the root of the discriminant - not an integer, when building it, it doesn’t really make sense for us to find the roots, but we can clearly see that we will have two points of intersection with the (oh) axis (since title = " Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}

So let's work out

Algorithm for constructing a parabola if it is given in the form

1) determine the direction of the branches (a>0 - up, a<0 – вниз)

2) find the coordinates of the vertex of the parabola by the formula , .

3) we find the point of intersection of the parabola with the axis (oy) by the free term, we build a point symmetrical to the given one with respect to the axis of symmetry of the parabola (it should be noted that it happens that it is unprofitable to mark this point, for example, because the value is large ... we skip this point ...)

4) At the found point - the top of the parabola (as at the point (0; 0) of the new coordinate system), we build a parabola. If title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;">, то парабола становится у’же по сравнению с , если , то парабола расширяется по сравнению с !}

5) We find the points of intersection of the parabola with the axis (oy) (if they themselves have not yet “surfaced”), solving the equation

Example 1

Example 2

Remark 1. If the parabola is initially given to us in the form , where are some numbers (for example, ), then it will be even easier to build it, because we have already been given the coordinates of the vertex . Why?

Let's take a square trinomial and select a full square in it: Look, here we got that , . We previously called the top of the parabola, that is, now,.

For example, . We mark the top of the parabola on the plane, we understand that the branches are directed downwards, the parabola is expanded (relatively). That is, we perform steps 1; 3; 4; 5 from the algorithm for constructing a parabola (see above).

Remark 2. If the parabola is given in a form similar to this (that is, represented as a product of two linear factors), then we immediately see the points of intersection of the parabola with the (x) axis. In this case - (0;0) and (4;0). For the rest, we act according to the algorithm, opening the brackets.

Lesson: how to build a parabola or a quadratic function?

THEORETICAL PART

A parabola is a graph of a function described by the formula ax 2 +bx+c=0.
To build a parabola, you need to follow a simple algorithm of actions:

1) Parabola formula y=ax 2 +bx+c,
If a>0 then the branches of the parabola are directed up,
and then the branches of the parabola are directed down.
free member c this point intersects the parabola with the OY axis;

2) , it is found by the formula x=(-b)/2a, we substitute the found x into the parabola equation and find y;

3)Function zeros or in other words, the points of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots, we equate the equation to 0 ax2+bx+c=0;

Types of equations:

a) The complete quadratic equation is ax2+bx+c=0 and is solved by the discriminant;
b) Incomplete quadratic equation of the form ax2+bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0:
ax2+bx=0,
x(ax+b)=0,
x=0 and ax+b=0;
c) Incomplete quadratic equation of the form ax2+c=0. To solve it, you need to move the unknown to one side, and the known to the other. x =±√(c/a);

4) Find some additional points to build the function.

PRACTICAL PART

And so now, with an example, we will analyze everything by actions:
Example #1:
y=x 2 +4x+3
c=3 means the parabola intersects OY at the point x=0 y=3. The branches of the parabola look up because a=1 1>0.
a=1 b=4 c=3 x=(-b)/2a=(-4)/(2*1)=-2 y= (-2) 2 +4*(-2)+3=4- 8+3=-1 the top is at the point (-2;-1)
Find the roots of the equation x 2 +4x+3=0
We find the roots by the discriminant
a=1 b=4 c=3
D=b 2 -4ac=16-12=4
x=(-b±√(D))/2a
x1=(-4+2)/2=-1
x2=(-4-2)/2=-3

Let's take some arbitrary points that are near the top x=-2

x -4 -3 -1 0
y 3 0 0 3

We substitute instead of x in the equation y \u003d x 2 + 4x + 3 values
y=(-4) 2 +4*(-4)+3=16-16+3=3
y=(-3) 2 +4*(-3)+3=9-12+3=0
y=(-1) 2 +4*(-1)+3=1-4+3=0
y=(0) 2 +4*(0)+3=0-0+3=3
It can be seen from the values ​​​​of the function that the parabola is symmetrical about the straight line x \u003d -2

Example #2:
y=-x 2 +4x
c=0 means the parabola intersects OY at the point x=0 y=0. The branches of the parabola look down because a=-1 -1 Find the roots of the equation -x 2 +4x=0
An incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0.
x(-x+4)=0, x=0 and x=4.

Let's take some arbitrary points that are near the vertex x=2
x 0 1 3 4
y 0 3 3 0
We substitute instead of x in the equation y \u003d -x 2 +4x values
y=0 2 +4*0=0
y=-(1) 2 +4*1=-1+4=3
y=-(3) 2 +4*3=-9+13=3
y=-(4) 2 +4*4=-16+16=0
It can be seen from the values ​​​​of the function that the parabola is symmetrical about the straight line x \u003d 2

Example #3
y=x 2 -4
c=4 means the parabola intersects OY at the point x=0 y=4. The branches of the parabola look up because a=1 1>0.
a=1 b=0 c=-4 x=(-b)/2a=0/(2*(1))=0 y=(0) 2 -4=-4 vertex is at point (0;-4 )
Find the roots of the equation x 2 -4=0
An incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknown to one side, and the known to the other. x =±√(c/a)
x2=4
x1=2
x 2 \u003d -2

Let's take some arbitrary points that are near the top x=0
x -2 -1 1 2
y 0 -3 -3 0
We substitute instead of x in the equation y \u003d x 2 -4 values
y=(-2) 2 -4=4-4=0
y=(-1) 2 -4=1-4=-3
y=1 2 -4=1-4=-3
y=2 2 -4=4-4=0
It can be seen from the values ​​of the function that the parabola is symmetrical about the straight line x=0

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The point is called the focus of the parabola, the straight line is the directrix of the parabola, the middle of the perpendicular dropped from the focus to the directrix is ​​the vertex of the parabola, the distance from the focus to the directrix is ​​the parameter of the parabola, and the distance from the vertex of the parabola to its focus is the focal length (Fig. 3.45, a) . The straight line perpendicular to the directrix and passing through the focus is called the axis of the parabola (the focal axis of the parabola). The line segment connecting an arbitrary point of the parabola with its focus is called the focal radius of the point. The line segment connecting two points of the parabola is called the chord of the parabola.

For an arbitrary point of the parabola, the ratio of the distance to the focus to the distance to the directrix is ​​equal to one. Comparing the directory properties of the ellipse, hyperbola and parabola, we conclude that parabola eccentricity is by definition equal to one.

The geometric definition of a parabola, expressing its directorial property, is equivalent to its analytical definition - the line given by the canonical equation of the parabola:

(3.51)

Indeed, let's introduce a rectangular coordinate system (Fig.3.45,6). Let's take the top of the parabola as the origin of the coordinate system; the straight line passing through the focus perpendicular to the directrix, we will take as the abscissa axis (positive direction on it from point to point); a straight line perpendicular to the abscissa axis and passing through the top of the parabola, we will take as the ordinate axis (the direction on the ordinate axis is chosen so that the rectangular coordinate system is right).

Let us compose the equation of a parabola using its geometric definition, which expresses the directorial property of a parabola. In the selected coordinate system, we determine the focus coordinates and the directrix equation. For an arbitrary point belonging to a parabola, we have:

where is the orthogonal projection of the point onto the directrix. We write this equation in coordinate form:

We square both sides of the equation: . Bringing like terms, we get canonical parabola equation

those. the chosen coordinate system is canonical.

By reasoning in reverse order, it can be shown that all points whose coordinates satisfy equation (3.51), and only they, belong to the locus of points, called a parabola. Thus, the analytic definition of a parabola is equivalent to its geometric definition, which expresses the directory property of a parabola.

We give the following properties of a parabola:

Property 10.10.

The parabola has an axis of symmetry.

Proof

The variable y enters the equation only to the second power. Therefore, if the coordinates of the point M (x; y) satisfy the parabola equation, then the coordinates of the point N (x; - y) will satisfy it. Point N is symmetrical to point M with respect to the axis Ox . Therefore, the axis Ox is the axis of symmetry of the parabola in the canonical coordinate system.

The axis of symmetry is called the axis of the parabola. The point of intersection of the parabola with the axis is called the vertex of the parabola. The vertex of the parabola in the canonical coordinate system is at the origin.

Property 10.11.

The parabola is located in the half-plane x ≥ 0.

Proof

Indeed, since the parameter p is positive, only points with non-negative abscissas, that is, points of the half-plane x ≥ 0, can satisfy the equation.

When changing the coordinate system, the point A with coordinates specified in the condition will have new coordinates determined from the relations. Thus, the point A will have coordinates in the canonical system. This point is called the focus of the parabola and is denoted by the letter F.

The straight line l, given in the old coordinate system by the equation in the new coordinate system, will have seen, omitting the hatching,

This straight line in the canonical coordinate system is called the directrix of the parabola. The distance from it to the focus is called the focal parameter of the parabola. Obviously, it is equal to p . The eccentricity of a parabola, by definition, is assumed to be equal to one, that is, ε = k = 1.

Now the property through which we defined the parabola can be formulated in new terms as follows: any point of the parabola is equidistant from its focus and directrix.

The form of the parabola in the canonical coordinate system and the location of its directrix are shown in fig. 10.10.1.

Figure 10.10.1.

Over the field P, there is a linear operator if 1) for any vectors2) for any vector and any.

1) Linear operator matrix: Let φ-L.O. vector space V over a field P and one of the bases of V: Let Then the L.O.φ matrix: 2) Relationship between the matrices of a linear operator in different bases: M(φ) - L.O. matrix φ in the old basis. M1(φ) - L.O. matrix φ in the new basis. T is the transition matrix from the older basis to the new basis. 2) Actions on linear operators: Let φ and f be different L.O. vector space V. Then φ+f is the sum of linear operators φ and f. k·φ - multiplication L.O. to a scalar k. φ f is the product of linear operators φ and f. I am also L.O. vector space V.

4) The core of the linear operator: d(φ) - dimension of the L.O. φ (defect). 5) The image of a linear operator: ranφ - L.O. rank φ (dimension Jmφ). 6) Eigenvectors and eigenvalues ​​of a linear vector:  Let φ be L.O. vector space V over the field P and uIf then λ is an eigenvalue - eigenvector φ corresponding to λ.  Characteristic equation of L.O. φ:  The set of eigenvectors corresponding to the eigenvalue λ:  L.O. vector space are called L.O. with simple spectrum if φ if φ has exactly n eigenvalues.  If φ - L.O. with a simple spectrum, then it has a basis of eigenvectors, with respect to which the L.O. φ is diagonal.

2) The position of a straight line in space is completely determined by setting any of its fixed points M 1 and a vector parallel to this line.

A vector parallel to a straight line is called guiding the vector of this line.

So let the straight l passes through a point M 1 (x 1 , y 1 , z 1 ) lying on a straight line parallel to the vector .

Consider an arbitrary point M(x,y,z) on a straight line. It can be seen from the figure that .

The vectors are collinear, so there is a number t, what , where is the multiplier t can take any numeric value depending on the position of the point M on a straight line. Factor t is called a parameter. Denoting the radius vectors of points M 1 And M respectively through and, we get. This equation is called vector straight line equation. It shows that each parameter value t corresponds to the radius vector of some point M lying on a straight line.

We write this equation in coordinate form. Note that, from here

The resulting equations are called parametric straight line equations.

When changing the parameter t coordinates change x, y And z and dot M moves in a straight line.

CANONICAL EQUATIONS DIRECT

Let M 1 (x 1 , y 1 , z 1 ) is a point lying on a line l, And is its direction vector. Again, take an arbitrary point on a straight line M(x,y,z) and consider the vector .

It is clear that the vectors and are collinear, so their respective coordinates must be proportional, hence

– canonical straight line equations.

Remark 1. Note that the canonical equations of the line could be obtained from the parametric equations by eliminating the parameter t. Indeed, from the parametric equations we obtain or .

Example. Write the equation of a straight line in a parametric way.

Denote , hence x = 2 + 3t, y = –1 + 2t, z = 1 –t.

Remark 2. Let the line be perpendicular to one of the coordinate axes, for example, the axis Ox. Then the direction vector of the line is perpendicular Ox, hence, m=0. Consequently, the parametric equations of the straight line take the form

Eliminating the parameter from the equations t, we obtain the equations of the straight line in the form

However, in this case too, we agree to formally write the canonical equations of the straight line in the form . Thus, if the denominator of one of the fractions is zero, then this means that the line is perpendicular to the corresponding coordinate axis.

Similarly, the canonical equations corresponds to a straight line perpendicular to the axes Ox And Oy or parallel axis Oz.

Examples.

Canonical equations: .

Parametric equations:

Write the equations of a straight line passing through two points M 1 (-2;1;3), M 2 (-1;3;0).

Let us compose the canonical equations of the straight line. To do this, we find the direction vector . Then l:.

GENERAL EQUATIONS A DIRECT LINE AS A LINE OF INTERCEPTION OF TWO PLANES

Through each straight line in space passes an infinite number of planes. Any two of them, intersecting, define it in space. Therefore, the equations of any two such planes, considered together, are the equations of this line.

In general, any two non-parallel planes given by the general equations

determine their line of intersection. These equations are called general equations straight.

Examples.

Construct a straight line given by equations

To construct a line, it is enough to find any two of its points. The easiest way is to choose the points of intersection of the line with the coordinate planes. For example, the point of intersection with the plane xOy we obtain from the equations of a straight line, assuming z= 0:

Solving this system, we find the point M 1 (1;2;0).

Similarly, assuming y= 0, we get the point of intersection of the line with the plane xOz:

From the general equations of a straight line, one can proceed to its canonical or parametric equations. To do this, you need to find some point M 1 on the straight line and the direction vector of the straight line.

Point coordinates M 1 we obtain from this system of equations by giving one of the coordinates an arbitrary value. To find the direction vector, note that this vector must be perpendicular to both normal vectors and. Therefore, for the directing vector l you can take the cross product of normal vectors:

.

Example. Give the general equations of the straight line to the canonical form.

Find a point on a straight line. To do this, we choose arbitrarily one of the coordinates, for example, y= 0 and solve the system of equations:

The normal vectors of the planes that define the line have coordinates. Therefore, the directing vector of the line will be

. Hence, l: .

1) Let and be two bases in R n .

Definition. transition matrix from basis to basis the matrix C is called, the columns of which are the coordinates of the vectors in basis :

The transition matrix is ​​invertible since the basis vectors are linearly independent and hence

The vector is linearly expressed in terms of the vectors of both bases. The relationship of vector coordinates in different bases is established in the following theorem.

Theorem. If

then the coordinates vectors in the basis , and its coordinates in basis related by the relations

Where - transition matrix from the basis to basis , - column vectors of vector coordinates in bases And respectively.

2)Mutual arrangement of two straight lines

If the lines are given by equations, then they are:

1) parallel (but not the same)

2) match

3) intersect

4) interbreed

If then cases 1 - 4 occur when (- negation sign of the condition):

3)

4)

Distance between two parallel lines

In coordinates

Distance between two intersecting lines

In coordinates

Angle between two lines

Necessary and sufficient condition for two lines to be perpendicular

Or

Mutual arrangement of a straight line and a plane

Plane and line

1) intersect

2) the line lies in a plane

3) parallel

If then cases 1 - 3 occur when:

1)

Necessary and sufficient condition for parallelism of a line and a plane

Angle between line and plane

Point of intersection of a line with a plane

In coordinates:

Equations of a line passing through a point perpendicular to the plane

In coordinates:

1) Obviously, the system of linear equations can be written as:

x 1 + x 2 + … + x n

Proof.

1) If a solution exists, then the column of free terms is a linear combination of the columns of matrix A, which means that adding this column to the matrix, i.e. transition АА * do not change the rank.

2) If RgA = RgA * , then this means that they have the same basic minor. The column of free members is a linear combination of the columns of the basis minor, those notation given above is correct.

2) plane in space.

We first obtain the equation of the plane passing through the point M 0 (X 0 ,y 0 , z 0 ) perpendicular to the vector n = {A, B, C), called the normal to the plane. For any point in the plane M(x, y,z) vector M 0 M = {x - x 0 , y - y 0 , z - z 0 ) is orthogonal to the vector n , therefore, their scalar product is equal to zero:

A(x - x 0 ) + B(y - y 0 ) + C(z - z 0 ) = 0. (8.1)

An equation is obtained that is satisfied by any point of a given plane - the equation of a plane passing through a given point perpendicular to a given vector.

After reducing similar ones, equation (8.1) can be written in the form:

Ax + By + Cz + D = 0, (8.2)

Where D=-Ax 0 - By 0 -Cz 0 . This linear equation in three variables is called the general equation of the plane.

Incomplete plane equations.

If at least one of the numbers A, B, C,D equals zero, equation (8.2) is called incomplete.

Consider the possible types of incomplete equations:

1) D= 0 - plane Ax + By + cz= 0 passes through the origin.

2) A = 0 – n = {0,B, C} Ox, hence the plane By + cz + D= 0 is parallel to the axis Oh.

3) IN= 0 - plane Ax + cz + D = 0 is parallel to the axis OU.

4) WITH= 0 - plane Ax + By + D= 0 is parallel to the axis ABOUTz.

5) A = B= 0 - plane cz + D Ohu(since it is parallel to the axes Oh And OU).

6) A = C= 0 - plane Wu +D= 0 parallel to the coordinate plane Ohz.

7) B = C= 0 - plane Ax + D= 0 parallel to the coordinate plane OUz.

8) A =D= 0 - plane By + cz= 0 passes through the axis Oh.

9) B = D= 0 - plane Ah + Cz= 0 passes through the axis OU.

10) C = D= 0 - plane Ax + By= 0 passes through the axis Oz.

11) A = B = D= 0 - equation WITHz= 0 specifies the coordinate plane Ohu.

12) A = C = D= 0 – we get Wu= 0 is the equation of the coordinate plane Ohz.

13) B = C = D= 0 - plane Oh= 0 is the coordinate plane OUz.

If the general equation of the plane is complete (that is, none of the coefficients is equal to zero), it can be reduced to the form:

called plane equation in segments. The conversion method is shown in lecture 7. Parameters A,b And With are equal to the values ​​of the segments cut off by the plane on the coordinate axes.

1) Homogeneous systems of linear equations

Homogeneous system of linear equations AX = 0 always together. It has non-trivial (non-zero) solutions if r= rank A< n .

For homogeneous systems, the basis variables (the coefficients at which form the basis minor) are expressed in terms of free variables by relations of the form:

Then n - r linearly independent vector solutions will be:

and any other solution is their linear combination. Decision-vector form a normalized fundamental system.

In a linear space, the set of solutions of a homogeneous system of linear equations forms a subspace of dimension n - r; is the basis of this subspace.