Level III
3.1. Hyperbole touches lines 5 x – 6y – 16 = 0, 13x – 10y– – 48 = 0. Write down the equation of the hyperbola provided that its axes coincide with the coordinate axes.
3.2. Write equations for tangents to a hyperbola
1) passing through a point A(4, 1), B(5, 2) and C(5, 6);
2) parallel to straight line 10 x – 3y + 9 = 0;
3) perpendicular to straight line 10 x – 3y + 9 = 0.
Parabola is the geometric locus of points in the plane whose coordinates satisfy the equation
Parabola parameters:
Dot F(p/2, 0) is called focus parabolas, magnitude p – parameter , point ABOUT(0, 0) – top . In this case, the straight line OF, about which the parabola is symmetrical, defines the axis of this curve.
Magnitude Where M(x, y) – an arbitrary point of a parabola, called focal radius , straight D: x = –p/2 – headmistress (it does not intersect the interior region of the parabola). Magnitude is called the eccentricity of the parabola.
The main characteristic property of a parabola: all points of the parabola are equidistant from the directrix and focus (Fig. 24).
There are other forms of the canonical parabola equation that determine other directions of its branches in the coordinate system (Fig. 25):
For parametric definition of a parabola as a parameter t the ordinate value of the parabola point can be taken:
Where t is an arbitrary real number.
Example 1. Determine the parameters and shape of a parabola using its canonical equation:
Solution. 1. Equation y 2 = –8x defines a parabola with vertex at point ABOUT Oh. Its branches are directed to the left. Comparing this equation with the equation y 2 = –2px, we find: 2 p = 8, p = 4, p/2 = 2. Therefore, the focus is at the point F(–2; 0), directrix equation D: x= 2 (Fig. 26).
2. Equation x 2 = –4y defines a parabola with vertex at point O(0; 0), symmetrical about the axis Oy. Its branches are directed downwards. Comparing this equation with the equation x 2 = –2py, we find: 2 p = 4, p = 2, p/2 = 1. Therefore, the focus is at the point F(0; –1), directrix equation D: y= 1 (Fig. 27).
Example 2. Determine parameters and type of curve x 2 + 8x – 16y– 32 = 0. Make a drawing.
Solution. Let's transform the left side of the equation using the complete square extraction method:
x 2 + 8x– 16y – 32 =0;
(x + 4) 2 – 16 – 16y – 32 =0;
(x + 4) 2 – 16y – 48 =0;
(x + 4) 2 – 16(y + 3).
As a result we get
(x + 4) 2 = 16(y + 3).
This is the canonical equation of a parabola with the vertex at the point (–4, –3), the parameter p= 8, branches pointing upward (), axis x= –4. Focus is on point F(–4; –3 + p/2), i.e. F(–4; 1) Headmistress D given by the equation y = –3 – p/2 or y= –7 (Fig. 28).
Example 4. Write an equation for a parabola with its vertex at the point V(3; –2) and focus at the point F(1; –2).
Solution. The vertex and focus of a given parabola lie on a straight line parallel to the axis Ox(same ordinates), the branches of the parabola are directed to the left (the abscissa of the focus is less than the abscissa of the vertex), the distance from the focus to the vertex is p/2 = 3 – 1 = 2, p= 4. Hence, the required equation
(y+ 2) 2 = –2 4( x– 3) or ( y + 2) 2 = = –8(x – 3).
Tasks for independent solution
I level
1.1. Determine the parameters of the parabola and construct it:
1) y 2 = 2x; 2) y 2 = –3x;
3) x 2 = 6y; 4) x 2 = –y.
1.2. Write the equation of a parabola with its vertex at the origin if you know that:
1) the parabola is located in the left half-plane symmetrically relative to the axis Ox And p = 4;
2) the parabola is located symmetrically relative to the axis Oy and passes through the point M(4; –2).
3) the directrix is given by equation 3 y + 4 = 0.
1.3. Write an equation for a curve all points of which are equidistant from the point (2; 0) and the straight line x = –2.
Level II
2.1. Determine the type and parameters of the curve.
I suggest that the rest of the readers significantly expand their school knowledge about parabolas and hyperbolas. Hyperbola and parabola - are they simple? ...Can't wait =)
The general structure of the presentation of the material will resemble the previous paragraph. Let's start with the general concept of a hyperbola and the task of constructing it.
The canonical equation of a hyperbola has the form , where are positive real numbers. Please note that, unlike ellipse, the condition is not imposed here, that is, the value of “a” may be less than the value of “be”.
I must say, quite unexpectedly... the equation of the “school” hyperbola does not even closely resemble the canonical notation. But this mystery will still have to wait for us, but for now let’s scratch our heads and remember what characteristic features the curve in question has? Let's spread it on the screen of our imagination graph of a function ….
A hyperbola has two symmetrical branches.
Not bad progress! Any hyperbole has these properties, and now we will look with genuine admiration at the neckline of this line:
Example 4
Construct the hyperbola given by the equation
Solution: in the first step, we bring this equation to canonical form. Please remember the standard procedure. On the right you need to get “one”, so we divide both sides of the original equation by 20:
Here you can reduce both fractions, but it is more optimal to do each of them three-story:
And only after that carry out the reduction:
Select the squares in the denominators:
Why is it better to carry out transformations this way? After all, the fractions on the left side can be immediately reduced and obtained. The fact is that in the example under consideration we were a little lucky: the number 20 is divisible by both 4 and 5. In the general case, such a number does not work. Consider, for example, the equation . Here everything is sadder with divisibility and without three-story fractions no longer possible:
So, let's use the fruit of our labors - the canonical equation:
There are two approaches to constructing a hyperbola - geometric and algebraic.
From a practical point of view, drawing with a compass... I would even say utopian, so it is much more profitable to once again use simple calculations to help.
It is advisable to adhere to the following algorithm, first the finished drawing, then the comments:
In practice, a combination of rotation by an arbitrary angle and parallel translation of the hyperbola is often encountered. This situation is discussed in class Reducing the 2nd order line equation to canonical form.
It's finished! She's the one. Ready to reveal many secrets. The canonical equation of a parabola has the form , where is a real number. It is easy to notice that in its standard position the parabola “lies on its side” and its vertex is at the origin. In this case, the function specifies the upper branch of this line, and the function – the lower branch. It is obvious that the parabola is symmetrical about the axis. Actually, why bother:
Example 6
Construct a parabola
Solution: the vertex is known, let’s find additional points. Equation determines the upper arc of the parabola, the equation determines the lower arc.
In order to shorten the recording of the calculations, we will carry out the calculations “with one brush”:
For compact recording, the results could be summarized in a table.
Before performing an elementary point-by-point drawing, let’s formulate a strict
A parabola is the set of all points in the plane that are equidistant from a given point and a given line that does not pass through the point.
The point is called focus parabolas, straight line - headmistress (spelled with one "es") parabolas. The constant "pe" of the canonical equation is called focal parameter, which is equal to the distance from the focus to the directrix. In this case. In this case, the focus has coordinates , and the directrix is given by the equation .
In our example:
The definition of a parabola is even simpler to understand than the definitions of an ellipse and a hyperbola. For any point on a parabola, the length of the segment (the distance from the focus to the point) is equal to the length of the perpendicular (the distance from the point to the directrix):
Congratulations! Many of you have made a real discovery today. It turns out that a hyperbola and a parabola are not graphs of “ordinary” functions at all, but have a pronounced geometric origin.
Obviously, as the focal parameter increases, the branches of the graph will “raise” up and down, approaching infinitely close to the axis. As the “pe” value decreases, they will begin to compress and stretch along the axis
The eccentricity of any parabola is equal to unity:
The parabola is one of the most common lines in mathematics, and you will have to build it really often. Therefore, please pay special attention to the final paragraph of the lesson, where I will discuss typical options for the location of this curve.
! Note : as in the cases with previous curves, it is more correct to talk about rotation and parallel translation of coordinate axes, but the author will limit himself to a simplified version of the presentation so that the reader has a basic understanding of these transformations.
A function of the form where is called quadratic function.
Graph of a quadratic function – parabola.
Let's consider the cases:
That is , ,
To construct, fill out the table by substituting the x values into the formula:
Mark the points (0;0); (1;1); (-1;1), etc. on the coordinate plane (the smaller the step we take the x values (in this case, step 1), and the more x values we take, the smoother the curve will be), we get a parabola:
It is easy to see that if we take the case , , , that is, then we get a parabola that is symmetrical about the axis (oh). It’s easy to verify this by filling out a similar table:
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):!}
In the first picture (see above) it is clearly visible 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 reason similarly in the cases of pictures 2 and 3.
And when the parabola “becomes wider” than the parabola:
Let's summarize:
1)The sign of the coefficient determines 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” and “compression” of the parabola. The larger , the narrower the parabola; the smaller |a|, the wider the parabola.
Now let's introduce into the game (that is, consider the case when), we will consider parabolas of the form . It is not difficult to guess (you can always refer to the table) that the parabola will shift up or down along the axis depending on the sign:
When will the parabola “break away” from the axis and finally “walk” along the entire coordinate plane? When will it cease 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 we can already do. If we are dealing with the case, then from the vertex we put 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 vertex we put one unit segment to the right, two - upward, 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 pattern, because in our case.
When constructing a parabola after finding the coordinates of the vertex veryIt is convenient to consider the following points:
1) parabola will definitely pass through the point . Indeed, substituting x=0 into the formula, we obtain that . That is, the ordinate of the point of intersection of the parabola with the axis (oy) is . In our example (above), the parabola intersects the ordinate at point , 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 it symmetrical relative to the symmetry axis of the parabola, 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 (oh). 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, our root of the discriminant is not an integer; when constructing, it doesn’t make much sense for us to find the roots, but we clearly see that we will have two points of intersection with the axis (oh) (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 it out
1) determine the direction of the branches (a>0 – up, a<0 – вниз)
2) we find the coordinates of the vertex of the parabola using the formula , .
3) we find the point of intersection of the parabola with the axis (oy) using the free term, construct a point symmetrical to this point relative to the symmetry axis 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 vertex of the parabola (as at the point (0;0) of the new coordinate system) we construct 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 have not yet “surfaced”) by solving the equation
Example 1
Example 2
Note 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 construct it, because we have already been given the coordinates of the vertex . Why?
Let's take a quadratic trinomial and isolate the complete square in it: Look, we got that , . You and I previously called the vertex of a parabola, that is, now,.
For example, . We mark the vertex of the parabola on the plane, we understand that the branches are directed downward, the parabola is expanded (relative to ). That is, we carry out points 1; 3; 4; 5 from the algorithm for constructing a parabola (see above).
Note 2. If the parabola is given in a form similar to this (that is, presented as a product of two linear factors), then we immediately see the points of intersection of the parabola with the axis (ox). In this case – (0;0) and (4;0). For the rest, we act according to the algorithm, opening the brackets.
Definition: A parabola is the locus of points on a plane for which the distance to some fixed point F of this plane is equal to the distance to some fixed straight line. Point F is called the focus of the parabola, and the fixed line is called the directrix of the parabola.
To derive the equation, we construct:
WITH according to the definition:
Since 2 >=0, the parabola lies in the right half-plane. As x increases from 0 to infinity
. The parabola is symmetrical about Ox. The point of intersection of a parabola with its axis of symmetry is called the vertex of the parabola.
There are 8 types of KVP:
1.ellipses
2.hyperboles
3.parabolas
Curves 1,2,3 are canonical sections. If we intersect the cone with a plane parallel to the axis of the cone, we obtain a hyperbola. If the plane is parallel to the generatrix, then it is a parabola. All planes do not pass through the vertex of the cone. If it is any other plane, then it is an ellipse.
4. pair of parallel lines y 2 +a 2 =0, a0
5. pair of intersecting lines y 2 -k 2 x 2 =0
6.one straight line y 2 =0
7.one point x 2 + y 2 =0
8.empty set - empty curve (curve without points) x 2 + y 2 +1=0 or x 2 + 1=0
Theorem (main theorem about KVP): Equation of the form
a 11 x 2 + 2 a 12 x y + a 22 y 2 + 2 a 1 x + 2a 2 y+a 0 = 0
can only represent a curve of one of these eight types.
Idea of proof is to move to a coordinate system in which the KVP equation will take the simplest form, when the type of curve it represents becomes obvious. The theorem is proven by rotating the coordinate system through an angle at which the term with the product of coordinates disappears. And with the help of parallel transfer of the coordinate system in which either the term with the x variable or the term with the y variable disappears.
Transition to a new coordinate system: 1. Parallel transfer
2. Rotate
P VP - a set of points whose rectangular coordinates satisfy the 2nd degree equation: (1)
It is assumed that at least one of the coefficients of squares or products is different from 0. The equation is invariant with respect to the choice of coordinate system.
Theorem Any plane intersects the PVP along the CVP, with the exception of a special case when the entire plane is in the section. (The PVP can be a plane or a pair of planes).
There are 15 types of PVP. Let us list them, indicating the equations by which they are specified in suitable coordinate systems. These equations are called canonical (the simplest). Construct geometric images corresponding to canonical equations using the method of parallel sections: Intersect the surface with coordinate planes and planes parallel to them. The result is sections and curves that give an idea of the shape of the surface.
1. Ellipsoid.
If a=b=c then we get a sphere.
2. Hyperboloids.
1). Single-sheet hyperboloid:
Section of a single-sheet hyperboloid by coordinate planes: XOZ:
- hyperbole.
YOZ:
- hyperbole.
XOY plane:
- ellipse.
2). Two-sheet hyperboloid.
The origin is a point of symmetry.
Coordinate planes are planes of symmetry.
Plane z
=
h intersects a hyperboloid along an ellipse
, i.e. plane z
=
h begins to intersect the hyperboloid at | h
|
c. Section of a hyperboloid by planes x
= 0
And y
= 0
- these are hyperboles.
The numbers a, b, c in equations (2), (3), (4) are called the semi-axes of ellipsoids and hyperboloids.
3. Paraboloids.
1). Elliptical paraboloid:
Plane section z
=
h There is
, Where
. From the equation it is clear that z 0 is an infinite bowl.
Intersection of planes y
=
h And x=
h
- this is a parabola and in general
2). Hyperbolic paraboloid:
Obviously, the XOZ and YOZ planes are planes of symmetry, the z axis is the axis of the paraboloid. Intersection of a paraboloid with a plane z
=
h– hyperboles:
,
. Plane z=0
intersects a hyperbolic paraboloid along two axes
which are asymptotes.
4. Cone and cylinders of the second order.
1). A cone is a surface
. The cone is formed by straight lines passing through the origin 0 (0, 0, 0). The cross section of a cone is an ellipse with semi-axes
.
2). Second order cylinders.
This is an elliptical cylinder
.
Whatever line we take that intersects the ellipses and is parallel to the Oz axis satisfies this equation. By moving this straight line around the ellipse we obtain a surface.
G hyperbolic cylinder:
On the XOU plane it is a hyperbola. We move the straight line intersecting the hyperbola parallel to Oz along the hyperbola.
Parabolic cylinder:
N and the XOU plane is a parabola.
Cylindrical surfaces are formed by a straight line (generative) moving parallel to itself along a certain straight line (guide).
10. Pair of intersecting planes
11.Pair of parallel planes
12.
- straight
13. Straight line - a “cylinder” built on one point
14.One point
15.Empty set
The main theorem about PVP: Each PVP belongs to one of the 15 types discussed above. There are no other PVP.
Surfaces of rotation. Let the PDSC Oxyz be given and in the Oyz plane the line e defined by the equation F(y,z)=0 (1). Let's create an equation for the surface obtained by rotating this line around the Oz axis. Let's take a point M(y,z) on line e. When the plane Oyz rotates around Oz, point M will describe a circle. Let N(X,Y,Z) be an arbitrary point of this circle. It is clear that z=Z.
.
Substituting the found values of z and y into equation (1) we obtain the correct equality:
those. the coordinates of point N satisfy the equation
. Thus, any point on the surface of revolution satisfies equation (2). It is not difficult to prove that if a point N(x 1 ,y 1 ,z 1) satisfies equation (2) then it belongs to the surface in question. Now we can say that equation (2) is the desired equation for the surface of revolution.
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A parabola is the locus of points in the plane equidistant from a given point F
and a given straight line dd not passing through a given point. This geometric definition expresses directorial property of a parabola.
Directorial property of parabolas
Point F is called the focus of the parabola, straight line d is the directrix of the parabola, the midpoint O of the perpendicular lowered 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 (focal axis of the parabola). The segment FM connecting an arbitrary point M of a parabola with its focus is called the focal radius of the point
M. The segment connecting two points of a parabola is called a chord of the parabola.
For an arbitrary point of a parabola, the ratio of the distance to the focus to the distance to the directrix is equal to one. Comparing the directorial properties of the ellipse, hyperbola and parabola, we conclude that parabola eccentricity by definition equal to one
Geometric definition of a parabola, expressing its directorial property, is equivalent to its analytical definition - the line given by the canonical equation of a parabola:
Properties
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 = 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 definition of the function (X).
In this case, x is called the argument, and y is the value of the function. The set X is called the domain of definition of the function, Y is the set of values of the function.
This rule is often given by a formula; for example, y = 2x + 5. This method of specifying a function using a formula is called analytical.
A function can also be specified by a graph - The graph of a function y - f(x) is a set of points in the plane whose coordinates are x and satisfy the relation y = f(x).