Curves of the second order on the plane are called lines defined by equations in which the variable coordinates x and y contained in the second degree. These include ellipse, hyperbola, and parabola.

The general view of the equation of the second-order curve is as follows:

where A, B, C, D, E, F- numbers and at least one of the coefficients A, B, C is not zero.

When solving problems with curves of the second order, the canonical equations of an ellipse, hyperbola and parabola are most often considered. It is easy to pass to them from general equations; Example 1 of problems with ellipses will be devoted to this.

Ellipse given by the canonical equation

Definition of an ellipse. An ellipse is the set of all points of the plane, such for which the sum of the distances to the points, called foci, is a constant value and greater than the distance between the foci.

Focuses are indicated as in the figure below.

The canonical equation of the ellipse is:

where a and b (a > b) - the lengths of the semiaxes, that is, half the lengths of the segments cut off by the ellipse on the coordinate axes.

The straight line passing through the foci of the ellipse is its axis of symmetry. Another axis of symmetry of the ellipse is a straight line passing through the middle of a segment perpendicular to this segment. Point O the intersection of these lines serves as the center of symmetry of the ellipse or simply the center of the ellipse.

The abscissa axis intersects the ellipse at points ( a, O) and (- a, O), and the ordinate axis is at the points ( b, O) and (- b, O). These four points are called the vertices of the ellipse. The segment between the vertices of the ellipse on the abscissa axis is called its major axis, and on the ordinate axis - the minor axis. Their segments from the top to the center of the ellipse are called semiaxes.

If a = b, then the equation of the ellipse takes the form. This is the equation of a circle of radius a, and a circle is a special case of an ellipse. An ellipse can be obtained from a circle of radius a if you compress it into a/b times along the axis Oy .

Example 1. Check if the line given by the general equation , an ellipse.

Solution. We make transformations of the general equation. We apply the transfer of the free term to the right side, term-by-term division of the equation by the same number and reduction of fractions:

Answer. The equation obtained as a result of the transformations is the canonical equation of the ellipse. Therefore, this line is an ellipse.

Example 2. Write the canonical equation of an ellipse if its semiaxes are 5 and 4, respectively.

Solution. We look at the formula for the canonical equation of the ellipse and substitute: the major semiaxis is a= 5, the minor semiaxis is b= 4. We get the canonical equation of the ellipse:

Points and, marked in green on the major axis, where

are called tricks.

called eccentricity ellipse.

Attitude b/a characterizes the "flattening" of the ellipse. The smaller this ratio, the more the ellipse is elongated along the major axis. However, the degree of elongation of an ellipse is more often expressed in terms of eccentricity, the formula for which is given above. For different ellipses, the eccentricity varies from 0 to 1, always remaining less than one.

Example 3. Write the canonical equation of an ellipse if the distance between the foci is 8 and the major axis is 10.

Solution. We make simple conclusions:

If the major axis is 10, then its half, that is, the semiaxis a = 5 ,

If the distance between the foci is 8, then the number c of focus coordinates is 4.

Substitute and calculate:

The result is the canonical equation of the ellipse:

Example 4. Write the canonical equation of an ellipse if its major axis is 26 and the eccentricity.

Solution. As follows from both the size of the major axis and the eccentricity equation, the major semiaxis of the ellipse a= 13. From the equation of eccentricity, we express the number c required to calculate the length of the minor semiaxis:

.

We calculate the square of the length of the minor semiaxis:

We compose the canonical equation of the ellipse:

Example 5. Determine the foci of the ellipse given by the canonical equation.

Solution. Find the number c defining the first coordinates of the focuses of the ellipse:

.

We get the focuses of the ellipse:

Example 6. Ellipse foci are located on the axis Ox symmetric about the origin. Write the canonical equation of an ellipse if:

1) the distance between the foci is 30 and the major axis is 34

2) the minor axis is 24, and one of the focuses is at the point (-5; 0)

3) eccentricity, and one of the focuses is at the point (6; 0)

We continue to solve problems on the ellipse together

If is an arbitrary point of the ellipse (in the drawing it is indicated in green in the upper right part of the ellipse) and is the distance to this point from the focuses, then the formulas for the distances are as follows:

For each point belonging to the ellipse, the sum of the distances from the foci is a constant value equal to 2 a.

Straight lines defined by equations

are called directors ellipse (in the drawing - red lines at the edges).

From the two above equations it follows that for any point of the ellipse

,

where and are the distances of this point to the directrix and.

Example 7. An ellipse is given. Make an equation for its directors.

Solution. We look at the directrix equation and find that it is required to find the eccentricity of the ellipse, i.e. All the data for this is there. We calculate:

.

We get the equation for the directrix of the ellipse:

Example 8. Write the canonical equation of an ellipse if its focuses are points and directrixes are straight lines.

11.1. Basic concepts

Consider the lines defined by equations of the second degree with respect to the current coordinates

The coefficients of the equation are real numbers, but at least one of the numbers A, B, or C is nonzero. Such lines are called lines (curves) of the second order. It will be established below that equation (11.1) defines a circle, ellipse, hyperbola or parabola on the plane. Before proceeding to this statement, let us study the properties of the curves listed.

11.2. Circle

The simplest second-order curve is a circle. Recall that a circle of radius R centered at a point is the set of all points Μ of the plane that satisfy the condition. Let a point in a rectangular coordinate system have coordinates x 0, y 0 and - an arbitrary point of the circle (see Fig. 48).

Then from the condition we obtain the equation

(11.2)

Equation (11.2) is satisfied by the coordinates of any point of the given circle and the coordinates of any point not lying on the circle do not satisfy.

Equation (11.2) is called the canonical equation of the circle

In particular, setting and, we obtain the equation of a circle centered at the origin .

The equation of the circle (11.2) after simple transformations will take the form. When comparing this equation with the general equation (11.1) of the curve of the second order, it is easy to see that two conditions are satisfied for the equation of the circle:

1) the coefficients at x 2 and y 2 are equal to each other;

2) there is no term containing the product xy of the current coordinates.

Consider the inverse problem. Putting the values ​​and in equation (11.1), we obtain

Let's transform this equation:

(11.4)

Hence it follows that equation (11.3) defines a circle under the condition ... Its center is at the point and the radius

.

If , then equation (11.3) has the form

.

It is satisfied with the coordinates of a single point ... In this case, they say: “the circle has degenerated into a point” (has a zero radius).

If , then equation (11.4), and hence the equivalent equation (11.3), will not define any line, since the right side of equation (11.4) is negative, and the left is not negative (say: “imaginary circle”).

11.3. Ellipse

Canonical Ellipse Equation

Ellipse called the set of all points of the plane, the sum of the distances from each of which to two given points of this plane, called tricks , there is a constant value greater than the distance between the foci.

We denote the focuses by F 1 and F 2, the distance between them is 2 c, and the sum of the distances from an arbitrary point of the ellipse to the foci - after 2 a(see fig. 49). By definition 2 a > 2c, i.e. a > c.

To derive the equation of the ellipse, we choose a coordinate system so that the foci F 1 and F 2 lay on the axis, and the origin coincided with the midpoint of the segment F 1 F 2... Then the foci will have the following coordinates: and.

Let be an arbitrary point of the ellipse. Then, according to the definition of an ellipse, i.e.

This, in essence, is the equation of the ellipse.

We transform equation (11.5) to a simpler form as follows:

Because a>with, then . We put

(11.6)

Then the last equation takes the form or

(11.7)

It can be proved that equation (11.7) is equivalent to the original equation. It's called the canonical ellipse equation .

The ellipse is a second-order curve.

Study of the shape of an ellipse by its equation

Let us establish the shape of the ellipse using its canonical equation.

1. Equation (11.7) contains x and y only in even powers, therefore, if a point belongs to an ellipse, then points ,, also belong to it. It follows that the ellipse is symmetrical about the axes and, as well as about a point called the center of the ellipse.

2. Find the points of intersection of the ellipse with the coordinate axes. Putting, we find two points and, at which the axis intersects the ellipse (see Fig. 50). Putting in equation (11.7), we find the points of intersection of the ellipse with the axis: and. Points A 1 , A 2 , B 1, B 2 are called the vertices of the ellipse... Segments A 1 A 2 and B 1 B 2, as well as their lengths 2 a and 2 b are named accordingly big and small axles ellipse. The numbers a and b are called, respectively, large and small semi-axes ellipse.

3. From equation (11.7) it follows that each term on the left-hand side does not exceed unity, i.e. inequalities and or and take place. Consequently, all points of the ellipse are inside the rectangle formed by straight lines.

4. In equation (11.7) the sum of non-negative terms is equal to one. Consequently, with an increase in one term, the other will decrease, that is, if it increases, then it decreases, and vice versa.

From what has been said it follows that the ellipse has the shape shown in Fig. 50 (oval closed curve).

Learn more about the ellipse

The shape of the ellipse depends on the ratio. When the ellipse turns into a circle, the ellipse equation (11.7) takes the form. The ratio is often used as a characteristic of the shape of an ellipse. The ratio of half the distance between the foci to the semi-major axis of the ellipse is called the eccentricity of the ellipse and o6o is denoted by the letter ε ("epsilon"):

and 0<ε< 1, так как 0<с<а. С учетом равенства (11.6) формулу (11.8) можно переписать в виде

From this it can be seen that the less the eccentricity of the ellipse, the less flattened the ellipse; if we put ε = 0, then the ellipse turns into a circle.

Let M (x; y) be an arbitrary point of an ellipse with foci F 1 and F 2 (see Fig. 51). The lengths of the segments F 1 M = r 1 and F 2 M = r 2 are called the focal radii of the point Μ. Obviously,

The following formulas are valid

Straight lines are called

Theorem 11.1. If is the distance from an arbitrary point of the ellipse to some focus, d is the distance from the same point to the directrix corresponding to this focus, then the ratio is a constant value equal to the eccentricity of the ellipse:

Equality (11.6) implies that. If, then, equation (11.7) defines an ellipse, the major axis of which lies on the Oy axis, and the minor axis on the Ox axis (see Fig. 52). The foci of such an ellipse are at the points and, where .

11.4. Hyperbola

Canonical Hyperbola Equation

Hyperbole is called the set of all points of the plane, the modulus of the difference between the distances from each of which to two given points of this plane, called tricks , there is a constant value less than the distance between the foci.

We denote the focuses by F 1 and F 2 the distance between them through 2c, and the modulus of the difference between the distances from each point of the hyperbola to the foci through 2a... A-priory 2a < 2c, i.e. a < c.

To derive the hyperbola equation, we choose a coordinate system so that the foci F 1 and F 2 lay on the axis, and the origin coincided with the midpoint of the segment F 1 F 2(see fig. 53). Then the focuses will have coordinates and

Let be an arbitrary point of the hyperbola. Then, according to the definition of hyperbola or, that is .. After simplifications, as was done when deriving the equation of the ellipse, we get canonical hyperbole equation

(11.9)

(11.10)

The hyperbola is a line of the second order.

Study of the shape of a hyperbola by its equation

Let us establish the form of the hyperbola using its kakonical equation.

1. Equation (11.9) contains x and y only in even powers. Consequently, the hyperbola is symmetric about the axes and, as well as about a point called center of hyperbole.

2. Find the points of intersection of the hyperbola with the coordinate axes. Putting in equation (11.9), we find two points of intersection of the hyperbola with the axis: and. Putting in (11.9), we get what cannot be. Consequently, the hyperbola does not intersect the Oy axis.

Points and are called peaks hyperbole, and the segment

real axis , section - real semiaxis hyperbole.

The segment connecting the points is called imaginary axis , number b - imaginary semiaxis ... Rectangle with sides 2a and 2b called the main rectangle of the hyperbola .

3. From equation (11.9) it follows that the value to be reduced is not less than one, that is, that or. This means that the points of the hyperbola are located to the right of the straight line (the right branch of the hyperbola) and to the left of the straight line (the left branch of the hyperbola).

4. From the equation (11.9) of the hyperbola it can be seen that when it increases, then it also increases. This follows from the fact that the difference remains constant, equal to one.

It follows from what has been said that the hyperbola has the shape shown in Figure 54 (a curve consisting of two unbounded branches).

Hyperbola asymptotes

The line L is called the asymptote an unbounded curve K if the distance d from a point M of a curve K to this straight line tends to zero at an unbounded distance of a point M along a curve K from the origin. Figure 55 illustrates the concept of an asymptote: the line L is the asymptote for the curve K.

Let us show that the hyperbola has two asymptotes:

(11.11)

Since the straight lines (11.11) and the hyperbola (11.9) are symmetric with respect to the coordinate axes, it is sufficient to consider only those points of the indicated lines that are located in the first quarter.

Take on a straight line a point N having the same abscissa x as a point on the hyperbola (see Fig. 56), and find the difference ΜΝ between the ordinates of the line and the branch of the hyperbola:

As you can see, as x increases, the denominator of the fraction increases; the numerator is a constant. Therefore, the length of the segment ΜΝ tends to zero. Since ΜΝ is greater than the distance d from the point Μ to the straight line, then d even more tends to zero. So, the straight lines are the asymptotes of the hyperbola (11.9).

When constructing the hyperbola (11.9), it is advisable to first construct the main rectangle of the hyperbola (see Fig. 57), draw straight lines passing through the opposite vertices of this rectangle, the asymptotes of the hyperbola, and mark the vertices and, hyperbolas.

Equilateral hyperbola equation.

whose asymptotes are the coordinate axes

A hyperbola (11.9) is called equilateral if its semiaxes are equal (). Her canonical equation

(11.12)

The asymptotes of the equilateral hyperbola have equations and, therefore, are the bisectors of the coordinate angles.

Consider the equation of this hyperbola in the new coordinate system (see Fig. 58), obtained from the old one by rotating the coordinate axes by an angle. We use the formulas for the rotation of the coordinate axes:

Substitute the values ​​of x and y into equation (11.12):

The equation of an equilateral hyperbola, for which the axes Ox and Oy are asymptotes, will have the form.

Learn more about hyperbole

Eccentricity hyperbola (11.9) is called the ratio of the distance between the foci to the magnitude of the real axis of the hyperbola, denoted by ε:

Since for hyperbola, the eccentricity of the hyperbola is greater than one:. Eccentricity characterizes the shape of the hyperbola. Indeed, equality (11.10) implies that that is, and .

Hence, it is clear that the less the eccentricity of the hyperbola, the lower the ratio of its semiaxes, and hence the more elongated its main rectangle.

The eccentricity of an equilateral hyperbola is. Really,

Focal Radii and for the points of the right branch, the hyperbolas have the form and, and for the left branch, and .

Straight lines are called hyperbole directrixes. Since for the hyperbola ε> 1, then. This means that the right directrix is ​​located between the center and the right vertex of the hyperbola, the left one is between the center and the left vertex.

Hyperbola directrixes have the same property as ellipse directrixes.

The curve defined by the equation is also a hyperbola, the real axis 2b of which is located on the Oy axis, and the imaginary axis 2 a- on the Ox axis. In Figure 59, it is shown with a dotted line.

It is obvious that hyperbolas and have common asymptotes. Such hyperboles are called conjugate.

11.5. Parabola

Canonical parabola equation

A parabola is the set of all points of the plane, each of which is equally distant from a given point, called the focus, and a given straight line, called the directrix. The distance from the focus F to the directrix is ​​called the parameter of the parabola and is denoted by p (p> 0).

To derive the parabola equation, we choose the Oxy coordinate system so that the Ox axis passes through the focus F perpendicular to the directrix in the direction from the directrix to F, and the origin of coordinates O is located in the middle between the focus and the directrix (see Fig. 60). In the chosen system, the focus F has coordinates, and the directrix equation has the form, or.

1. In equation (11.13) the variable y is included in an even power, which means that the parabola is symmetric about the Ox axis; the Ox axis is the axis of symmetry of the parabola.

2. Since ρ> 0, it follows from (11.13) that. Consequently, the parabola is located to the right of the Oy axis.

3. For, we have y = 0. Consequently, the parabola passes through the origin.

4. With an unbounded increase in x, the module у also increases unboundedly. The parabola has the form (shape) shown in Figure 61. The point O (0; 0) is called the vertex of the parabola, the segment FM = r is called the focal radius of the point M.

Equations,, ( p> 0) also define parabolas, they are shown in Figure 62

It is easy to show that the graph of a square trinomial, where, B and C are any real numbers, is a parabola in the sense of its definition above.

11.6. General equation of second-order lines

Equations of curves of the second order with axes of symmetry parallel to the coordinate axes

Let us first find the equation of an ellipse centered at a point, the axes of symmetry of which are parallel to the coordinate axes Ox and Oy, and the semiaxes are, respectively, equal a and b... We place in the center of the ellipse O 1 the origin of the new coordinate system, the axes of which and the semiaxes a and b(see fig. 64):

Finally, the parabolas shown in Figure 65 have corresponding equations.

The equation

The equations of an ellipse, hyperbola, parabola and the equation of a circle after transformations (open the brackets, move all the terms of the equation in one direction, bring similar terms, introduce new designations for the coefficients) can be written using a single equation of the form

where the coefficients A and C are not equal to zero at the same time.

The question arises: does any equation of the form (11.14) determine one of the curves (circle, ellipse, hyperbola, parabola) of the second order? The answer is given by the following theorem.

Theorem 11.2... Equation (11.14) always defines: either a circle (for A = C), or an ellipse (for A C> 0), or a hyperbola (for A C< 0), либо параболу (при А×С= 0). При этом возможны случаи вырождения: для эллипса (окружности) - в точку или мнимый эллипс (окружность), для гиперболы - в пару пересекающихся прямых, для параболы - в пару параллельных прямых.

General second order equation

Consider now a general equation of the second degree with two unknowns:

It differs from equation (11.14) by the presence of a term with the product of coordinates (B¹ 0). It is possible, by rotating the coordinate axes through the angle a, to transform this equation so that there is no term with the product of coordinates in it.

Using the axes rotation formulas

we express the old coordinates in terms of the new ones:

We choose the angle a so that the coefficient at x "· y" vanishes, that is, so that the equality

Thus, when the axes are rotated through the angle a, satisfying the condition (11.17), the equation (11.15) is reduced to the equation (11.14).

Output: the general second-order equation (11.15) defines the following curves on the plane (except for the cases of degeneration and decay): a circle, an ellipse, a hyperbola, a parabola.

Note: If A = C, then the equation (11.17) loses its meaning. In this case cos2α = 0 (see (11.16)), then 2α = 90 °, i.e., α = 45 °. So, when A = C, the coordinate system should be rotated 45 °.

Transcript

Chapter 1 SECOND ORDER LINES ON A PLANE. 1. Ellipse, Hyperbola, Parabola Definition. An ellipse is the set of all points of the plane for which the sum of the distances to two given points F 1 and F is a constant value a exceeding the distance between F 1 and. M (, x) F 1 О F x Fig. Points F 1 and F are called the focal points of the ellipse, and the distance FF 1 between them is the focal distance, which is denoted c. Let the point M belong to the ellipse. The segments F1 M and F M are called the focal radii of the point M. Let F1F = c. By definition, a> c. Consider a rectangular Cartesian coordinate system Ox, in which the foci F 1 and F are located on the abscissa axis symmetrically relative to the origin. In this coordinate system, the ellipse is described by the canonical equation: x + = 1, a b 1

2. where b = a c The parameters a and b are called, respectively, the major and minor semiaxes of the ellipse. The eccentricity of an ellipse is the number ε equal to the ratio of half of its focal distance c to the semi-major axis, i.e. ε =. The eccentricity of the ellipse a satisfies the inequalities 0 ε< 1. Случай c = 0 соответствует окружности, эксцентриситет окружности равен нулю. Фокальные радиусы точки M(x,) эллипса могут быть найдены по формулам r 1 = a ε x, r = a+ ε x. Нормальное уравнение окружности имеет вид (x c) + (d) = R. Определение. Гиперболой называется множество всех точек плоскости, для которых абсолютная величина разности расстояний до данных точек F 1 и F есть величина постоянная, равная a. Точки F 1 и F называются фокусами гиперболы, а расстояние между ними фокальным расстоянием, которое обозначается c. Отрезки F1 M и F M называются фокальными радиусами точки M (x,) гиперболы. Рассмотрим прямоугольную декартову систему координат Ox, в которой фокусы F 1 и F расположены на оси абсцисс симметрично относительно начала координат. M (x,) F 1 F x Рис. 3

3 The canonical hyperbola equation has the form x a = b 1 ,. where b = c a The numbers a and b are called, respectively, the real and imaginary semiaxes of the hyperbola. There is no hyperbola inside the region defined by the inequality. x a b Definition. Asymptotes of hyperbola are straight lines, b b given by equations = x, = x. a a The focal radii of the point M (x,) of the hyperbola can be found by the formulas r 1 = ε x a, r = ε x + a. The eccentricity of the hyperbola, as for the ellipse, is determined by the formula ε =. It is easy to verify that the inequality ε a> 1 is true for the eccentricity of the hyperbola. Definition. A parabola is the set of all points of the plane for which the distance to a given point F is equal to the distance to a given straight line d that does not pass through point F. Point F is called the focus of the parabola, and the straight line d is called the directrix. The distance from the focus to the directrix is ​​called the parameter of the parabola and is denoted by p. d M (x,) F x Fig. 4 3

4 Select the origin O of the Cartesian coordinate system at the middle of the segment FD, which is a perpendicular dropped from point F to line d. In this coordinate system, the focus F has coordinates F p p; 0, and the directrix d is given by the equation x + = 0. The canonical equation of the parabola is: = px. The parabola is symmetrical about the OF axis, called the parabola axis. The point O of the intersection of this axis with the parabola is called the vertex of the parabola. The focal radius of the point M (x,) i.e. its p distance to focus is found by the formula r = x +. 10B .. General equation of a second-order line A second-order line is the set of points of the plane whose coordinates x and which satisfy the equation ax + a x + a + a x + a + a = 0, 11 1 where a11, a1, a, a10, a0, a00 some real numbers, and a, a, a are not equal to zero at the same time. This equation is called the general equation of the second-order curve and can also be written in the vector form rr rr (Ax, x) + (b, x) + a = 0, where 00 a11 a1 rr A =, a1 ab = (a10; a0) , x = (x;). T Since A = A, then A is a matrix of quadratic form r r r f (x) = (Ax, x) = a x + a x + a Ellipse, hyperbola and parabola are examples of second-order curves in the plane. In addition to these curves, there are other types of curves of the second order, which are associated with x lines. So, for example, the equation = 0, where a 0, b 0, a b 4

5 defines a pair of intersecting lines on the plane. Coordinate systems in which the equation of the curve takes the simplest form are called canonical. Using the composition of transformations: rotation of the axes through the angle α, parallel translation of the origin to the point (x0; 0) and reflection about the abscissa axis, the equation of the second-order curve is reduced to one of the canonical equations, the main ones of which were listed above. 11BExamples 1. Make the canonical equation of an ellipse with a center at the origin and foci located on the abscissa axis, if it is known that its eccentricity ε = and the point N (3;) lies on the 3rd ellipse. x a b Ellipse equation: + = 1. We have that =. a b a 3 9 From this we calculate that a = b. Substituting the coordinates of the point N (3;) into the equation, we get + = 1 and then b = 9 and a b 81 a = = 16 ,. Consequently, the canonical equation of the ellipse is 5 x + = 1. 16, 9. Write the canonical equation of the hyperbola with the center at the origin and the foci located on the abscissa if the point M 1 (5; 3) of the hyperbola and the eccentricity ε = are given. x Canonical equation of hyperbola = 1. From the equality a b a + b = we have b = a 5 9. Hence = 1 and a = 16. Therefore, the canonical equation of the ellipse = a a a x 16 5

6 3. Find the points on the parabola = 10x whose focal radius is 1.5. Note that the parabola is located in the right half-plane. If M (x; lies on a parabola, then x 0. Parameter p = 5. Let (;)) M x be the required point, F focus, () the directrix of the parabola. Then F, 5; 0, d: x =, 5. Since FM = ρ (M, d), then x +, 5 = 1.5, 10 Answer: () 1 10; 10 x =, = 100, = ± 10. So, we got two points. M 10; 10 M, () 4. On the right branch of the hyperbola given by the equation x = 1, find a point whose distance from the right focus is 16 9 two times less than its distance from the left focus. For the right branch of the hyperbola, the focal radii are determined by the formulas r 1 = ε x a and r = ε x + a. Therefore, we obtain the equation ε x + a = (ε x a). For a given hyperbola, a = 4, 5, c = 5, and ε =. Therefore, x = 9.6. Hence we have = ± x 16 = ± d Answer: two points M 1 (9.6; 0.6 119), (9.6; 0.6 119) M. 5. Find the equation of the line, for any point of which the distance ratio to the point F (3; 0) to the distance to the straight line 1 x 8 = 0 is ε =. Specify the name of the line and its parameters. M x; the desired line, the equality is true: For an arbitrary point () FM (x 3) + 1 = =. ρ (Ml,) x 8 6

7 Hence we have [(x 3) +] = (x 8). Expanding the brackets and rearranging the terms, we obtain (x +) + = 50, i.e. (x +) + = Answer: the required line is an ellipse with the center at a point and semiaxes a = 5 and b = Find the equation of the hyperbola Old coordinates of coordinates O () x; 0; ;,;. C (; 0) = 8 in the new system (x;) and new ones (zt;) are related by the matrix equality 1 1 x z 1 z + t = 1 1 t = z t. Hence, the equation x = 8 z + t z t = 8, zt = 4. Answer: zt = 4. γ: 4x 4x + 8x + 4+ 3 = 0 to cano- 7. Bring the curve to a new form. in new coordinates has the form Consider the quadratic form () q x, = 4x 4x +. The matrix of the form q has eigenvalues ​​5 and 0 and the corresponding orthonormal vectors and Let us pass to the new coordinate system: 7

8 z 1 1 x. t = 5 1 Let's express the old coordinates (x;) in terms of the new ones (zt); : 1 1 z + tx 1 z = 1 t =, 1 zt means, x = z + t, = z + t Substituting these expressions into the equation of the curve γ, we obtain 0 = 4x 4x + 8x = x = z + 1 t, = 1 z + t ( ) () () () = 5z 4 5z + 3 = z 5 4 z 5 + 3 = z 5 1 z 5 3. Hence, in the new coordinates the curve γ is given by the equation 1 3 γ: zz =. Setting = z, x = t, we obtain γ: =, 1 whence we find the canonical equation of the curve γ: = 0 in canonical coordinates = 5 x 1 1 x Note that the curve γ is a pair of parallel lines. 1B Appendices to economic and financial problems 8. Let Anya, Boris and Dmitry have 150 rubles each for the purchase of fruit. It is known that 1 kg of pears costs 15 currency units, and 1 kg of apples costs 10 currency units. Moreover, each of the three 8

9 has its own utility function, for which it wants to maximize when buying. Let x1 kg of pears and x kg of apples be bought. These utility functions are as follows: u = x + x for Ani, 1 A 1 x u B = + x for Boris, and ud = x1 x for Dmitry. It is required to find a purchase plan (x1, x) for Anya, Boris and Dmitry, in which they provide the maximum of their utility function. x Fig. 5 The problem under consideration can be solved geometrically. To solve this problem, the concept of a level line should be introduced. x x 1 Fig. 6 The level line of the function z = f (x,) is the set of all points on the plane on which the function retains a constant value equal to h. x 9

10 In this case, the solution will also use the initial concepts of geometric domains on the plane, specified by linear inequalities (see subsection 1.4). x x 1 Fig. 7 The level lines of the functions ua, u B and u D represent straight lines, ellipses and hyperbolas for Ani, Boris and Dmitry, respectively. In the sense of the problem, we assume that x1 0, x 0. On the other hand, the budget constraint is written as the inequality 15x1 + 10x 150. Dividing the last inequality by 10, we get 3x1 + x 30, or + 1. It is easy to see that x1 x is the domain of solutions of this inequality together with the nonnegativity conditions is a triangle bounded by the lines x1 = 0, x = 0 and 3x1 + x =

11 X * X * Fig. 8 Fig. 9 Based on the geometric patterns, it is now easy to establish that uamax = ua (0.15) = 15, ubmax = ub (0.15) = 5 and udmax = ud (Q). The coordinates of the point Q of the touching of the hyperbola of the level of the side of the budget triangle must already be calculated analytically. To do this, note that the point Q satisfies three equations: xx 1 = h, 3x1 + x = 30, h 3 x "= =. X1 X * Fig

12 Eliminating h from the equations, we obtain the coordinates of the point Q = (x, x) = (5; 7.5). 1 Answer: Q = (x1, x) = (5; 7.5). 9. Non-linear model of costs and profits of the firm. Let a firm produce multipurpose equipment of two types, A and B, in the amount of x and units of production, respectively. In this case, the firm's income for the year is expressed by the income function Rx (,) = 4x +, and the production costs are expressed by the cost function 1 1 Cx (,) = 7.5+ x + 4 which the firm receives the maximum profit. Determine the production plan (x, ) at 3

13 The profit function is compiled as the difference between the income function and the cost function: 1 1 Π (x,) = R (x,) C (x,) = 4x + 7.5 x. 4 Having made the transformations, the last expression is reduced to the form 1 1 Π (x,) = 9 (x 8) (1). 4 The level lines for the profit function are (x 8) (1) = h. 4 Each level line 0 h 9 is an ellipse centered at the origin. From the obtained expression it is easy to see that the maximum of the profit function is 9 and is achieved at x = 8, = 1. Answer: x = 8, = 1. 13BExercises and test questions.1. Write the normal equation for a circle. Find the coordinates of the center and the radius of the circle: a) x + + 8x 6 = 0; b) x x = 0 ... Make the equation of the circle passing through the points M 1 (1;), M (0; 1), M 3 (3; 0) .. 3. Define an ellipse and write its canonical equation. Write the canonical equation of an ellipse, if 1 its eccentricity is ε =, and the semi-major axis is equal to Make the equation of an ellipse, the focuses of which lie on the ordinate axis symmetrically about the origin, knowing, in addition, that the distance between its focuses is c = 4 and the eccentricity ε = Give determination of the eccentricity of the ellipse. Find the eccentricity of an ellipse if its major axis is four times the minor axis. 33

14 .6. Define hyperbola and write its canonical equation. A straight line is drawn through the point M (0; 0.5) and the right vertex of the hyperbola, given by the equation = 1. Find the coordinates of the second point of intersection of the straight line and the hyperbola Give a definition of the eccentricity of the hyperbola. Write its canonical equation if a = 1, b = 5. What is the eccentricity of this hyperbola? .8. Write the equations for the asymptotes of the hyperbola given by its canonical equation. Make the equation of the hyperbola, 3 if its asymptotes are given by the equations = ± x and the hyperbola 5 passes through the point M (10; 3 3) .. 9. Define a parabola and write its canonical equation. Make the canonical equation of a parabola, if the abscissa axis is its axis of symmetry, its vertex lies at the origin and the length of the chord of the parabola perpendicular to the Ox axis is 8, and the distance of this chord from the vertex is On the parabola = 1x, find a point whose focal radius is Proposition and the demand for a certain product is given by the functions p = 4q 1, p = +. Find a market equilibrium point. 1 q Generate graphs ... 1. Andrey, Katya and Nikolay are going to buy oranges and bananas. Buy x1 kg of oranges and x kg of bananas. Each of the three has its own utility function, which shows how useful he considers his purchase. These utility functions are as follows: u = x + x for Andrey, 1 4 A 4 1 u K = x + x for Katya, and un = x1 x for Nikolay. a) Plot the lines of the level of the utility function for the values ​​of the level h = 1, 3. b) For each, arrange in the order of purchase preference rrr r = (4,1), s = (3,8), t = (1,1 ). 34

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In Cartesian coordinates, an equation of the first degree defines a certain straight line.

Lines that are determined by an equation of the first degree in Cartesian coordinates are called first-order lines. Therefore, each straight line is a line of the first order.

General equation of the straight line(as a general equation of the first degree) is determined by an equation of the form:

Oh + Woo + WITH = 0.

Consider the incomplete equations of the straight line.

1. WITH= 0. The equation of the straight line has the form: Ah + Wu = 0; the straight line passes through the origin.

2. V = 0 (A¹ 0). The equation has the form Oh + WITH= 0 or NS =a, where a= Line passes through point A(a; 0), it is parallel to the axis OU... Number a Oh(fig. 1).

Rice. 1

If a= 0, then the line coincides with the axis OU. The equation of the ordinate axis Oy has the form: NS = 0.

3. A = 0 (V¹ 0). The equation is: Woo + WITH= 0 or at = b, where b=. The straight line goes through the point V(0; b), it is parallel to the axis Oh... Number b is the value of the segment that is cut off by the straight line on the axis OU(fig. 2).

Rice. 2

If b = 0, then the straight line coincides with the abscissa axis Ox. The equation of the abscissa axis Ox has the form: y = 0.

Equation of a straight line in line segments on the axes is determined by the equation:

Where are the numbers a and b are the values ​​of the segments cut off by the straight line on the coordinate axes (Fig. 3).

(NS 0 ;at 0)perpendicular to normal vector = {A; B), is determined by the formula:

A(NS – NS 0) + V(at – at 0) = 0.

Equation of a straight line passing through a given point M(NS 0 ; at 0) parallel to the direction vector = {l; m), has the form:

Equation of a straight line passing through two given points M 1 (NS 1 ; at 1) and M 2 (NS 2 ; at 2) is determined by the equation:

The slope of the straight line k called the tangent of the angle of inclination of the straight line to the axis Oh, which is counted from the positive direction of the axis to the straight line counterclockwise, k= tgα.

Equation of a straight line with slope k looks like:

y = kx + b,

where k= tgα, b- the size of the segment cut off by a straight line on the axis OU(fig. 4).

Equation of a straight line passing through a given point M(NS 0 ;at 0)in this direction(slope k known), is determined by the formula:

y - y 0 = k(NS –NS 0).

The equation of a bundle of straight lines passing through a given point M(NS 0 ;at 0) (slope k unknown), is determined by the formula:

y - y 0 = k(NS –NS 0).

Equation of a bundle of straight lines passing through the point of intersection of straight lines

A 1 NS + V 1 at + WITH 1 = 0 and A 2 NS + V 2 at + WITH 2 = 0, is determined by the formula:

α( A 1 NS + V 1 at + WITH 1) + β ( A 2 NS + V 2 at + WITH 2) = 0.

Injection j counted counterclockwise from the straight line y = k 1 NS + b 1 to straight y = k 2 NS + b 2 is determined by the formula (Fig. 5):

For straight lines given by general equations A 1 NS + V 1 at + WITH 1 = 0 and A 2 NS + V 2 at + WITH 2 = 0, the angle between two straight lines is determined by the formula:

The parallelism condition for two straight lines has the form: k 1 = k 2 or.

The condition of perpendicularity of two straight lines has the form: or A 1 A 2 + V 1 V 2 = 0.

The normal equation of the straight line has the form:

x cosα + y sinα - p = 0,

where p - the length of the perpendicular dropped from the origin to the straight line, α is the angle of inclination of the perpendicular to the positive direction of the axis Oh(fig. 6).

To give the general equation of the straight line Oh + Woo + WITH= 0 to normal form, you need to multiply all its terms by normalizing factor μ=, taken with the opposite sign of the free term WITH.

Distance from point M(NS 0 ;at 0)to straight Ah + Woo + WITH= 0 is determined by the formula:

Equations of bisectors of angles between straight lines A 1 NS + V 1 at + WITH 1 = 0 and A 2 NS + V 2 at + WITH 2 = 0 have the form:

Example 4... The vertices of the triangle are given ABC: A (–5; –7), V (7; 2), WITH(–6; 8). Find: 1) side length AB; 2) side equations AB and AS and their slopes; 3) inner corner V; 4) median equation AE; 5) equation and length of height CD; 6) the equation of the bisector AK; 7) the equation of a straight line passing through a point E parallel to the side AB; 8) point coordinates M located symmetrically to the point A relatively straight CD.

1. Distance d between two points A(NS 1 ; at 1) and V(NS 2 ; at 2) is determined by the formula:

Find the side length AB as the distance between two points A(–7; –8) and V(8; –3):

2. Equation of a straight line passing through points A(NS 1 ; at 1) and V(NS 2 ;y 2), has the form:

Substituting the coordinates of the points A and V, we obtain the side equation AB:

3(NS+ 5) = 4(at+ 7); 3NS– 4at– 13 = 0 (AB).

To find the slope k AB straight ( AB) we solve the resulting equation with respect to at:

4y= 3x– 13;

- the equation of the straight line ( AB) with slope,

Similarly, substituting the coordinates of the points V and WITH, we obtain the equation of the straight line ( Sun):

6NS– 42 = –13at+ 26; 6x + 13y– 68 = 0 (BC).

Let us solve the equation of the straight line ( Sun)relatively at: .

3. The tangent of the angle j between two straight lines, the slopes of which are equal k 1 and k 2, is determined by the formula:

Inner corner V formed by straight lines ( AB) and ( Sun), and this is an acute angle through which the straight line must be turned Sun in the positive direction (counterclockwise) until it coincides with the straight line ( AB). Therefore, we substitute in the formula k 1 = , k 2 = :

Ð V= arctg = arctan 1.575 "57.59 °.

4. To find the median equation ( AE), we first determine the coordinates of the point E, which is the middle of the side Sun. To do this, we apply the formulas for dividing a segment into two equal parts:

Hence the point E has coordinates: E(0,5; 5).

Substituting into the equation a straight line passing through two points, the coordinates of the points A and E, we find the median equation ( AE):

24NS – 11at + 43 = 0 (AE).

5. Since the height CD perpendicular to the side AB, then the straight line ( AB) is perpendicular to the straight line ( CD). To find the slope of the height CD, we will use the condition of perpendicularity of two straight lines:

Equation of a straight line passing through a given point M(NS 0 ; at 0) in a given direction (slope k known), has the form:

y - y 0 = k (x - x 0).

Substituting into the last equation the coordinates of the point WITH(–6; 8) and, we obtain the height equation CD:

at – 8 = (NS -(–6)), 3at – 24 = – 4NS– 24, 4NS + 3at = 0 (CD).

Distance from point M(NS 0 ; at 0) to straight Ax + By + C = 0 is determined by the formula:

Length of height CD find as the distance from the point WITH(–6; 8) to the straight line ( AB): 3NS – 4at- 13. Substituting the required quantities in the formula, we find the length CD:

6. Equations of bisectors of angles between straight lines Ax + By + C = 0 and
A 1 x + B 1 y + C 1 = 0 are determined by the formula:

Bisector equation AK as one of the equations of the bisectors of the angles between the straight lines ( AB)and ( AS).

Let us compose the equation of the straight line ( AS) as the equation of a straight line passing through two points A(–5; –7) and WITH (–6; 8):

We transform the last equation:

15(NS+ 5) = – (at+ 7); 15x + y + 82 = 0 (AC).

Substituting the coefficients from the general equations of the straight lines ( AB)and ( AS), we obtain the equations for the bisectors of the angles:

We transform the last equation:

; (3NS – 4at- 13) = ± 5 (15 x + y + 82);

3 NS - 4 at- 13 = ± (75 NS +5at + 410).

Consider two cases:

1) 3 NS - 4 at – 13 = 75NS +5at+ 410.y l AB.

Triangle ABC, height CD, median AE, bisector AK, straight l and point M plotted in coordinate system Ooh(fig. 7).

Lines of the second order.
Ellipse and its canonical equation. Circle

After a thorough study straight lines on the plane we continue to study the geometry of the two-dimensional world. The stakes are doubled, and I invite you to visit the picturesque gallery of ellipses, hyperbolas, parabolas, which are typical representatives of lines of the second order... The tour has already begun, and first, brief information about the entire exposition on different floors of the museum:

The concept of an algebraic line and its order

A line on a plane is called algebraic, if in affine coordinate system its equation has the form, where is a polynomial consisting of terms of the form (- a real number, - non-negative integers).

As you can see, the equation of an algebraic line does not contain sines, cosines, logarithms and other functional beau monde. Only "x" and "games" in non-negative integers degrees.

Line order is equal to the maximum value of the terms included in it.

According to the corresponding theorem, the concept of an algebraic line, as well as its order, do not depend on the choice affine coordinate system, therefore, for the ease of being, we assume that all subsequent calculations take place in Cartesian coordinates.

General Equation the second-order line has the form, where - arbitrary real numbers (it is customary to write with a multiplier - "two"), and the coefficients are not equal to zero at the same time.

If, then the equation is simplified to , and if the coefficients are not simultaneously equal to zero, then this is exactly general equation of a "flat" straight line which is first order line.

Many have understood the meaning of the new terms, but, nevertheless, in order to 100% assimilate the material, we stick our fingers into the socket. To determine the order of the line, you need to iterate all terms its equations and for each of them find sum of degrees incoming variables.

For example:

the term contains "x" in the 1st degree;
the term contains "game" in the 1st degree;
there are no variables in the term, so the sum of their powers is zero.

Now let's figure out why the equation sets the line second order:

the term contains "x" in the 2nd degree;
the summand has the sum of the degrees of the variables: 1 + 1 = 2;
the term contains "game" in the 2nd degree;
all other terms - lesser degree.

Maximum value: 2

If we additionally add, say, to our equation, then it will already determine third order line... Obviously, the general form of the third-order line equation contains a "complete set" of terms, the sum of the powers of the variables in which is equal to three:
, where the coefficients are not equal to zero at the same time.

In the event that we add one or more suitable terms that contain , then we will talk about 4th order lines, etc.

We will have to deal with algebraic lines of the 3rd, 4th and higher orders more than once, in particular, when getting acquainted with polar coordinate system.

However, let's return to the general equation and recall its simplest school variations. As examples, a parabola suggests itself, the equation of which can be easily reduced to a general form, and a hyperbola with an equivalent equation. However, not everything is so smooth….

A significant drawback of the general equation is that it is almost always unclear which line it sets. Even in the simplest case, you will not immediately realize that this is a hyperbole. Such layouts are good only at a masquerade, so a typical problem is considered in the course of analytical geometry reducing the equation of the second order line to the canonical form.

What is the canonical form of the equation?

This is a generally accepted standard form of an equation, when it becomes clear in a matter of seconds which geometric object it defines. In addition, the canonical view is very convenient for solving many practical tasks. So, for example, according to the canonical equation "Flat" straight, firstly, it is immediately clear that this is a straight line, and secondly, the point belonging to it and the direction vector can be easily seen.

Obviously, any 1st order line is a straight line. On the second floor, however, not a watchman is waiting for us, but a much more diverse company of nine statues:

Classification of lines of the second order

With the help of a special set of actions, any equation of a second-order line is reduced to one of the following types:

(and are positive real numbers)

1) - the canonical equation of the ellipse;

2) - the canonical hyperbola equation;

3) - the canonical equation of the parabola;

4) – imaginary ellipse;

5) - a pair of intersecting straight lines;

6) - pair imaginary intersecting lines (with the only valid intersection point at the origin);

7) - a pair of parallel straight lines;

8) - pair imaginary parallel lines;

9) - a pair of coincident straight lines.

Some readers may get the impression that the list is incomplete. For example, in point 7, the equation sets the pair direct parallel to the axis, and the question arises: where is the equation that determines the straight lines parallel to the ordinate? Answer: it not considered canonical... The straight lines represent the same standard case, rotated 90 degrees, and the additional entry in the classification is redundant, since it does not carry anything fundamentally new.

Thus, there are nine and only nine different types of 2nd order lines, but in practice, the most common are ellipse, hyperbola and parabola.

Let's look at an ellipse first. As usual, I focus on those points that are of great importance for solving problems, and if you need a detailed derivation of formulas, proofs of theorems, please refer, for example, to the textbook of Bazylev / Atanasyan or Aleksandrov.

Ellipse and its canonical equation

Spelling ... please do not repeat the mistakes of some Yandex users who are interested in "how to build an ellipsis", "the difference between an ellipse and an oval" and "eccentricity of an elebsis".

The canonical equation of the ellipse has the form, where are positive real numbers, and. I will formulate the very definition of an ellipse later, but for now it's time to take a break from the talking shop and solve a common problem:

How do I build an ellipse?

Yes, take it and just draw it. The task is often encountered, and a significant part of the students do not quite competently cope with the drawing:

Example 1

Construct the ellipse given by the equation

Solution: first we bring the equation to the canonical form:

Why lead? One of the advantages of the canonical equation is that it allows you to instantly determine ellipse vertices that are in points. It is easy to see that the coordinates of each of these points satisfy the equation.

In this case :


Section are called major axis ellipse;
section – minor axis;
number are called semi-major axis ellipse;
number – semi-minor axis.
in our example:.

To quickly imagine what this or that ellipse looks like, it is enough to look at the values ​​"a" and "bs" of its canonical equation.

Everything is fine, foldable and beautiful, but there is one caveat: I made the drawing using the program. And you can complete the drawing using any application. However, in the harsh reality, there is a checkered piece of paper on the table, and mice are dancing in circles on our hands. People with artistic talent, of course, can argue, but you also have mice (though smaller). It is not for nothing that mankind has invented a ruler, compasses, protractor and other simple devices for drawing.

For this reason, we are unlikely to be able to accurately draw an ellipse, knowing only the vertices. Still all right, if the ellipse is small, for example, with semiaxes. Alternatively, you can reduce the scale and, accordingly, the dimensions of the drawing. But in the general case, it is highly desirable to find additional points.

There are two approaches to constructing an ellipse - geometric and algebraic. I do not like the construction with the help of a compass and a ruler because of the not the shortest algorithm and the significant clutter of the drawing. In case of emergency, please refer to the textbook, but in reality it is much more rational to use the tools of algebra. From the equation of the ellipse on the draft, quickly express:

Further, the equation breaks down into two functions:
- defines the upper arc of the ellipse;
- defines the lower arc of the ellipse.

The ellipse specified by the canonical equation is symmetric about the coordinate axes, as well as about the origin. And that's great - symmetry is almost always a harbinger of freebies. Obviously, it is enough to deal with the 1st coordinate quarter, so we need the function ... Finding additional points with abscissas suggests itself ... We hit three sms on the calculator:

Of course, it is also nice that if a serious error is made in the calculations, it will immediately become clear during the construction.

Mark the points on the drawing (red), symmetrical points on the remaining arcs (blue) and carefully connect the whole company with a line:


It is better to draw the initial sketch thinly and thinly, and only then give pressure to the pencil. The result should be a decent ellipse. By the way, would you like to know what this curve is?

Definition of an ellipse. Ellipse foci and ellipse eccentricity

An ellipse is a special case of an oval. The word "oval" should not be understood in a philistine sense ("a child drew an oval", etc.). This is a mathematical term that has a detailed formulation. The purpose of this lesson is not to consider the theory of ovals and their various types, which are almost overlooked in the standard course of analytic geometry. And, in line with more relevant needs, we jump straight to the strict definition of an ellipse:

Ellipse Is the set of all points of the plane, the sum of the distances to each of which is from two given points, called tricks ellipse, - is a constant value, numerically equal to the length of the major axis of this ellipse:.
In this case, the distance between the focuses is less than this value:.

Now everything will become clearer:

Imagine that the blue dot is "driving" an ellipse. So, no matter what point of the ellipse we take, the sum of the lengths of the segments will always be the same:

Let's make sure that in our example the value of the sum is really equal to eight. Mentally place the point "em" at the right vertex of the ellipse, then:, which is what you wanted to check.

Another way of drawing it is based on the definition of an ellipse. Higher mathematics, at times, is the cause of tension and stress, so it's time to have another unloading session. Please take a Whatman paper or a large piece of cardboard and pin it to the table with two studs. These will be tricks. Tie a green thread to the protruding nail heads and pull it all the way with a pencil. The neck of the pencil will be at some point that belongs to the ellipse. Now start tracing your pencil across the sheet of paper, keeping the green thread taut. Continue the process until you return to the starting point ... excellent ... the drawing can be submitted to the doctor for checking the teacher =)

How do I find the focuses of an ellipse?

In the given example, I have depicted "ready-made" focal points, and now we will learn how to extract them from the depths of geometry.

If an ellipse is given by a canonical equation, then its foci have coordinates , where is it distance from each focus to the center of symmetry of the ellipse.

Calculations are easier than a steamed turnip:

! Concrete coordinates of foci cannot be identified with the meaning "tse"! I repeat that this is DISTANCE from each focus to the center(which in the general case does not have to be located exactly at the origin).
And, therefore, the distance between the foci cannot be tied to the canonical position of the ellipse either. In other words, the ellipse can be moved to another place and the value will remain unchanged, while the focuses will naturally change their coordinates. Please keep this in mind as you explore the topic further.

Eccentricity of an ellipse and its geometric meaning

The eccentricity of an ellipse is a ratio that can take values ​​within.

In our case:

Let's find out how the shape of the ellipse depends on its eccentricity. For this fix the left and right vertices the considered ellipse, that is, the value of the semi-major axis will remain constant. Then the eccentricity formula will take the form:.

Let's start bringing the eccentricity value closer to unity. This is only possible if. What does it mean? ... remember the magic tricks ... This means that the foci of the ellipse will "move apart" along the abscissa axis to the lateral vertices. And, since the "green segments are not rubber", the ellipse will inevitably begin to flatten, turning into a thinner and thinner sausage, strung on an axis.

Thus, the closer the value of the eccentricity of the ellipse to unity, the more elongated the ellipse.

Now let's simulate the opposite process: the foci of the ellipse went towards each other, approaching the center. This means that the value of "tse" becomes less and less and, accordingly, the eccentricity tends to zero:.
In this case, the "green segments" will, on the contrary, "become crowded" and they will begin to "push" the ellipse line up and down.

Thus, the closer the eccentricity value is to zero, the more the ellipse looks like... look at the extreme case where the foci are successfully reunited at the origin:

A circle is a special case of an ellipse

Indeed, in the case of equality of the semiaxes, the canonical equation of the ellipse takes the form, which is reflexively transformed to the well-known from school equation of a circle with a center at the origin of coordinates of radius "a".

In practice, the recording with the "speaking" letter "er" is more often used:. The radius is the length of a segment, with each point of the circle removed from the center by the distance of the radius.

Note that the definition of the ellipse remains completely correct: the focuses coincide, and the sum of the lengths of the coinciding segments for each point of the circle is a constant value. Since the distance between the foci, then the eccentricity of any circle is zero.

A circle is built easily and quickly, it is enough to arm yourself with a compass. Nevertheless, sometimes it is necessary to find out the coordinates of some of its points, in this case we go the familiar way - we bring the equation to a brisk Matan form:

- function of the upper semicircle;
- the function of the lower semicircle.

Then we find the required values, differentiate, integrate and doing other good things.

The article, of course, is for reference only, but how can one live without love in the world? Creative task for independent solution

Example 2

Write the canonical equation of an ellipse if one of its focuses and the semi-minor axis are known (the center is at the origin). Find vertices, additional points and draw a line in the drawing. Calculate eccentricity.

Solution and drawing at the end of the lesson

Let's add an action:

Rotating and parallel translation of an ellipse

Let's return to the canonical equation of the ellipse, namely, to the condition, the riddle of which has been tormenting inquisitive minds since the first mention of this curve. Here we examined the ellipse , but isn’t in practice the equation ? After all, here, however, it seems to be like an ellipse too!

Such an equation is rare, but it does come across. And it really defines an ellipse. Let's dispel mysticism:

As a result of construction, our native ellipse is obtained, rotated by 90 degrees. That is, - this is non-canonical notation ellipse . Record!- the equation does not define any other ellipse, since there are no points (foci) on the axis that would satisfy the definition of an ellipse.