Curves of the second order. Hyperbola and its canonical equation

Chapter III. Curves of the second order

§ 40. Hyperbole.

Hyperbole is called a set plane points, for each of which the modulus of the difference in distances to two given points of the plane is constant and less distance between these points.

These points are called tricks hyperbolas, and the distance between them is focal distance.

Denote the foci of the hyperbola by the letters F 1 and F 2 .
Let focal length| F 1 F 2 | = 2 With.

If M is an arbitrary point of the hyperbola (Fig. 112), then by the definition of the hyperbola, the modulus of the difference | F 1 M | - | F 2 M | constant. Denoting it through 2 a, we get

| | F 1 M | - | F 2 M | | = 2 a. (1)

Note that by definition of hyperbola 2 a< 2With, i.e. a< с .

Equality (1) is the equation of a hyperbola.

We choose a coordinate system so that the abscissa axis passes through the foci of the hyperbola; draw the y-axis through the middle of the segment F 1 F 2 perpendicular to it (Fig. 113).

Then the foci of the hyperbola will be the points F 1 (- c; 0) and F 2 ( c; 0).

Let M( X; at) is any point of the hyperbola, then

| F 1 M | = √( x+c) 2 + y 2 and | F 2 M | = √( x-c) 2 + y 2 .

Substituting values | F 1 M | and | F 2 M | into equation (1), we get

| √(x+c) 2 + y 2 - √(x-c) 2 + y 2 | = 2a. (2)

The equation we have obtained is the equation of a hyperbola in the chosen coordinate system. This equation can be reduced to a simpler form.

Let X > 0, then equation (2) can be written without the modulus sign as follows:

√(x+c) 2 + y 2 - √(x-c) 2 + y 2 = 2a,

√(x+c) 2 + y 2 =2a + √(x-c) 2 + y 2 (3)

Let's square both sides of the resulting equality:

(x + c) 2 + at 2 = 4a 2 + 4a √(x-c) 2 + y 2 + (x - c) 2 + at 2 .

After appropriate simplifications and transformations:

√(x-c) 2 + y 2 = c / a x - a, (4)

(x - c) 2 + at 2 = (c / a x - a) 2 ,

we come to the equation

(5)

By definition of a hyperbole a< With, that's why With 2 - a 2 - positive number. Let's denote it by b 2 , i.e. put b 2 = With 2 - a 2. Then equation (5) takes the form

Divided term by term into b 2 , we get the equation

If a X < 0, то уравнение (2) переписывается без знака модуля следующим образом:

√(x-c) 2 + y 2 - √(x+c) 2 + y 2 = 2a,

and in the same way as in the case X > 0 is converted to the form (6).

Equation (6) is called the canonical equation of the hyperbola.

Comment. Squaring both parts of Eqs. (3) and (4) did not violate the equivalence of the equations. Both parts of equation (3) are obviously non-negative for all values X and at. The left side of equation (4) is also always non-negative. At X > a right part equation (4) is positive, since

c / a x - a > c / a a - a = c - a > 0

So, extraneous points could appear only under the condition 0 < X< а , but from equation (6) it follows that x 2 /a 2 > 1, i.e. | x | > a.

Task 1. Write canonical equation hyperbola passing through a point
M (-5; 9/4) if the focal length of the hyperbola is 10.

Since |F 1 F 2 |= 10, then With= 5. Let us write the canonical equation of the hyperbola

By condition, the point M (-5; 9/4) belongs to the hyperbola, therefore,

The second equation to determine a 2 and b 2 gives the ratio

b 2 = With 2 -a 2 = 25 -a 2 .

Having solved the system

find a 2 =16, b 2 = 9. The desired equation will be the equation

Task 2. Prove that the equation

20x 2 - 29y 2 = 580

is the equation of a hyperbola. Find coordinates of tricks.

Dividing both sides of the equation by 580, we get

This is the hyperbolic equation for which a 2 = 29, b 2 = 20.
From the relation c 2 = a 2 + b 2 find c 2 = 29 + 20 = 49, With= 7. Therefore, the foci of the hyperbola are at the points F 1 (-7; 0) and F 2 (7; 0).

1. General equation of curves of the second order.

Any equation of the second degree with respect to x and y, that is, an equation of the form

where - given constant coefficients, and
, defines a line on the plane, which is usually called a curve of the second order. The reverse is also true. There are four types of second order curves: circle, ellipse, hyperbola and parabola. All of them can be obtained by cutting a cone with a plane and therefore they are also called horses.

Curve equations can be derived from their geometric properties as some locus of points that satisfies certain conditions.

2. Circle. The circle is called geometric place points of the plane equidistant from a given point, called the center.

If r is the radius of the circle, and the point C () is its center, then the equation of the circle has the form:

. (12.2)

If the center of the circle coincides with the origin, then the circle equation has the simplest canonical form: .

Example 14. Write an equation for a circle passing through the points
A(5; 0) and B(1; 4), if its center lies on the line x - y - 3 = 0.

Find the coordinates of the point M - the middle of the chord AB:

, that is, M(3; 2).

The center of the circle is on the perpendicular restored from the middle of the segment AB. Let's compose the equation of the straight line AB:

, or x + y - 5 = 0.

The slope of the line AB is -1, hence the slope of the perpendicular . Perpendicular equation

y - 2 \u003d 1 (x - 3), or x - y - 1 \u003d 0.

The center of the circle C lies on the line x + y - 3 = 0 according to the condition of the problem, as well as on the perpendicular x - y - 1 = 0, that is, the coordinates of the center satisfy the system of equations:

x - y - 3 = 0

x - y - 1 \u003d 0.

Hence x = 2, y = 1, and the point C(2; 1).

Circle radius equal to length segment CA:

Circle equation: (x - 2) 2 + (y-1) 2 \u003d 10.

3. Ellipse. An ellipse is the locus of points in a plane, the sum of the distances of which to two given points, called foci, is a constant value equal to greater than the distance between the foci. The canonical equation of an ellipse is:

. (12.3)

Here - semi-major axis ellipse, is the minor semiaxis, and if the distance between the foci is 2c, then . Value is called the eccentricity of the ellipse and characterizes the measure of compression. Since with< , то < 1. Расстояния от некоторой точки М, расположенной на эллипсе, до фокусов называются фокальными радиус-векторами этой точки. Фокальные радиус-векторы выражаются через абсциссу точки эллипса по формулам: .

Direct and are called directrixes of the ellipse. Directrixes of an ellipse have the following property: if r is the focal radius vector of the point M, d is the distance from this point to the one-sided directrix with focus, then .

Example15. Write an equation for an ellipse whose foci lie on the x-axis, symmetrical about the origin, knowing that its major axis is 8 and the distance between directrixes is 16.

By the condition of the problem Directrix equation ; directrix distance , hence ; because , then , that is, c = 2.

Because , then .

Ellipse equation: .

Note: if in the canonical equation of an ellipse , then the foci of the ellipse lie on the y-axis and ; directrix equations: ; focal radius vectors are determined by the formulas: .

Example 16 Write an equation for an ellipse whose foci lie symmetrically on the y-axis with respect to the origin, knowing that the distance between the foci is 2c = 24, the eccentricity .

The canonical equation of an ellipse is: .

By the condition of the problem c = 12. since , then , that is .

Because , then .

Ellipse equation: .

4. Hyperbola. A hyperbola is the locus of points in a plane for which absolute value the difference in distances to two fixed points of the same plane, called foci, is a constant value equal to , less than the distance between the foci ( ).

The canonical equation of a hyperbola has the form:

, (12.4)

where .

The hyperbola consists of two branches and is located symmetrically about the coordinate axes. points and called the vertices of the hyperbola. Line segment is called the real axis of the hyperbola, and the segment connecting points and , - imaginary axis. A hyperbola has two asymptotes whose equations are . Attitude is called the eccentricity of the hyperbola. straight, given by equations are called directrixes of a hyperbola. Focal radius vectors of the right branch of the hyperbola: .

Focal radius vectors of the left branch of the hyperbola: .

The equation is also an equation of a hyperbola, but the real axis of this hyperbola is a segment of the OY axis of length . points and serve as vertices of the hyperbola. The branches of the hyperbola are located at the top and bottom coordinate plane. Two hyperbolas and are called conjugate hyperbolas.

Example17. The eccentricity of the hyperbola is . Compose the simplest equation of a hyperbola passing through the point M( ).

By definition of eccentricity, we have , or .

But , Consequently . Since point M( ) is on a hyperbola, then . From here .

Thus, the equation of the desired hyperbola has the form: .

Example 18. The angle between the asymptotes of the hyperbola is 60°. Calculate the eccentricity of the hyperbola.

Slope of hyperbola asymptote
. Eccentricity of a hyperbola
.

Substituting value slope, we get

Example 19. Write an equation for a hyperbola passing through a point
M(9; 8) if the asymptotes of the hyperbola are given by the equations .

From the asymptote equation we have . Since the point M(9; 8) belongs to the hyperbola, its coordinates satisfy the equation of the hyperbola, i.e. .

To find the semiaxes of the hyperbola, we have the system:

Solving the system, we get The desired equation of the hyperbola has the form: .

5. Parabola. A parabola is the locus of points in a plane that are equidistant from a given point, called the focus, and from a given line, called the directrix. If the directrix is given by the equation , and the focus is at the point F(), then the parabola equation has the form:

. (12.5)

This parabola is located symmetrically about the x-axis.

The equation is the equation of a parabola symmetric about the y-axis.

The length of the focal radius vector of the parabola is determined by the formula .

Example 20. Compose the equation of a parabola with a vertex at the origin, symmetrical about the OY axis and cutting off a chord of length 8 on the bisector of the first and third coordinate angles.

The desired parabola equation has the form .

Bisector equation y \u003d x. Let's define the intersection points of the parabola and the bisector:

Having solved the system, we obtain O(0; 0) and M(2p; 2p).

Chord length OM = .

By condition, we have: OM \u003d 8, whence 2p \u003d 8.

The desired parabola equation .

Plane equation

AT Cartesian coordinates each plane is defined by a first-degree equation in the unknowns x, y, and z, and each first-degree equation in three unknowns defines a plane.

Let's take an arbitrary vector with the beginning at the point . Let us derive the equation of the locus of points M(x, y, z), for each of which the vector perpendicular to the vector. Let us write down the condition of perpendicularity of vectors:

The resulting equation is linear with respect to x, y, z, therefore, it defines a plane passing through the point perpendicular to the vector . Vector is called the normal vector of the plane. Expanding the brackets in the resulting plane equation and denoting the number
letter D, we represent it in the form:

Ax + By + Cz + D = 0. (13.2)

This equation is called the general equation of the plane. A, B, C and D are the coefficients of the equation, A 2 + B 2 + C 2 0.

1. Incomplete Equations planes.

If in the general equation of the plane one, two or three coefficients are equal to zero, then the equation of the plane is called incomplete. May introduce themselves following cases:

1) D = 0 - the plane passes through the origin;

2) A = 0 - the plane is parallel to the Ox axis;

3) B = 0 - the plane is parallel to the Oy axis;

4) C = 0 - the plane is parallel to the Oz axis;

5) A = B = 0 - the plane is parallel to the XOY plane;

6) A \u003d C \u003d 0 - the plane is parallel to the XOZ plane;

7) B = C = 0 - the plane is parallel to the YOZ plane;

8) A \u003d D \u003d 0 - the plane passes through the Ox axis;

9) B = D = 0 - the plane passes through the Oy axis;

10) C \u003d D \u003d 0 - the plane passes through the Oz axis;

11) A = B = D = 0 - the plane coincides with the XOY plane;

12) A = C = D = 0 - the plane coincides with the XOZ plane;

13) C \u003d B \u003d D \u003d 0 - the plane coincides with the YOZ plane.

2. Equation of a plane in segments.

If in the general equation of the plane D 0, then it can be transformed to the form

, (13.3)

which is called the equation of the plane in segments. - determine the length of the segments cut off by the plane on the coordinate axes.

3. Normal equation of the plane.

The equation

where are the direction cosines of the normal vector of the plane , called normal equation planes. To bring the general equation of the plane to normal form, it must be multiplied by the normalizing factor:
,

in this case, the sign in front of the root is chosen from the condition .

Distance d from point to the plane is determined by the formula: .

4. Equation of a plane passing through three points

Let's take an arbitrary point of the plane M(x,y,z) and connect the point M1 with each of the three remaining ones. We get three vectors . For three vectors to belong to the same plane, it is necessary and sufficient that they be coplanar. The condition for the complanarity of three vectors is the equality to zero of their mixed product, that is .

Writing this equality in terms of the coordinates of the points, we obtain the desired equation:

. (13.5)

5. Angle between planes.

The planes can be parallel, coincide or intersect, forming dihedral angle. Let two planes be given general equations and . For the planes to coincide, it is necessary that the coordinates of any point that satisfies the first equation would also satisfy the second equation.

This will take place if
.

If a , then the planes are parallel.

The angle formed by two intersecting planes, equal to the angle formed by their normal vectors. The cosine of the angle between vectors is determined by the formula:

If , then the planes are perpendicular.

Example 21. Write an equation for a plane that passes through two points and perpendicular to the plane.

We write the desired equation in general view: . Since the plane must pass through the points and , the coordinates of the points must satisfy the equation of the plane. Substituting the coordinates of the points and , we obtain: and .

From the condition of perpendicularity of the planes we have: . Vector located in the desired plane and, therefore, perpendicular to the normal vector: .

Combining the obtained equations, we have:

Solving the system, we get: , , , .

The desired equation has the form: .

The second way. normal vector given plane has coordinates . Vector . The normal vector of the required plane is perpendicular to the vector and the vector , i.e. collinear to the vector product . Compute vector product:
.

Vector
. Let's write the equation of the plane passing through the point perpendicular to the vector:

Or the desired equation.

Definition . A hyperbola is a locus of points, the difference from each of which to two given points, called foci, is a constant value

Let's take a coordinate system so that the foci lie on the abscissa axis, and the origin of coordinates divides the segment F 1 F 2 in half (Fig. 30). Denote F 1 F 2 = 2c. Then F 1 (c; 0); F2 (-c; 0)

MF 2 \u003d r 2, MF 1 \u003d r 1 - focal radii hyperbole.

According to the definition of a hyperbola, r 1 - r 2 = const.

Let's denote it by 2a

Then r 2 - r 1 = ±2a so:

=> canonical equation of a hyperbola

Since the equation of the hyperbola x and y is in even powers, then if the point M 0 (x 0; y 0) lies on the hyperbola, then the points M 1 (x 0; -y 0) M 2 (-x 0; -x 0; -y 0) M 3 (-x 0; -y 0).

Therefore, the hyperbola is symmetrical about both coordinate axes.

When y \u003d 0 x 2 \u003d a 2 x \u003d ± a. The vertices of the hyperbola will be points A 1 (a; 0); A 2 (-a; 0).

. Due to symmetry, the study is carried out in the first quarter

1) at
y has an imaginary value, hence the points of the hyperbola with abscissas
does not exist

2) at x = a; y \u003d 0 A 1 (a; 0) belongs to a hyperbola

3) for x > a; y > 0. Moreover, with an unlimited increase in x, the branch of the hyperbola goes to infinity.

It follows that a hyperbola is a curve consisting of two infinite branches.

P 6. Asymptotes of a hyperbola

Consider together with the equation
straight line equation

To the curve will lie below the straight line (Fig. 31). Consider points N (x, Y) and M (x, y) whose abscissas are the same, and Y - y \u003d MN. Consider the length of the segment MN

Let's find

So, if the point M, moving along the hyperbola in the first quarter, moves away to infinity, then its distance from the straight line
decreases and tends to zero.

Due to symmetry, the straight line has the same property.
.

Definition. Direct lines to which
the curve approaches indefinitely are called asymptotes.

And
so, the equation of the asymptotes of the hyperbola
.

The asymptotes of the hyperbola are located along the diagonals of a rectangle, one side of which is parallel to the x-axis and is equal to 2a, and the other is parallel to the y-axis and is equal to 2b, and the center lies at the origin (Fig. 32).

P 7. Eccentricity and directrixes of a hyperbola

r 2 – r 1 = ± 2a sign + refers to the right branch of the hyperbola

sign - refers to the left branch of the hyperbola

Definition. The eccentricity of a hyperbola is the ratio of the distance between the foci of this hyperbola to the distance between its vertices.

. Since c > a, ε > 1

We express the focal radii of the hyperbola in terms of the eccentricity:

Definition . Let's call the lines
, perpendicular to the focal axis of the hyperbola and located at a distancefrom its center by the directrix of the hyperbola corresponding to the right and left foci.

T
like for hyperbole
consequently, the directrixes of the hyperbola are located between its vertices (Fig. 33). Let us show that the ratio of the distances of any point of the hyperbola to the focus and the corresponding directrix is a constant and equal to ε.

P. 8 Parabola and its equation

O
definition. A parabola is the locus of points that are equidistant from a given point, called the focus, and from a given line, called the directrix.

To compose the equation of a parabola, let's take the x-axis as a straight line passing through the focus F 1 perpendicular to the directrix and consider the x-axis directed from the directrix to the focus. For the origin of coordinates, we take the midpoint O of the segment from the point F to the given straight line, the length of which we denote by p (Fig. 34). The quantity p will be called the parameter of the parabola. Focus coordinate point
.

Let M(x, y) be an arbitrary point of the parabola.

By definition

at 2 = 2px is the canonical equation of the parabola

To determine the type of parabola, we transform its equation
this implies . Therefore, the vertex of the parabola is at the origin and the axis of symmetry of the parabola is x. The equation y 2 \u003d -2px with positive p is reduced to the equation y 2 \u003d 2px by replacing x with -x and its graph looks like (Fig. 35).

At
the equation x 2 \u003d 2py is the equation of a parabola with a vertex at the point O (0; 0) whose branches are directed upwards.

X
2 \u003d -2ru - the equation of a parabola centered at the origin is symmetrical about the y-axis, the branches of which are directed downwards (Fig. 36).

A parabola has one axis of symmetry.

If x is to the first power and y is to the second power, then the axis of symmetry is x.

If x is to the second power and y is to the first power, then the axis of symmetry is the y-axis.

Remark 1. The directrix equation of a parabola has the form
.

Remark 2. Since for a parabola , thenε parabola is 1.ε = 1 .

Given the equation of an ellipse.

Solution:

We write the equation of the ellipse in the canonical form:
.

From here
. Using the relation
, we find
. Consequently,
.

According to the formula find .

Directrix equations
look like
, the distance between them
.

According to the formula
find the abscissa of the points, the distance from which to the point equals 12:

. Substituting value x into the equation of an ellipse, we find the ordinates of these points:

Thus, the point A(7;0) satisfies the condition of the problem.

Problem 56.

Write an equation for an ellipse passing through the points.

Solution:

We are looking for the ellipse equation in the form
.

Since the ellipse passes through the points
, then their coordinates satisfy the ellipse equation:
. Multiplying the second equality by (-4) and adding to the first, we find
.

Substituting the found value into the first equation, we find
. Thus, the desired equation
.

Problem 57.

;
.

Hyperbola

Hyperbole a line is called, consisting of all points of the plane, the modulus of the difference in distances from which to two given points and is a constant value (not equal to zero and less than the distance between the points and ).

points and called tricks hyperbole. Let the distance between the foci be
. Module of distances from hyperbola points to foci and denote by . By condition,
.

where
- coordinates arbitrary point hyperbole,

The equation
called canonical equation hyperbole.

The hyperbole has two asymptotes
.

eccentricity hyperbola is called a number . For any hyperbole
.

The focal radii of a hyperbola point called the line segments connecting this point with the foci and . Their lengths and are given by the formulas:

Direct
are called directrixes of the hyperbola. As in the case of an ellipse, the points of a hyperbola are characterized by the relation .

Problem 58.

Find the distance between the foci and the eccentricity of the hyperbola
.

Answer:
.

Problem 59.

Write the canonical equation of the hyperbola if (
). Determine the eccentricity of the hyperbola.

Answer:
.

Problem 60.

Write the canonical equation of a hyperbola symmetric about the coordinate axes if it passes through a point
, and the eccentricity is
.

Answer:
.

Task 61.

Find the equations of a hyperbola whose vertices are at foci and whose foci are at the vertices of an ellipse
.

Answer:
.

Problem 62.

Determine the locus of points
, the distances from which to the straight line
half as much as before the point
.

Answer:
.

Problem 63.

Write the equation of a hyperbola symmetric with respect to the coordinate system if it passes through points
,
.

Answer:
.

Task 64.

Write an equation for a hyperbola if its asymptotes are given by the equation
, and the hyperbola passes through the point
.

Answer:
.

Problem 65.

How are the points located on the plane, the coordinates of which satisfy the conditions:

Parabola

parabola called a line consisting of all points of the plane equidistant from a given point
(focus) and given line (directors).

To derive the canonical equation of the parabola, the axis
pass through the focus
perpendicular to the directrix in the direction from the directrix to the focus; the origin of coordinates is taken in the middle of the segment between the focus
and dot
axis intersection
with the headmistress . If denoted by focus distance from the directrix, then
and the directrix equation will look like
.