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 - arbitrary point hyperbolas (Fig. 112), then, by definition of a 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).

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 of an arbitrary point of the hyperbola,

.

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.

Define geometric place 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
.

In the chosen coordinate system, the parabola equation has the form:
. This equation is called the canonical equation of the parabola.

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 .