A straight line is the goal of a hyperbola. Hyperbola and its canonical equation

Class 10 . Second order curves.

10.1. Ellipse. Canonical equation. Semi-axes, eccentricity, graph.

10.2. Hyperbola. Canonical equation. Semi-axes, eccentricity, asymptotes, graph.

10.3. Parabola. Canonical equation. Parabola parameter, graph.

Second-order curves on a plane are lines whose implicit definition has the form:

Where
- given real numbers,
- coordinates of the curve points. The most important lines among second-order curves are the ellipse, hyperbola, and parabola.

10.1. Ellipse. Canonical equation. Semi-axes, eccentricity, graph.

Definition of an ellipse.An ellipse is a plane curve whose sum of distances from two fixed points is
plane to any point

(those.). Points
are called the foci of the ellipse.

Canonical ellipse equation:
. (2)

(or axis
) goes through tricks
, and the origin is the point - is located in the center of the segment
(Fig. 1). Ellipse (2) is symmetrical with respect to the coordinate axes and the origin (the center of the ellipse). Permanent
,
are called semi-axes of the ellipse.

If the ellipse is given by equation (2), then the foci of the ellipse are found like this.

1) First, we determine where the foci lie: the foci lie on the coordinate axis on which the major semi-axes are located.

2) Then the focal length is calculated (distance from foci to origin).

At
foci lie on the axis
;
;
.

Eccentricity ellipse is called the quantity: (at
);(at
).

The ellipse always
.

Eccentricity serves as a characteristic of the compression of the ellipse.

,
If the ellipse (2) is moved so that the center of the ellipse hits the point

, then the equation of the resulting ellipse has the form

10.2. Hyperbola. Canonical equation. Semi-axes, eccentricity, asymptotes, graph.Definition of hyperbole.
plane to any point
A hyperbola is a plane curve in which the absolute value of the difference in distances from two fixed points is this curve has constant
(those.). , independent of the point
Points

are called the foci of a hyperbola.:
Canonical hyperbola equation
. (3)

or
(or axis
) goes through tricks
, and the origin is the point - is located in the center of the segment
This equation is obtained if the coordinate axis
,
are called ..

Hyperbolas (3) are symmetrical about the coordinate axes and the origin. Permanent

semi-axes of the hyperbola
foci lie on the axis
:
The foci of a hyperbole are found like this.

semi-axes of the hyperbola
foci lie on the axis
:
At the hyperbole

(Fig. 2.a). - focal length (distance from the foci to the origin). It is calculated by the formula:
.

Eccentricity hyperbola is the quantity:

(For
);(For
).

Hyperbole always has
.

Asymptotes of hyperbolas(3) are two straight lines:
. Both branches of the hyperbola approach the asymptotes without limit with increasing .

The construction of a hyperbola graph should be carried out as follows: first along the semi-axes
we build an auxiliary rectangle with sides parallel to the coordinate axes; then draw straight lines through the opposite vertices of this rectangle, these are the asymptotes of the hyperbola; finally we depict the branches of the hyperbola, they touch the midpoints of the corresponding sides of the auxiliary rectangle and get closer with growth to asymptotes (Fig. 2).

If hyperbolas (3) are moved so that their center hits the point
, and the semi-axes will remain parallel to the axes
,
, then the equation of the resulting hyperbolas will be written in the form

,
.

10.3. Parabola. Canonical equation. Parabola parameter, graph.

Definition of a parabola.A parabola is a plane curve for which, for any point
this curve is the distance from
to a fixed point plane (called the focus of the parabola) is equal to the distance from
to a fixed straight line on the plane(called the directrix of the parabola) .

Canonical parabola equation:
, (4)

Where - a constant called parameter parabolas.

Dot
parabola (4) is called the vertex of the parabola. Axis
is the axis of symmetry. The focus of the parabola (4) is at the point
, directrix equation
.
Parabola graphs (4) with meanings
And

are shown in Fig. 3.a and 3.b respectively.
The equation
also defines a parabola on the plane
,
, whose axes, compared to parabola (4),

switched places.
If parabola (4) is moved so that its vertex hits the point
, and the axis of symmetry will remain parallel to the axis

, then the equation of the resulting parabola has the form

Let's move on to examples. Example 1
. The second order curve is given by the equation
.

. Give a name to this curve. Find its foci and eccentricity. Draw a curve and its foci on a plane
Solution. This curve is an ellipse centered at the point
and axle shafts
. This can be easily verified by replacing
. This transformation means a transition from a given Cartesian coordinate system to a new one Cartesian system
coordinates
, whose axis
,
parallel to the axes
. This coordinate transformation is called a system shift exactly Cartesian system
. IN new system the equation of the curve is converted to
canonical equation

ellipse
, its graph is shown in Fig. 4.
Let's find tricks.
, so the tricks
:
ellipse located on the axis
.. In the coordinate system
foci have coordinates.

Example 2. Give the name of the second-order curve and provide its graph.

Solution. Let us select perfect squares based on terms containing variables Parabola graphs (4) with meanings .

Now, the equation of the curve can be rewritten as follows:

Therefore, the given curve is an ellipse centered at the point
Solution. This curve is an ellipse centered at the point
. The information obtained allows us to draw its graph.

Example 3. Give a name and graph of the line
.

Solution. .
Solution. This curve is an ellipse centered at the point
.

This is the canonical equation of an ellipse centered at the point
Because the, , we conclude: behind given equation
defines on the plane

the lower half of the ellipse (Fig. 5). Example 4
. Give the name of the second order curve

. Find its focuses, eccentricity. Give a graph of this curve.
.

- canonical equation of a hyperbola with semi-axes

Focal length. , its graph is shown in Fig. 4.
The minus sign precedes the term with
hyperbolas lie on the axis
.

The branches of the hyperbola are located above and below the axis

- eccentricity of the hyperbola.

Asymptotes of a hyperbola: . The construction of a graph of this hyperbola is carried out in accordance with the procedure outlined above: we build an auxiliary rectangle, draw asymptotes of the hyperbola, draw branches of the hyperbola (see Fig. 2.b).
Example 5

. Find out the type of curve given by the equation
and plot it.

- hyperbola with center at a point
and axle shafts.
Because , we conclude: the given equation determines that part of the hyperbola that lies to the right of the straight line
.
It is better to draw a hyperbola in an auxiliary coordinate system

, obtained from the coordinate system shift

, and then highlight the desired part of the hyperbola with a bold line Example 6. Find out the type of curve and draw its graph. :

Solution. Let's highlight

perfect square
by terms with variable
Let's rewrite the equation of the curve. This is the equation of a parabola with its vertex at the point. Using a shift transformation, the parabola equation is brought to the canonical form
, from which it is clear
what-parameter
parabolas. Focus

parabolas in the system.

has coordinates
,, and in the system

(according to shift transformation). The parabola graph is shown in Fig. 7.
Homework

1. Draw ellipses given by the equations:
Find their semi-axes, focal length, eccentricity and indicate on the graphs of ellipses the locations of their foci.

2. Draw hyperbolas given by the equations:
defines the 2nd order part of the curve. Find the canonical equation of this curve, write down its name, plot its graph and highlight on it that part of the curve that corresponds to the original equation.

A hyperbole is called locus points for which the difference in distances from two fixed points of the plane, called foci, is a constant value; the indicated difference is taken according to absolute value and is usually denoted by 2a. The foci of a hyperbola are denoted by the letters F 1 and F 2, the distance between them by 2c. By definition of hyperbola 2a

Let a hyperbole be given. If the axis is Cartesian rectangular system coordinates are chosen so that the foci of a given hyperbola are located on the x-axis symmetrically relative to the origin of coordinates, then in this coordinate system the equation of the hyperbola has the form

x 2 /a 2 + y 2 /b 2 = 1, (1)

where b = √(c 2 - a 2). An equation of type (I) is called the canonical equation of a hyperbola. With the specified choice of coordinate system, the coordinate axes are the axes of symmetry of the hyperbola, and the origin is its center of symmetry (Fig. 18). The axes of symmetry of a hyperbola are simply called its axes, the center of symmetry is the center of the hyperbola. The hyperbola intersects one of its axes; the intersection points are called the vertices of the hyperbola. In Fig. The 18 vertices of a hyperbola are points A" and A.

A rectangle with sides 2a and 2b, located symmetrically relative to the axes of the hyperbola and touching it at the vertices, is called the main rectangle of the hyperbola.

The segments of length 2a and 2b connecting the midpoints of the sides of the main rectangle of the hyperbola are also called its axes. The diagonals of the main rectangle (extended indefinitely) are asymptotes of the hyperbola; their equations are:

y = b/a x, y = - b/a x

The equation

X 2 /a 2 + y 2 /b 2 = 1 (2)

defines a hyperbola that is symmetrical with respect to coordinate axes with focuses on the ordinate axis; equation (2), like equation (1), is called the canonical hyperbola equation; in this case, the constant difference in distances from an arbitrary point of the hyperbola to the foci is equal to 2b.

Two hyperbolas, which are defined by the equations

x 2 /a 2 - y 2 /b 2 = 1, - x 2 /a 2 + y 2 /b 2 = 1

in the same coordinate system are called conjugate.

A hyperbola with equal half-axis (a = b) is called equilateral; its canonical equation has the form

x 2 - y 2 = a 2 or - x 2 + y 2 = a 2.

where a is the distance from the center of the hyperbola to its vertex, called the eccentricity of the hyperbola. Obviously, for any hyperbola ε > 1. If M(x; y) - arbitrary point hyperbolas, then the segments F 1 M and F 2 M (see Fig. 18) are called the focal radii of the point M. The focal radii of the points on the right branch of the hyperbola are calculated using the formulas

r 1 = εx + a, r 2 = εx - a,

focal radii of points of the left branch - according to the formulas

r 1 = -εх - a, r 2 = -εх + a

If the hyperbola is given by equation (1), then the lines defined by the equations

x = -a/ε, x = a/ε

are called its directrices (see Fig. 18). If the hyperbola is given by equation (2), then the directrixes are determined by the equations

x = -b/ε, x = b/ε

Each headmistress has the following property: if r is the distance from an arbitrary point of the hyperbola to a certain focus, d is the distance from the same point to the one-sided directrix with this focus, then the ratio r/d is a constant value equal to the eccentricity of the hyperbola:

515. Compose an equation of a hyperbola whose foci are located on the abscissa axis symmetrically relative to the origin, knowing in addition that:

1) its axes 2a = 10 and 2b = 8;

2) the distance between the foci 2c = 10 and the axis 2b = 8;

3) distance between foci 2с = 6 and eccentricity ε = 3/2;

4) axis 2a = 16 and eccentricity ε = 5/4;

5) equations of asymptotes y = ±4/3x and the distance between the foci 2c = 20;

6) the distance between the directrixes is 22 2/13 and the distance between the foci is 2c = 26; 39

7) the distance between the directrixes is 32/5 and the axis 2b = 6;

8) the distance between the directrixes is 8/3 and the eccentricity ε = 3/2;

9) the equation of asymptotes y = ± 3/4 x and the distance between the directrixes is 12 4/5.

516. Compose an equation of a hyperbola whose foci are located on the ordinate axis symmetrically with respect to the origin, knowing in addition that:

1) its semi-axes a = 6, b = 18 (by the letter a we denote the semi-axis of the hyperbola located on the x-axis);

2) the distance between the foci is 2c = 10 and the ecceitricity is ε = 5/3; very good 12

3) the equation of asymptotes y = ±12/5x and the distance between the vertices is 48;

4) the distance between the directrixes is 7 1/7 and the eccentricity ε = 7/5;

5) the equation of asymptotes y = ± 4/3x and the distance between the directrixes is 6 2/5.

517. Determine the semi-axes a and b for each of the following hyperbolas:

1) x 2 /9 - y 2 /4 = 1; 2) x 2 /16 - y 2 = 1; 3) x 2 - 4y 2 = 16;

4) x 2 - y 2 = 1; 5) 4x 2 - 9y 2 = 25; 6) 25x 2 -16y 2 = 1;

7) 9x 2 - 64y 2 = 1.

518. Given a hyperbola 16x 2 - 9y 2 = 144. Find: 1) semi-axes a and b; 2) tricks; 3) eccentricity; 4) equations of asymptotes; 5) directrix equations.

519. Given a hyperbola 16x 2 - 9y 2 = -144. Find: 1) semi-axes a and b; 2) tricks; 3) eccentricity; 4) equations of asymptotes; 5) directrix equations.

520. Calculate the area of the triangle formed by the asymptotes of the hyperbola x 2 /4 - y 2 /9 = 1 and the line 9x + 2y - 24 = 0.

521. Establish which lines are determined by the following equations:

1) y = +2/3√(x 2 - 9); 2) y = -3√(x 2 + 1)

3) x = -4/3√(y 2 + 9); 4) +2/5√(x 2 + 25)

522. Given a point M 1 (l0; - √5) on a hyperbola - x 2 /80 - y 2 /20 = 1. Create equations of the lines on which the focal radii of the point M 1 lie.

523. Having made sure that point M 1 (-5; 9/4) lies on the gillerball x 2 /16 - y 2 /9 = 1, determine the focal radii of point M 1.

524. The eccentricity of the hyperbola is ε = 2, the focal radius of its point M, drawn from a certain focus, is equal to 16. Calculate the distance from the point M to the one-sided directrix with this focus.

525. The eccentricity of the hyperbola is ε = 3, the distance from the point M of the hyperbola to the directrix is 4. Calculate the distance from the point M to the focus, one-sided with this directrix.

526. The eccentricity of the hyperbola is ε = 2, its center lies at the origin, one of the foci F(12; 0). Calculate the distance from point M 1 of the hyperbola with abscissa equal to 13 to the directrix corresponding to the given focus.

527. The eccentricity of the hyperbola is ε = 3/2, its center lies at the origin, one of the directrixes is given by the equation x = -8. Calculate the distance from point M 1 of the hyperbola with abscissa equal to 10 to the focus corresponding to the given directrix.

528. Determine the points of the hyperbola - x 2 /64 - y 2 /36 = 1, the distance of which to the right focus is 4.5.

529. Determine the points of the hyperbola x 2 /9 - y 2 /16 = 1, the distance of which to the left focus is 7.

530. Through the left focus of the hyperbola x 2 /144 - y 2 /25 = 1 a perpendicular is drawn to its axis containing the vertices. Determine the distances from the foci to the points of intersection of this perpendicular with the hyperbola.

531. Using one compass, construct the foci of the hyperbola x 2 /16 - y 2 /25 = 1 (assuming that the coordinate axes are depicted and the scale unit is given).

532. Compose an equation of a hyperbola whose foci lie on the abscissa axis symmetrically with respect to the origin, if given:

1) points M 1 (6; -1) and M 2 (-8; 2√2) hyperbolas;

2) point M 1 (-5; 3) hyperbola and eccentricity ε = √2;

3) point M 1 (9/2;-l) hyperbola and equation of asymptotes y = ± 2.3x;

4) point M 1 (-3; 5.2) hyperbola and directrix equation x = ± 4/3;

5) equations of asymptotes y = ±-3/4x and equations of directrixes x = ± 16/5

533. Determine the eccentricity of an equilateral hyperbola.

534. Determine the eccentricity of a hyperbola if the segment between its vertices is visible from the foci of the conjugate hyperbola at an angle of 60°.

535. The foci of the hyperbola coincide with the foci of the ellipse x 2 /25 + y 2 /9 = 1. Write an equation for the hyperbola if its eccentricity ε = 2.

536. Write an equation for a hyperbola whose foci lie at the vertices of the ellipse x 2 /100 + y 2 /64 = 1, and the directrixes pass through the foci of this ellipse.

537. Prove that the distance from the focus of the hyperbola x 2 /a 2 - y 2 /b 2 = 1 to its asymptote is equal to b.

538. Prove that the product of distances from any point of the hyperbolax x 2 /a 2 - y 2 /b 2 = 1 to its two asymptotes is a constant value equal to a 2 b 2 /(a 2 + b 2)

539. Prove that the area of a parallelogram bounded by the asymptotes of the hyperbola x 2 /a 2 - y 2 /b 2 = 1 and the lines drawn through any of its points parallel to the asymptotes is a constant value equal to ab/2.

540. Write an equation for a hyperbola if its semi-axes a and b are known, the center C(x 0;y 0) and the foci are located on a straight line: 1) parallel to the Ox axis; 2) parallel to the Oy axis.

541. Establish that each of the following equations defines a hyperbola, and find the coordinates of its center C, semi-axis, eccentricity, equations of asymptotes and equations of directrixes:

1) 16x 2 - 9у 2 - 64x - 54у - 161 =0;

2) 9x 2 - 16y 2 + 90x + 32y - 367 = 0;

3) 16x 2 - 9y 2 - 64x - 18y + 199 = 0.

542. Establish which lines are determined by the following equations:

1) y = - 1 + 2/3√(x 2 - 4x - 5);

2) y = 7 - 3/2√(x 2 - 6x + 13);

3) x = 9 - 2√(y 2 + 4y + 8);

4) X = 5 + 3/4√(y 2 + 4y - 12).

Draw these lines on the drawing.

543. Create an equation for a hyperbola, knowing that:

1) the distance between its vertices is 24 and the foci are F 1 (-10; 2), F 2 (16; 2);

2) the foci are F 1 (3; 4), F 2 (-3; -4) and the distance between the directrixes is 3.6;

3) the angle between the asymptotes is 90° and the foci are F 1 (4; -4), F 1 (- 2; 2).

544. Write an equation for a hyperbola if its eccentricity ε = 5/4, focus F (5; 0) and the equation of the corresponding directrix 5x - 16 = 0 are known.

545. Write an equation for a hyperbola if its eccentricity e - focus F(0; 13) and the equation of the corresponding directrix 13y - 144 = 0 are known.

546. Point A (-3; - 5) lies on a hyperbola whose focus is F (-2;-3), and the corresponding directrix is given by the equation x + 1 = 0. Write an equation for this hyperbola.

547. Write an equation for a hyperbola if its eccentricity ε = √5, focus F(2;-3) and the equation of the corresponding directrix Zx - y + 3 = 0 are known.

548. Point M 1 (1; 2) lies on a hyperbola whose focus is F(-2; 2), and the corresponding directrix is given by the equation 2x - y - 1 = 0. Write an equation for this hyperbola.

549. The equation of an equilateral hyperbola x 2 - y 2 = a 2 is given. Find its equation in the new system, taking its asymptotes as the coordinate axes.

550. Having established that each of the following equations defines a hyperbola, find for each of them the center, semi-axes, equations of asymptotes and plot them on the drawing: 1) xy = 18; 2) 2xy - 9 = 0; 3) 2xy + 25 = 0.

551. Find the intersection points of the straight line 2x - y - 10 = 0 and the hyperbola x 2 /20 - y 2 /5 = 1.

552. Find the intersection points of the straight line 4x - 3y - 16 = 0 and the hyperbola x 2 /25 - y 2 /16 = 1.

553. Find the intersection points of the straight line 2x - y + 1 = 0 and the hyperbola x 2 /9 - y 2 /4 = 1.

554.V following cases determine how the line is located relative to the hyperbola: whether it intersects, touches or passes outside it:

1) x - y - 3 = 0, x 2 /12 - y 2 /3 = l;

2) x - 2y + 1 = 0, x 2 /16 - y 2 /9 = l;

555. Determine at what values of m the straight line y = 5/2x + m

1) intersects the hyperbola x 2 /9 - y 2 /36 = 1; 2) touches her;

3) passes outside this hyperbole.

556. Derive the condition under which the straight line y = kx + m touches the hyperbola x 2 /a 2 - y 2 /b 2 = 1.

557. Write an equation for the tangent to the hyperbola x 2 /a 2 - y 2 /b 2 = 1 at its point Af, (*,; #i).

558. Prove that the tangents to a hyperbola drawn at the ends of the same diameter are parallel.

559. Compose equations of tangents to the hyperbola x 2 /20 - y 2 /5 = 1, perpendicular to the line 4x + 3y - 7 = 0.

560. Compose equations of tangents to the hyperbola x 2 /16 - y 2 /64 = 1, parallel to the line 10x - 3y + 9 = 0.

561. Draw tangents to the hyperbola x 2 /16 - y 2 /8 = - 1 parallel to the straight line 2x + 4y - 5 = 0 and calculate the distance d between them.

562. On the hyperbola x 2 /24 - y 2 /18 = 1, find the point M 1 closest to the line 3x + 2y + 1 = O, and calculate the distance d from the point M x to this line.

563. Create an equation for the tangents to the hyperbola x 2 - y 2 = 16 drawn from point A(- 1; -7).

564. From point C(1;-10) tangents are drawn to the hyperbola x 2 /8 - y 2 /32 = 1. Create an equation for the chord connecting the points of tangency.

565. From point P(1; -5) tangents are drawn to the hyperbola x 2 /3 - y 2 /5 = 1. Calculate the distance d from point P to the chord of the hyperbola connecting the points of tangency.

566. A hyperbola passes through point A(√6; 3) and touches the line 9x + 2y - 15 == 0. Write an equation for this hyperbola, provided that its axes coincide with the coordinate axes.

567. Write an equation for a hyperbola tangent to two straight lines: 5x - 6y - 16 = 0, 13x - 10y - 48 = 0, provided that its axes coincide with the coordinate axes.

568. Having made sure that the intersection points of the ellipse x 2 /3 - y 2 /5 = 1 and the hyperbola x 2 /12 - y 2 /3 = 1 are the vertices of the rectangle, compose the equations of its sides.

569. Given hyperbolas x 2 /a 2 - y 2 /b 2 = 1 and some of its tangents: P is the point of intersection of the tangent with the Ox axis, Q is the projection of the point of tangency onto the same axis. Prove that OP OQ = a 2 .

570. Prove that the foci of a hyperbola are located along different sides from any of its tangents.

571. Prove that the product of the distances from the foci to any tangent to the hyperbola x 2 /a 2 - y 2 /b 2 = 1 is a constant value equal to b 2.

572. The straight line 2x - y - 4 == 0 touches the hyperbola, the foci of which are at points F 1 (-3; 0) and F 2 (3; 0). Write an equation for this hyperbola.

573. Compose an equation of a hyperbola, the foci of which are located on the x-axis symmetrically with respect to the origin, if the equation of the tangent to the hyperbola is known 15x + 16y - 36 = 0 and the distance between its vertices is 2a = 8.

574. Prove that the line tangent to the hyperbola at some point M is equal angles with focal radii F 1 M, F 2 M and passes inside the corner F 1 MF 2. X^

575. From the right focus of the hyperbola x 2 /5 - y 2 /4 = 1 at angle α(π

576. Prove that an ellipse and a hyperbola, having common foci, intersect at right angles.

577. The coefficient of uniform compression of the plane to the Ox axis is equal to 4/3. Determine the equation of the line into which the hyperbola x 2 /16 - y 2 /9 = 1 is transformed during this compression. Hint. See problem 509.

578. The coefficient of uniform compression of the plane to the Oy axis is equal to 4/5. Determine the equation of the line into which the hyperbola x 2 /25 - y 2 /9 = 1 is transformed during this compression.

579. Find the equation of the line into which the hyperbola x 2 - y 2 = 9 is transformed under two successive uniform compressions of the plane to the coordinate axes, if the coefficients of uniform compression of the plane to the Ox and Oy axes are respectively equal to 2/3 and 5/3.

580. Determine the coefficient q of uniform compression of the plane to the Ox axis, at which the hyperbola - x 2 /25 - y 2 /36 = 1 is transformed into the hyperbola x 2 /25 - y 2 /16 = 1.

581. Determine the coefficient q of uniform compression of the plane to the Oy axis, at which the hyperbola x 2 /4 - y 2 /9 = 1 is transformed into the hyperbola x 2 /16 - y 2 /9 = 1.

582. Determine the coefficients q 1 and q 2 of two successive uniform compressions of the plane to the Ox and Oy axes, at which the hyperbola x 2 /49 - y 2 /16 = 1 is transformed into the hyperbola x 2 /25 - y 2 /64 = 1.

A hyperbola is a set of points on a plane whose distances differ from two given points, foci, is a constant value and is equal to .

Similarly to the ellipse, we place the foci at points , (see Fig. 1).

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It is known that in a triangle the difference between two sides is less than the third side, so, for example, with we get:

Let’s bring both sides to the square and after further transformations we find:

Where . The hyperbola equation (1) is canonical hyperbola equation.

The hyperbola is symmetrical with respect to the coordinate axes, therefore, as for the ellipse, it is enough to plot its graph in the first quarter, where:

Range of values for the first quarter.

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Form and characteristics of a hyperbola

Let us examine equation (1) the shape and location of the hyperbola.

Variables and are included in equation (1) in pair powers. Therefore, if a point belongs to a hyperbola, then the points also belong to a hyperbola. This means that the figure is symmetrical about the and axes and the point, which is called the center of the hyperbola.
Let's find the points of intersection with the coordinate axes. Substituting into equation (1) we find that the hyperbola intersects the axis at points . Putting it, we get an equation that has no solutions. This means that the hyperbola does not intersect the axis. The points are called vertices of the hyperbola. The segment = and is called the real axis of the hyperbola, and the segment is called the imaginary axis of the hyperbola. The numbers and are called the real and imaginary semi-axes of the hyperbola, respectively. The rectangle created by the axes is called the principal rectangle of the hyperbola.
From equation (1) it turns out that , that is . This means that all points of the hyperbola are located to the right of the line (right branch of the hyperbola) and to the left of the line (left branch of the hyperbola).
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Asymptotes of a hyperbola

There are two asymptotes of a hyperbola. Let's find the asymptote to the branch of the hyperbola in the first quarter, and then use the symmetry. Consider the point in the first quarter, that is. In this case, , then the asymptote has the form: , where

This means that the straight line is the asymptote of the function. Therefore, due to symmetry, the asymptotes of a hyperbola are straight lines.

Using the established characteristics, we will construct a branch of the hyperbola, which is located in the first quarter, and use the symmetry:

Rice. 2

In the case when , that is, the hyperbola is described by the equation. In this hyperbola there are asymptotes, which are bisectors coordinate angles.

Examples of problems on constructing a hyperbola

Let's move on to examples.

Task

Find the axes, vertices, foci, eccentricity and equations of asymptotes of the hyperbola. Construct a hyperbola and its asymptotes.

Solution

Let's reduce the hyperbola equation to canonical form:

Comparing this equation with the canonical one (1) we find , , . Peaks, focuses and . Eccentricity; asptotes; We are building a parabola. (see Fig. 3)

Write the equation of the hyperbola:

Solution

By writing the asymptote equation in the form we find the ratio of the semi-axes of the hyperbola. According to the conditions of the problem, it follows that . Therefore, the Problem was reduced to solving a system of equations:

Substituting into the second equation of the system, we get:

where . Now we find it.

Therefore, the hyperbola has the following equation:

Answer

Hyperbola and its canonical equation updated: June 17, 2017 by: Scientific Articles.Ru

Definition . A hyperbola is the 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). Let us denote F 1 F 2 = 2c. Then F 1 (c; 0); F 2 (-c; 0)

MF 2 = r 2 , MF 1 = r 1 – focal radii of the hyperbola.

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 hyperbola equation

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) also lie on it. -y 0) M 3 (-x 0; -y 0).

Therefore, the hyperbola is symmetrical about both coordinate axes.

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

. Due to symmetry, we conduct research in the first quarter

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

2) for x = a; y = 0 A 1 (a; 0) belongs to the 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

Let us consider together with the equation
equation of a line

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

We'll find

So, if point M, moving along a 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 to which at
The curve approaches indefinitely and is 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 oy axis and is equal to 2b, and the center lies at the origin of coordinates (Fig. 32).

P 7. Eccentricity and directrixes of a hyperbola

r 2 – r 1 = ± 2a the + 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

Let us express the focal radii of the hyperbola in terms of eccentricity:

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

T
as for hyperbole
therefore, 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 value and equal to ε.

P. 8 Parabola and its equation

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

To compose the equation of a parabola, we take as the x axis a straight line passing through the focus F 1 perpendicular to the directrix and we assume that the x axis is directed from the directrix to the focus. For the origin of coordinates we take the middle O of the segment from point F to this straight line, the length of which we denote by p (Fig. 34). We will call the value p the parameter of the parabola. Focus coordinate point
.

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

According to definition

at 2 = 2рх – canonical equation of a 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 oh. The equation y 2 = -2px with positive p is reduced to the equation y 2 = 2px by replacing x with –x and its graph looks like (Fig. 35).

U
The equation x2 = 2py is the equation of a parabola with a vertex at point O (0; 0) whose branches are directed upward.

X
2 = -2ру – equation of a parabola with a center at the origin, symmetrical about the y-axis, the branches of which are directed downward (Fig. 36).

A parabola has one axis of symmetry.

If x is to the first power and y is to the second, 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.

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

Note 2. Since for a parabola , Thatε parabola is equal to 1.ε = 1 .

Hello, dear Argemona University students! Welcome to another lecture on the magic of functions and integrals.

Today we will talk about hyperbole. Let's start simple. The simplest type of hyperbole is:

This function, unlike the straight line in its standard forms, has a special feature. As we know, the denominator of a fraction cannot be zero, because you cannot divide by zero.
x ≠ 0
From here we conclude that the domain of definition is the entire number line, except for point 0: (-∞; 0) ∪ (0; +∞).

If x tends to 0 from the right (written like this: x->0+), i.e. becomes very, very small, but remains positive, then y becomes very, very large positive (y->+∞).
If x tends to 0 from the left (x->0-), i.e. becomes very, very small in absolute value, but remains negative, then y will also be negative, but in absolute value it will be very large (y->-∞).
If x tends to plus infinity (x->+∞), i.e. becomes a very large positive number, then y will become an increasingly smaller positive number, i.e. will tend to 0, remaining positive all the time (y->0+).
If x tends to minus infinity (x->-∞), i.e. becomes large in modulus, but a negative number, then y will also always be a negative number, but small in modulus (y->0-).

Y, like x, cannot take the value 0. It only tends to zero. Therefore, the set of values is the same as the domain of definition: (-∞; 0) ∪ (0; +∞).

Based on these considerations, we can schematically draw a graph of the function

It can be seen that the hyperbola consists of two parts: one is located in the 1st coordinate angle, where the x and y values are positive, and the second part is in the third coordinate angle, where the x and y values are negative.
If we move from -∞ to +∞, then we see that our function decreases from 0 to -∞, then there is a sharp jump (from -∞ to +∞) and the second branch of the function begins, which also decreases, but from +∞ to 0. That is, this hyperbole is decreasing.

If you change the function just a little: use the magic of minus,

(1")

Then the function will miraculously move from 1 to 3 coordinate quarters in the 2nd and 4th quarters and will become increasing.

Let me remind you that the function is increasing, if for two values x 1 and x 2 such that x 1<х 2 , значения функции находятся в том же отношении f(х 1) < f(х 2).
And the function will be decreasing, if f(x 1) > f(x 2) for the same values of x.

The branches of the hyperbola approach the axes, but never intersect them. Lines that the graph of a function approaches but never intersects are called asymptote this function.
For our function (1), the asymptotes are the straight lines x=0 (OY axis, vertical asymptote) and y=0 (OX axis, horizontal asymptote).

Now let's complicate the simplest hyperbola a little and see what happens to the graph of the function.

(2)

We just added the constant “a” to the denominator. Adding a number to the denominator as a term to x means moving the entire “hyperbolic construction” (along with the vertical asymptote) to (-a) positions to the right, if a - a negative number, and to (a) positions to the left, if a — positive number.

On the left graph, a negative constant is added to x (a<0, значит, -a>0), which causes the graph to move to the right, and on the right graph there is a positive constant (a>0), due to which the graph is moved to the left.

And what magic can affect the transfer of the “hyperbolic structure” up or down? Adding a constant term to a fraction.

(3)

Now our entire function (both branches and the horizontal asymptote) will rise b positions up if b is a positive number, and will go down b positions if b is a negative number.

Please note that the asymptotes move along with the hyperbola, i.e. the hyperbola (both of its branches) and both of its asymptotes must necessarily be considered as an inseparable structure that moves uniformly to the left, right, up or down. It’s a very pleasant feeling when by just adding a number you can make the entire function move in any direction. Isn’t it magic that you can master very easily and direct it at your discretion in the right direction?
By the way, you can control the movement of any function this way. In the next lessons we will consolidate this skill.

Before asking you homework, I want to draw your attention to this function

(4)

The lower branch of the hyperbola moves from the 3rd coordinate angle upward - to the second, to the angle where the value of y is positive, i.e. this branch is reflected symmetrically relative to the OX axis. And now we get an even function.

What means " even function"? The function is called even, if the condition is met: f(-x)=f(x)
The function is called odd, if the condition is met: f(-x)=-f(x)
In our case

(5)

Every even function is symmetrical about the OY axis, i.e. a parchment with a drawing of a graph can be folded along the OY axis, and the two parts of the graph will exactly coincide with each other.

As we can see, this function also has two asymptotes - horizontal and vertical. Unlike the functions discussed above, this function is increasing on one part and decreasing on the other.

Let's now try to manipulate this graph by adding constants.

(6)

Recall that adding a constant as a term to “x” causes the entire graph (along with the vertical asymptote) to move horizontally, along the horizontal asymptote (to the left or to the right, depending on the sign of this constant).

(7)

And adding the constant b as a term to a fraction causes the graph to move up or down. Everything is very simple!

Now try experimenting with this magic yourself.

Homework 1.

Everyone takes two functions for their experiments: (3) and (7).
a=the first digit of your LD
b=second digit of your LD
Try to get to the magic of these functions, starting with the simplest hyperbola, as I did in the lesson, and gradually adding your own constants. You can already model function (7) based on the final form of function (3). Indicate the domains of definition, the set of values, and asymptotes. How functions behave: decrease, increase. Even odd. In general, try to do the same research as you did in class. Perhaps you will find something else that I forgot to talk about.

By the way, both branches of the simplest hyperbola (1) are symmetrical with respect to the bisector of the 2nd and 4th coordinate angles. Now imagine that the hyperbola began to rotate around this axis. Let's get such a nice figure that can be used.

Task 2. Where can this figure be used? Try to draw a rotation figure for function (4) relative to its axis of symmetry and think about where such a figure could find application.

Remember how at the end of the last lesson we got a straight line with a punctured point? And here's the last one task 3.
Construct a graph of this function:

(8)

Coefficients a, b are the same as in task 1.
c=third digit of your LD or a-b if your LD is two-digit.
A little hint: first, the fraction obtained after substituting the numbers must be simplified, and then you will get an ordinary hyperbola, which you need to construct, but in the end you must take into account the domain of definition of the original expression.