Topic: Straight circular cone. Section of a cone by planes

TEXT CODE OF THE LESSON:

We continue to study the section of stereometry "Solids of revolution".

The bodies of revolution include: cylinders, cones, balls.

Let's remember the definitions.

Height is the distance from the top of the shape or body to the base of the shape (body). Otherwise - a line segment connecting the top and bottom of the figure and perpendicular to it.

Remember, to find the area of \u200b\u200ba circle, you need to multiply pi by the square of the radius.

The area of \u200b\u200bthe circle is.

Let's remember how to find the area of \u200b\u200ba circle, knowing the diameter? Because

substitute in the formula:

The cone is also a body of revolution.

A cone (more precisely, a circular cone) is a body that consists of a circle - the base of the cone, a point not lying in the plane of this circle - the vertex of the cone and all segments connecting the vertex of the cone with the base points.

Let's get acquainted with the formula for finding the volume of a cone.

Theorem. The volume of the cone is equal to one third of the product of the base area and the height.

Let us prove this theorem.

Given: cone, S - area of \u200b\u200bits base,

h - cone height

Prove: V \u003d

Proof: Consider a cone of volume V, base radius R, height h, and apex at point O.

Let's introduce the axis Оx through ОМ - the axis of the cone. An arbitrary section of the cone by a plane perpendicular to the Ox axis is a circle centered at the point

M1 - the point of intersection of this plane with the Ox axis. Let's denote the radius of this circle through R1, and the sectional area through S (x), where x is the abscissa of the point M1.

From the similarity of right-angled triangles ОМ1A1 and ОМА (ے ОМ1A1 \u003d ے ОМА - straight lines, ے MOA-common, therefore, triangles are similar in two angles) it follows that

The figure shows that ОМ1 \u003d х, OM \u003d h

or whence, by the property of proportion, we find R1 \u003d.

Since the section is a circle, then S (x) \u003d πR12, substitute the previous expression instead of R1, the sectional area is equal to the ratio of the product pi er of the square by the square x to the square of the height:

Let's apply the basic formula

calculating the volumes of bodies, for a \u003d 0, b \u003d h, we obtain the expression (1)

Since the base of the cone is a circle, the area S of the base of the cone will be equal to pi er square

in the formula for calculating the volume of a body, we replace the value of pi er square with the area of \u200b\u200bthe base and we obtain that the volume of the cone is equal to one third of the product of the area of \u200b\u200bthe base by the height

The theorem is proved.

Corollary from the theorem (formula for the volume of a truncated cone)

The volume V of the truncated cone, the height of which is h, and the areas of the bases S and S1, is calculated by the formula

Ve is equal to one-third ash multiplied by the sum of the areas of the bases and the square root of the product of the areas of the base.

Solving problems

A rectangular triangle with 3 cm and 4 cm legs rotates around the hypotenuse. Determine the volume of the resulting body.

By rotating the triangle around the hypotenuse, we get a cone. When solving this problem, it is important to understand that there are two possible cases. In each of them, we apply the formula to find the volume of the cone: the volume of the cone is equal to one third of the product of the base and the height

In the first case, the figure will look like this: a cone is given. Let the radius r \u003d 4, height h \u003d 3

The base area is π times the square of the radius

Then the volume of the cone is equal to one third of the product of π by the square of the radius and by the height.

We substitute the value in the formula, it turns out that the volume of the cone is 16π.

In the second case, like this: a cone is given. Let the radius r \u003d 3, height h \u003d 4

The volume of the cone is equal to one third of the product of the base area by the height:

The base area is equal to π times the square of the radius:

Then the volume of the cone is equal to one third of the product of π by the square of the radius and the height:

Substitute the value in the formula, it turns out that the volume of the cone is 12π.

Answer: The volume of the cone V is 16 π or 12 π

Problem 2. Given a straight circular cone with a radius of 6 cm, angle ВСО \u003d 45.

Find the volume of the cone.

Solution: A finished drawing is given for this task.

Let's write down the formula for finding the volume of the cone:

Let's express it through the base radius R:

We find h \u003d BO by construction, - rectangular, since angle BOC \u003d 90 (the sum of the angles of the triangle), the angles at the base are equal, so the triangle ΔBOC is isosceles and BO \u003d OC \u003d 6 cm.

The diagnostic work consists of two parts, including 19 tasks. Part 1 contains 8 tasks of a basic level of difficulty with a short answer. Part 2 contains 4 tasks of an increased difficulty level with a short answer and 7 tasks of an increased and high difficulty level with a detailed answer.
Diagnostic work in mathematics is given for 3 hours 55 minutes (235 minutes).
Answers to tasks 1-12 are written as an integer or final decimal fraction. Write the numbers in the answer fields in the text of the work, and then transfer them to answer form No. 1. When completing tasks 13-19, you need to write down the complete solution and answer in answer form No. 2.
All forms are filled in with bright black ink. The use of gel, capillary or fountain pens is allowed.
When completing assignments, you can use the draft. Draft entries do not count towards grading work.
The points received by you for completed tasks are summed up.
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Problem conditions

Find if
To obtain an enlarged image of a light bulb on the screen, a collecting lens with a main focal length \u003d 30 cm is used in the laboratory. from lens to screen - within the range from 75 to 100 cm. The image on the screen will be clear if the ratio is met. Specify at what maximum distance from the lens you can place the light bulb so that its image on the screen is clear. Express your answer in centimeters.
The motor ship goes along the river to its destination 300 km and after stopping returns to the point of departure. Find the speed of the current, if the speed of the ship in still water is 15 km / h, the stay lasts 5 hours, and the ship returns to the point of departure 50 hours after leaving it. Give your answer in km / h.
Find the smallest function value on the segment
a) Solve the equation b) Find all the roots of this equation belonging to the segment
Given a straight circular cone with apex M... The axial section of the cone is a triangle with an angle of 120 ° at the apex M... The generatrix of the cone is. Through point M the section of the cone is drawn, perpendicular to one of the generators.
a) Prove that the resulting triangle in the section is obtuse.
b) Find the distance from the center ABOUT the base of the cone to the section plane.
Solve the equation
Circle with center ABOUTtouches the side ABisosceles triangle ABC,side extensions ASand continuing the foundation Sunat the point N... Dot M- middle of the base Sun.
a) Prove that MN \u003d AC.
b) Find OS,if the sides of the triangle ABCare 5, 5 and 8.
Business project "A" assumes an increase in the amount invested in it by 34.56% annually during the first two years and by 44% annually over the next two years. Project "B" assumes growth by a constant integer n percent annually. Find the smallest value n, in which in the first four years project "B" will be more profitable than project "A".
Find all values \u200b\u200bof the parameter,, for each of which the system of equations has the only solution
Anya plays a game: two different natural numbers are written on the board and both are less than 1000. If both are natural, then Anya makes a move - replaces the previous ones with these two numbers. If at least one of these numbers is not natural, then the game is over.
a) Can the game continue for exactly three moves?
b) Are there two initial numbers such that the game will last at least 9 moves?
c) Anya made the first move in the game. Find the largest possible ratio of the product of the two numbers obtained to the product

Introduction

Relevance of the research topic.Conical sections were already known to the mathematicians of Ancient Greece (for example, Menechmu, 4th century BC); with the help of these curves, some construction problems (doubling a cube, etc.) were solved, which turned out to be inaccessible when using the simplest drawing tools - a compass and a ruler. In the first studies that have come down to us, Greek geometers obtained conical sections by drawing a cutting plane perpendicular to one of the generatrices, while, depending on the opening angle at the apex of the cone (i.e., the largest angle between the generatrices of one cavity), the intersection line turned out to be an ellipse, if this angle is acute, a parabola, if it is straight, and hyperbola, if it is obtuse. The most complete work on these curves was the "Conical Sections" by Apollonius of Perga (about 200 BC). Further advances in the theory of conic sections are associated with the creation in the 17th century. new geometric methods: projective (French mathematicians J. Desargues, B. Pascal) and, in particular, coordinate (French mathematicians R. Descartes, P. Fermat).

Interest in conical sections has always been supported by the fact that these curves are often found in various natural phenomena and in human activity. In science, conical sections acquired special significance after the German astronomer I. Kepler discovered from observations, and the English scientist I. Newton theoretically substantiated the laws of planetary motion, one of which claims that the planets and comets of the solar system move along conical sections, in one of which the Sun is located. The following examples relate to individual types of conical sections: a parabola is described by a projectile or a stone thrown obliquely to the horizon (the correct shape of the curve is somewhat distorted by air resistance); some mechanisms use elliptical cogwheels ("elliptical cogwheels"); hyperbole serves as a graph of inverse proportionality, often observed in nature (for example, Boyle's law - Marriott).

Objective:

Study of the theory of conic sections.

Research topic:

Conical sections.

Purpose of the study:

Theoretically study the features of conic sections.

Object of study:

Conical sections.

Subject of study:

Historical development of conic sections.

1. Formation of conic sections and their types

Tapered sections are lines that are formed in the section of a straight circular cone with different planes.

Note that a conical surface is a surface formed by the motion of a straight line passing all the time through a fixed point (the top of the cone) and intersecting all the time a fixed curve - a guide (in our case, a circle).

By classifying these lines by the nature of the location of the secant planes relative to the generatrices of the cone, curves of three types are obtained:

I. Curves formed by the section of a cone by planes not parallel to any of the generators. These curves will be various circles and ellipses. These curves are called elliptical curves.

II. Curves formed by the section of the cone by planes, each of which is parallel to one of the generatrices of the cone (Fig. 1 b). Only parabolas will be such curves.

III. Curves formed by the section of the cone by planes, each of which is parallel to some two generators (Fig. 1 c). such curves are hyperboles.

There can no longer be any type IV curves, since there cannot be a plane parallel to three generators of the cone at once, since no three generators of the cone themselves already lie in the same plane.

Note that the cone can be crossed by planes in such a way that two straight lines are obtained in the section. For this, the secant planes must be drawn through the apex of the cone.

2. Ellipse

Two theorems are important for studying the properties of conic sections:

Theorem 1. Let a straight circular cone be given, which is cut by planes b 1, b 2, b 3, perpendicular to its axis. Then all the segments of the generatrices of the cone between any pair of circles (obtained in the section with the given planes) are equal to each other, i.e. A 1 B 1 \u003d A 2 B 2 \u003d etc. and B 1 C 1 \u003d B 2 C 2 \u003d etc. Theorem 2. If a spherical surface is given and some point S outside it, then the segments of the tangents drawn from the point S to the spherical surface will be equal to each other, that is, SA 1 \u003d SA 2 \u003d SA 3, etc.

2.1 The main property of an ellipse

We dissect a straight circular cone with a plane intersecting all its generators. In section we get an ellipse. Draw a plane through the axis of the cone, perpendicular to the plane.

Let us inscribe two balls into the cone so that, being located on opposite sides of the plane and touching the conical surface, each of them touches the plane at some point.

Let one ball touch the plane at the point F 1 and touch the cone along the circle С 1, and the other - at the point F 2 and touch the cone around the circle С 2.

Take an arbitrary point P on the ellipse.

This means that all conclusions drawn about it will be valid for any point of the ellipse. Draw the generator of the OP of the cone and mark the points R 1 and R 2 at which it touches the constructed balls.

Let's connect point P with points F 1 and F 2. Then РF 1 \u003d РR 1 and РF 2 \u003d РR 2, since РF 1, РR 1 are tangents drawn from point Р to one ball, and РF 2, РR 2 are tangents drawn from point Р to another ball (Theorem 2 ). Adding both equalities term by term, we find

РF 1 + РF 2 \u003d РR 1 + РR 2 \u003d R 1 R 2 (1)

This relationship shows that the sum of the distances (РF 1 and РF 2) of an arbitrary point P of the ellipse to two points F 1 and F 2 is a constant value for a given ellipse (i.e., it does not depend on the position of point P on the ellipse).

Points F 1 and F 2 are called the focal points of the ellipse. The points at which the line F 1 F 2 intersects the ellipse are called the vertices of the ellipse. The segment between the vertices is called the major axis of the ellipse.

The length of the generatrix R 1 R 2 is equal to the major axis of the ellipse. Then the main property of the ellipse is formulated as follows: the sum of the distances of an arbitrary point P of the ellipse to its foci F 1 and F 2 is a constant value for a given ellipse, equal to the length of its major axis.

Note that if the foci of the ellipse coincide, then the ellipse is a circle, i.e. a circle is a special case of an ellipse.

2.2 Equation of an ellipse

To form the equation of an ellipse, we must consider the ellipse as a locus of points that have some property that characterizes this locus. Let's take the main property of an ellipse for its definition: An ellipse is a locus of points on a plane for which the sum of the distances to two fixed points F 1 and F 2 of this plane, called foci, is a constant value equal to the length of its major axis.

Let the length of the segment F 1 F 2 \u003d 2c, and the length of the major axis is 2a. To derive the canonical equation of the ellipse, we choose the origin O of the Cartesian coordinate system in the middle of the segment F 1 F 2, and direct the axes Ox and Oy as shown in Figure 5. (If the foci coincide, then O coincides with F 1 and F 2, and for the Ox axis, you can take any axis passing through O). Then in the selected coordinate system the points F 1 (s, 0) and F 2 (-s, 0). Obviously, 2a\u003e 2c, i.e. a\u003e c. Let M (x, y) be a point in the plane belonging to an ellipse. Let МF 1 \u003d r 1, МF 2 \u003d r 2. According to the definition of an ellipse, the equality

r 1 + r 2 \u003d 2a (2) is a necessary and sufficient condition for the location of the point M (x, y) on a given ellipse. Using the formula for the distance between two points, we get

r 1 \u003d, r 2 \u003d. Let's return to equality (2):

Let's move one root to the right side of the equality and square it:

Reducing, we get:

We give similar ones, reduce them by 4 and isolate the radical:

Squaring

Expand the brackets and reduce to:

whence we get:

(a 2 -c 2) x 2 + a 2 y 2 \u003d a 2 (a 2 -c 2). (3)

Note that a 2 -c 2\u003e 0. Indeed, r 1 + r 2 is the sum of the two sides of the triangle F 1 MF 2, and F 1 F 2 is its third side. Therefore, r 1 + r 2\u003e F 1 F 2, or 2а\u003e 2с, i.e. a\u003e c. Let us denote a 2 -c 2 \u003d b 2. Equation (3) will have the form: b 2 x 2 + a 2 y 2 \u003d a 2 b 2. Let's perform a transformation that brings the equation of the ellipse to the canonical (literally: taken as a sample) form, namely, divide both sides of the equation by a 2 b 2:

(4) - the canonical equation of the ellipse.

Since equation (4) is an algebraic consequence of equation (2 *), the coordinates x and y of any point M of the ellipse will also satisfy equation (4). Since during algebraic transformations associated with getting rid of radicals, "extra roots" could appear, it is necessary to make sure that any point M whose coordinates satisfy Eq. (4) is located on this ellipse. To do this, it suffices to prove that the values \u200b\u200br 1 and r 2 for each point satisfy relation (2). So, let the coordinates x and y of the point M satisfy equation (4). Substituting the value у 2 from (4) into the expression r 1, after simple transformations we find that r 1 \u003d. Since, then r 1 \u003d. In a completely similar way, we find that r 2 \u003d. Thus, for the considered point М r 1 \u003d, r 2 \u003d, i.e. r 1 + r 2 \u003d 2a, so point M is located on the ellipse. The quantities a and b are called, respectively, the major and minor semiaxes of the ellipse.

2.3 Study of the shape of an ellipse by its equation

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

1. Equation (4) contains x and y only in even powers, therefore if a point (x, y) belongs to an ellipse, then it also contains points (x, - y), (-x, y), (-x, - y). It follows that the ellipse is symmetric about the Ox and Oy axes, as well as about the point O (0,0), which is called the center of the ellipse.

2. Find the points of intersection of the ellipse with the coordinate axes. Putting y \u003d 0, we find two points A 1 (a, 0) and A 2 (-a, 0), at which the Ox axis intersects the ellipse. Putting in equation (4) x \u003d 0, we find the points of intersection of the ellipse with the axis Oy: B 1 (0, b) and. B 2 (0, - b) Points A 1, A 2, B 1, B 2 are called the vertices of the ellipse.

3. From equation (4) it follows that each term on the left-hand side does not exceed unity, i.e. inequalities and or and take place. Therefore, all points of the ellipse lie inside the rectangle formed by straight lines,.

4. In equation (4) the sum of non-negative terms and is equal to one. Consequently, as one term increases, the other will decrease, i.e. if x increases, then y decreases and vice versa.

From the above it follows that the ellipse has the shape shown in Fig. 6 (oval closed curve).

Note that if a \u003d b, then equation (4) will take the form x 2 + y 2 \u003d a 2. This is the equation of the circle. An ellipse can be obtained from a circle with a radius a, if it is compressed in times along the Oy axis. With this compression, the point (x; y) goes to the point (x; y 1), where. Substituting circles in the equation, we get the equation of the ellipse:.

Let us introduce one more quantity that characterizes the shape of the ellipse.

The eccentricity of an ellipse is the ratio of the focal length 2c to the length 2a of its major axis.

Eccentricity is usually denoted by e: e \u003d Since c< a, то. Заметив, что c 2 = a 2 - b 2 , находим: , отсюда.

From the last equality, it is easy to obtain a geometric interpretation of the eccentricity of the ellipse. For very small numbers a and b are almost equal, that is, the ellipse is close to a circle. If it is close to one, then the number b is very small compared to the number a, and the ellipse is strongly elongated along the major axis. Thus, the eccentricity of the ellipse characterizes the measure of the elongation of the ellipse.

3. Hyperbola

3.1 The main property of hyperbola

Investigating the hyperbola using constructions similar to those used to study the ellipse, we find that the hyperbola has properties similar to those of the ellipse.

We dissect a straight circular cone by plane b intersecting both of its planes, i.e. parallel to its two generators. In the section, you get a hyperbola. Let us draw plane АSB through the ST axis of the cone, perpendicular to plane b.

Let us inscribe two balls into the cone - one into one of its cavities, the other into the other, so that each of them touches the conical surface and the secant plane. Let the first ball touch the plane b at point F 1 and touch the conical surface along the circle UґVґ. Let the second ball touch the plane b at point F 2 and touch the conical surface along the circle UV.

Let us choose an arbitrary point M. on the hyperbola. Draw through it the generator of the cone MS and mark the points d and D at which it touches the first and second balls. Let's connect the point M with the points F 1, F 2, which we will call the foci of the hyperbola. Then МF 1 \u003d Md, since both segments are tangent to the first ball, drawn from the point M. Similarly, МF 2 \u003d MD. Subtracting the second equality term by term, we find

MF 1 -MF 2 \u003d Md-MD \u003d dD,

where dD is a constant value (as a generatrix of a cone with bases UґVґ and UV), independent of the choice of the point M on the hyperbola. Let P and Q denote the points at which the line F 1 F 2 intersects the hyperbola. These points P and Q are called the vertices of the hyperbola. The segment PQ is called the real axis of the hyperbola. In the course of elementary geometry, it is proved that dD \u003d PQ. Therefore, MF 1 -MF 2 \u003d PQ.

If point M will be on that branch of the hyperbola, near which the focus F 1 is located, then MF 2 -MF 1 \u003d PQ. Then we finally get MF 1 -MF 2 \u003d PQ.

The modulus of the difference between the distances of an arbitrary point M of the hyperbola from its foci F 1 and F 2 is a constant value equal to the length of the real axis of the hyperbola.

3.2 Equation of hyperbola

Let's take the main property of a hyperbola as its definition: A hyperbola is a locus of points on a plane, for which the modulus of the difference between distances to two fixed points F 1 and F 2 of this plane, called foci, is a constant value equal to the length of its real axis.

Let the length of the segment F 1 F 2 \u003d 2с, and the length of the real axis is equal to 2а. To derive the canonical hyperbola equation, we choose the origin O of the Cartesian coordinate system in the middle of the segment F 1 F 2, and direct the axes Ox and Oy as shown in Figure 5. Then, in the selected coordinate system, the points F 1 (c, 0) and F 2 ( -c, 0). Obviously, 2a<2с, т.е. а<с. Пусть М (х, у) - точка плоскости, принадлежащая гиперболе. Пусть МF 1 =r 1 , МF 2 =r 2 . Согласно определению гиперболы равенство

r 1 -r 2 \u003d 2a (5) is a necessary and sufficient condition for the location of the point M (x, y) on a given hyperbola. Using the formula for the distance between two points, we get

r 1 \u003d, r 2 \u003d. Let's return to equality (5):

Squaring both sides of the equality

(x + c) 2 + y 2 \u003d 4a 2 ± 4a + (x-c) 2 + y 2

Reducing, we get:

2 xs \u003d 4a 2 ± 4a-2 xs

± 4a \u003d 4a 2 -4 xs

a 2 x 2 -2a 2 xc + a 2 c 2 + a 2 y 2 \u003d a 4 -2a 2 xc + x 2 c 2

x 2 (s 2 -a 2) - a 2 y 2 \u003d a 2 (s 2 -a 2) (6)

Note that with 2 -а 2\u003e 0. Let us denote c 2 -а 2 \u003d b 2. Equation (6) will have the form: b 2 x 2 -a 2 y 2 \u003d a 2 b 2. Let us perform a transformation that reduces the hyperbola equation to the canonical form, namely, we divide both sides of the equation by a 2 b 2: (7) - the canonical equation of the hyperbola, the quantities a and b are the real and imaginary semiaxes of the hyperbola, respectively.

We must make sure that equation (7), obtained by algebraic transformations of equation (5 *), has not acquired new roots. For this, it is sufficient to prove that for each point M, the coordinates x and y of which satisfy equation (7), the values \u200b\u200br 1 and r 2 satisfy relation (5). Carrying out arguments similar to those that were made when deriving the ellipse formula, we find the following expressions for r 1 and r 2:

Thus, for the considered point M we have r 1 -r 2 \u003d 2a, and therefore it is located on the hyperbola.

3.3 Study of the hyperbola equation

Now let us try to get an idea of \u200b\u200bthe location of the hyperbola on the basis of considering equation (7).
1. First of all, equation (7) shows that the hyperbola is symmetric with respect to both axes. This is due to the fact that the equation of the curve includes only even powers of the coordinates. 2. Let us now mark the area of \u200b\u200bthe plane where the curve will lie. The hyperbola equation resolved with respect to y has the form:

It shows that y always exists when x 2? a 2. This means that for x? a and for x? - and the ordinate y will be real, and for - a

Further, with x increasing (and greater than a), the y ordinate will also grow all the time (in particular, this shows that the curve cannot be wavy, i.e. such that with the growth of the abscissa x the ordinate y either increases or decreases) ...

H. The center of a hyperbola is a point relative to which each point of the hyperbola has a point symmetric to itself on it. The point O (0,0), the origin of coordinates, as for the ellipse, is the center of the hyperbola given by the canonical equation. This means that each point of the hyperbola has a symmetric point on the hyperbola with respect to the point O. This follows from the symmetry of the hyperbola with respect to the axes Ox and Oy. Any chord of a hyperbola passing through its center is called the diameter of the hyperbola.

4. The points of intersection of the hyperbola with the line on which its foci lie are called the vertices of the hyperbola, and the segment between them is called the real axis of the hyperbola. In this case, the real axis is the Ox axis. Note that the real axis of the hyperbola is often called both the segment 2a and the straight line itself (the Ox axis) on which it lies.

Let's find the points of intersection of the hyperbola with the Oy axis. The Oy axis equation has the form x \u003d 0. Substituting x \u003d 0 into equation (7), we obtain that the hyperbola has no points of intersection with the Oy axis. This is understandable, since there are no hyperbola points in the 2a-wide strip covering the Oy axis.

A straight line perpendicular to the real axis of the hyperbola and passing through its center is called the imaginary axis of the hyperbola. In this case, it coincides with the Oy axis. So, the denominators of the terms with x 2 and y 2 in the hyperbola equation (7) are the squares of the real and imaginary semiaxes of the hyperbola.

5. The hyperbola intersects with the line y \u003d kx for k< в двух точках. Если k то общих точек у прямой и гиперболы нет.

Evidence

To determine the coordinates of the intersection points of the hyperbola and the straight line y \u003d kx, you need to solve the system of equations

Eliminating y, we get

or When b 2 -k 2 a 2 0 that is, when k is the resulting equation, and therefore the system of solutions does not have.

Lines with equations y \u003d and y \u003d - are called hyperbola asymptotes.

For b 2 -k 2 a 2\u003e 0 that is, for k< система имеет два решения:

Therefore, each straight line passing through the origin with slope k< пересекает гиперболу в двух точках. При k = 0 получаем точки пересечения (a; 0) и (- a; 0) - вершины гиперболы.

6. Optical property of hyperbola: optical rays emanating from one focus of the hyperbola, reflected from it, seem to be emanating from the second focus.

The eccentricity of a hyperbola is the ratio of the focal length 2c to the length 2a of its real axis? \u003d Since c\u003e a, then e\u003e 1, then the foci of the hyperbola, as in the case of an ellipse, is inside the curve,
those. from the side of its concavity.

3.4 Conjugate hyperbola

Along with the hyperbola (7), the so-called hyperbola conjugate with respect to it is considered. The conjugate hyperbola is defined by the canonical equation.

In fig. 10 shows the hyperbola (7) and the associated hyperbola. The conjugate hyperbola has the same asymptotes as this one, but F 1 (0, c),

4. Parabola

4.1 The main property of a parabola

Let us establish the main properties of the parabola. We cut a straight circular cone with apex S by a plane parallel to one of its generators. In the section we get a parabola. Let us draw plane АSB through the ST axis of the cone, perpendicular to the plane (Fig. 11). The generatrix SA lying in it will be parallel to the plane. Let us inscribe into the cone a spherical surface tangent to the cone along the circle UV and tangent to the plane at point F. Draw through the point F a straight line parallel to the generator SA. Let us denote the point of its intersection with the generator SB by P. Point F is called the focus of the parabola, point P is its apex, and the line PF passing through the vertex and focus (and parallel to the generator SA) is called the parabola axis. The parabola will not have the second vertex - the point of intersection of the PF axis with the generatrix SA: this point “goes to infinity”. Let us call the directrix (translated as “guide”) the line q 1 q 2 of intersection of the plane with the plane in which the circle UV lies. Take an arbitrary point M on the parabola and connect it with the vertex of the cone S. Line MS touches the ball at point D lying on the circle UV. Connect point M with focus F and drop the perpendicular MK to the directrix from point M. Then it turns out that the distances of an arbitrary point M of the parabola to the focus (MF) and to the directrix (MK) are equal to each other (the main property of the parabola), i.e. MF \u003d MK.

Proof: МF \u003d MD (as tangents to the ball from one point). Let us denote the angle between any of the generatrices of the cone and the ST axis through c. Let's design the segments MD and MK onto the ST axis. The segment MD forms a projection on the ST axis, equal to MDcosc, since MD lies on the generatrix of the cone; the MK segment forms a projection on the ST axis, equal to MKsots, since the MK segment is parallel to the generatrix SA. (Indeed, the directrix q 1 q 1 is perpendicular to the plane ASB. Consequently, the straight line РF intersects the directrix at point L. At the right angle. But the lines MK and PF lie in the same plane, and MK is also perpendicular to the directrix). The projections of both segments MK and MD on the ST axis are equal to each other, since one of their ends - point M - is common, and the other two D and K lie in a plane perpendicular to the ST axis (Fig.). Then MDcosts \u003d MKsots or MD \u003d MK. Therefore, MF \u003d MK.

Property 1. (Focal property of a parabola).

The distance from any point of the parabola to the middle of the main chord is equal to its distance to the directrix.

Evidence.

Point F is the intersection point of the line QR and the main chord. This point lies on the axis of symmetry Oy. Indeed, triangles RNQ and ROF are equal, as rectangular

triangles with wound legs (NQ \u003d OF, OR \u003d RN). Therefore, no matter what point N we take, the line QR constructed from it will intersect the main chord in its middle F. Now it is clear that the triangle FMQ is isosceles. Indeed, the segment MR is both the median and the height of this triangle. Hence it follows that MF \u003d MQ.

Property 2. (Optical property of a parabola).

Any tangent to a parabola makes equal angles with a focal radius drawn to the point of tangency and a ray passing from the point of tangency and co-directional with the axis (or, rays leaving a single focus, reflected from the parabola, will go parallel to the axis).

Evidence. For a point N lying on the parabola itself, the equality | FN | \u003d | NH | is true, and for a point N "lying in the inner region of the parabola, | FN" |<|N"H"|. Если теперь провести биссектрису l угла FМК, то для любой отличной от М точки M" прямой l найдём:

| FM "| \u003d | M" K "|\u003e | M" K "|, that is, point M" lies in the outer region of the parabola. So, the whole line l, except for the point M, lies in the outer region, that is, the inner region of the parabola lies on one side of l, which means that l is tangent to the parabola. This gives proof of the optical property of the parabola: angle 1 is equal to angle 2, since l is the bisector of the angle FMK.

4.2 Parabola equation

Based on the main property of a parabola, let us formulate its definition: 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 \u200b\u200bcalled the parameter of the parabola and is denoted by p (p\u003e 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 (Fig. 12). In the chosen system, the focus is F (, 0), and the directrix equation has the form x \u003d -, or x + \u003d 0 Let m (x, y) be an arbitrary point of the parabola. Let's connect the point M with F. Let's draw the segment MH perpendicular to the directrix. According to the definition of the parabola MF \u003d MH. Using the formula for the distance between two points, we find:

Therefore, Squaring both sides of the equation, we get

those. (8) Equation (8) is called the canonical equation of the parabola.

4.3 Study of the shape of a parabola by its equation

1. In equation (8) 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 c\u003e 0, it follows from (8) that x\u003e 0. Consequently, the parabola is located to the right of the Oy axis.

3. Let x \u003d 0, then y \u003d 0. Therefore, the parabola passes through the origin.

4. With an unbounded increase in x, the module у also increases unboundedly. The parabola y 2 \u003d 2 px has the form (shape) shown in Figure 13. Point O (0; 0) is called the apex of the parabola, the segment FM \u003d r is called the focal radius of point M. Equations y 2 \u003d -2 px, x 2 \u003d - 2 py, x 2 \u003d 2 py (p\u003e 0) also define parabolas.

1.5. Directory property of conic sections .

Here we will prove that each non-circular (non-degenerate) conic section can be defined as a set of points M, the ratio of the distance MF of which from a fixed point F to the distance MP from a fixed straight line d not passing through the point F is equal to a constant value e: where F is the focus of the conical section, line d is the directrix, and the ratio e is the eccentricity. (If a point F belongs to a line d, then the condition defines a set of points, which is a pair of lines, that is, a degenerate conical section; for e \u003d 1, this pair of lines merges into one line. For the proof, consider the cone formed by the rotation of the line l around the intersecting its at point O of the straight line p, making angle b< 90є; пусть плоскость р не проходит через вершину конуса и образует с его осью p угол в < 90є (если в = 90є, то плоскость р пересекает конус по окружности).

We inscribe a ball K into the cone, tangent to the plane p at point F and tangent to the cone along the circle S. The line of intersection of the plane p with the plane at the circle S is denoted by d.

Now we connect an arbitrary point M lying on the line L of the intersection of the plane p and the cone with the vertex O of the cone and with the point F and drop from M the perpendicular MP to the line d; we also denote by E the intersection point of the generator MO of the cone with the circle S.

Moreover, MF \u003d ME, as the segments of two tangent lines of the ball K drawn from one point M.

Further, the segment ME forms a constant angle b with the axis p of the cone (that is, not depending on the choice of the point M), and the segment MP forms a constant angle b; therefore, the projections of these two segments on the p axis are respectively equal to ME cos b and MP cos c.

But these projections coincide, since the segments ME and MP have a common origin M, and their ends lie in the y plane, perpendicular to the p axis.

Therefore, ME cos b \u003d MP cos c, or, since ME \u003d MF, MF cos b \u003d MP cos c, whence it follows that

It is also easy to show that if the point M of the plane p does not belong to the cone, then. Thus, each section of a right circular cone can be described as a set of points of the plane for which. On the other hand, by changing the values \u200b\u200bof the angles b and c, we can give the eccentricity any value e\u003e 0; further, from considerations of similarity, it is easy to understand that the distance FQ from the focus to the directrix is \u200b\u200bdirectly proportional to the radius r of the ball K (or the distance d of the plane p from the vertex O of the cone). It can be shown that, thus, by choosing a suitable distance d, we can give the distance FQ any value. Therefore, each set of points M, for which the ratio of the distances from M to a fixed point F and to a fixed straight line d is constant, can be described as a curve obtained in the section of a straight circular cone by a plane. This proves that (non-degenerate) conic sections can also be defined by the property referred to in this subsection.

This property of conical sections is called them directory property... It is clear that if c\u003e b, then e< 1; если в = б, то е = 1; наконец, если в < б, то е > 1. On the other hand, it is easy to see that if c\u003e b, then the plane p intersects the cone along a closed bounded line; if c \u003d b, then the plane p intersects the cone along an unbounded line; if in< б, то плоскость р пересекает обе полы конуса и, следовательно, линия пересечения этой плоскости и конуса состоит из двух (неограниченных) частей или ветвей (рис. 17).

Conical section for which e< 1, называется эллипсом; коническое сечение с эксцентриситетом е = 1 называется параболой; коническое сечение, для которого е > 1 is called a hyperbola. Ellipses also include a circle, which cannot be specified by a directory property; since for a circle the ratio turns to 0 (since in this case b \u003d 90є), it is conventionally considered that the circle is a conical section with an eccentricity of 0.

6. Ellipse, hyperbola and parabola as conical sections

conical section ellipse hyperbola

The ancient Greek mathematician Menekhm, who discovered the ellipse, hyperbola and parabola, defined them as sections of a circular cone by a plane perpendicular to one of the generators. He called the resulting curves sections of acute-angled, rectangular and obtuse-angled cones, depending on the axial angle of the cone. The first, as we will see below, is an ellipse, the second is a parabola, and the third is one branch of a hyperbola. The names "ellipse", "hyperbola" and "parabola" were introduced by Apollonius. Almost completely (7 out of 8 books) of Apollonius's composition "On conical sections" has come down to us. In this work, Apollonius examines both sides of the cone and intersects the cone with planes not necessarily perpendicular to one of the generatrices.

Theorem. The section of any straight round cone by a plane (not passing through its vertex) defines a curve that can only be a hyperbola (Fig. 4), a parabola (Fig. 5) or an ellipse (Fig. 6). Moreover, if the plane intersects only one plane of the cone and along a closed curve, then this curve is an ellipse; if the plane intersects only one plane along an open curve, then this curve is a parabola; if the cutting plane intersects both planes of the cone, then a hyperbola is formed in the section.

An elegant proof of this theorem was proposed in 1822 by Dandelen using spheres, which are now commonly called Dandelen spheres. Consider this proof.

Let us inscribe into the cone two spheres tangent to the plane of the section П from different sides. Let F1 and F2 denote the tangency points of this plane with the spheres. Let us take an arbitrary point M. on the line of section of the cone by plane P. Note on the generatrix of the cone passing through M, points P1 and P2, lying on the circle k1 and k2, along which the spheres touch the cone.

It is clear that МF1 \u003d МР1 as segments of two tangents to the first sphere going out from М; similarly, МF2 \u003d МР2. Therefore, MF1 + MF2 \u003d MP1 + MP2 \u003d P1P2. The length of the segment P1P2 is the same for all points M of our section: this is the generatrix of the truncated cone bounded by parallel planes 1 and 11, in which the circles k1 and k2 lie. Consequently, the line of section of the cone by plane P is an ellipse with foci F1 and F2. The validity of this theorem can also be established proceeding from the general position that the intersection of a second-order surface by a plane is a second-order line.

Literature

1. Atanasyan L.S., Bazylev V.T. Geometry. In 2 hours. Part 1. Textbook for students of physics and mathematics. ped. in - comrade-M .: Education, 1986.

2. Bazylev V.T. and others. Geometry. Textbook. manual for 1st year students nat. - mat. facts - tov ped. in. - Comrade-M .: Education, 1974.

3. Pogorelov A.V. Geometry. Textbook. for 7-11 cl. Wednesday shk. - 4th ed.-M .: Education, 1993.

4. The history of mathematics from ancient times to the beginning of the 19th century. A.P. Yushkevich - Moscow: Nauka, 1970.

5. Boltyansky V.G. Optical properties of ellipse, hyperbola and parabola. // Quant. - 1975. - No. 12. - from. 19 - 23.

6. Efremov N.V. A short course in analytical geometry. - M: Science, 6th edition, 1967. - 267 p.

Topic: Straight circular cone. Section of a cone by planes

Problem conditions

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