Consider a line on the plane and a point not lying on this line. AND ellipse, And hyperbola can be defined in a unified way as the geometric locus of points for which the ratio of the distance to a given point to the distance to a given straight line is a constant value

rank ε. At 0 1 - hyperbola. The parameter ε is eccentricity of both ellipse and hyperbola. Of the possible positive values one parameter ε, namely ε = 1, turns out to be unused. This value corresponds to the geometric locus of points equidistant from a given point and from a given line.

Definition 8.1. The locus of points in a plane equidistant from a fixed point and from a fixed line is called parabola.

The fixed point is called focus of the parabola, and the straight line - directrix of a parabola. At the same time, it is believed that parabola eccentricity equal to one.

From geometric considerations it follows that the parabola is symmetrical with respect to the straight line perpendicular to the directrix and passing through the focus of the parabola. This straight line is called the axis of symmetry of the parabola or simply the axis of the parabola. A parabola intersects its axis of symmetry at a single point. This point is called the vertex of the parabola. It is located in the middle of the segment connecting the focus of the parabola with the point of intersection of its axis with the directrix (Fig. 8.3).

Parabola equation. To derive the equation of a parabola, we choose on the plane origin at the vertex of the parabola, as x-axis- the axis of the parabola, the positive direction on which is specified by the position of the focus (see Fig. 8.3). This coordinate system is called canonical for the parabola in question, and the corresponding variables are canonical.

Let us denote the distance from the focus to the directrix by p. He is called focal parameter of the parabola.

Then the focus has coordinates F(p/2; 0), and the directrix d is described by the equation x = - p/2. The locus of points M(x; y), equidistant from the point F and from the line d, is given by the equation

Let us square equation (8.2) and present similar ones. We get the equation

which is called canonical parabola equation.

Note that squaring in this case is an equivalent transformation of equation (8.2), since both sides of the equation are non-negative, as is the expression under the radical.

Type of parabola. If the parabola y 2 = x, the form of which we consider known, is compressed with a coefficient 1/(2р) along the abscissa axis, then we get a parabola general view, which is described by equation (8.3).

Example 8.2. Let us find the coordinates of the focus and the equation of the directrix of a parabola if it passes through a point whose canonical coordinates are (25; 10).

In canonical coordinates, the equation of the parabola has the form y 2 = 2px. Since the point (25; 10) is on the parabola, then 100 = 50p and therefore p = 2. Therefore, y 2 = 4x is the canonical equation of the parabola, x = - 1 is the equation of its directrix, and the focus is at the point (1; 0 ).

Optical property of a parabola. The parabola has the following optical property. If a light source is placed at the focus of the parabola, then all light rays after reflection from the parabola will be parallel to the axis of the parabola (Fig. 8.4). The optical property means that at any point M of the parabola normal vector the tangent makes equal angles with the focal radius MF and the abscissa axis.

Lectures on algebra and geometry. Semester 1.

Lecture 17. Parabola.

Chapter 17. Parabola.

clause 1. Basic definitions.

Definition. A parabola is the GMT of a plane equidistant from one fixed point of the plane, called the focus, and one fixed line, called the directrix.

Definition. The distance from an arbitrary point M of the plane to the focus of the parabola is called the focal radius of the point M.

Designations: F – focus of the parabola, r – focal radius points M,d– distance from point M to directrix D.

By the definition of a parabola, a point M is a point of a parabola if and only if
.

By the definition of a parabola, its focus and directrix are fixed objects, therefore the distance from the focus to the directrix is ​​a constant value for a given parabola.

Definition. The distance from the focus of a parabola to its directrix is ​​called the focal parameter of the parabola.

Designation:
.

Let us introduce a coordinate system on this plane, which we will call canonical for the parabola.

Definition. The axis drawn through the focus of the parabola perpendicular to the directrix is ​​called the focal axis of the parabola.

Let's construct a canonical PDSC for the parabola, see Fig. 2.

As the abscissa axis, we select the focal axis, the direction on which we select from the directrix to the focus.

The ordinate axis is drawn through the middle of the segment FN perpendicular to the focal axis. Then the focus has coordinates
.

clause 2. Canonical equation of a parabola.

Theorem. In the coordinate system canonical for a parabola, the equation of the parabola has the form:

. (1)

Proof. We carry out the proof in two stages. At the first stage, we will prove that the coordinates of any point lying on the parabola satisfy equation (1). At the second stage we will prove that any solution to equation (1) gives the coordinates of a point lying on the parabola. It will follow from this that equation (1) is satisfied by the coordinates of those and only those points of the coordinate plane that lie on the parabola.

From this and from the definition of the equation of a curve it will follow that equation (1) is the equation of a parabola.

1) Let the point M(x, y) be a point of a parabola, i.e.

.

Let's use the formula for the distance between two points on the coordinate plane and use this formula to find the focal radius of a given point M:

.

From Figure 2 we see that a parabola point cannot have a negative abscissa, because in this case
. That's why
And
. From here we get the equality

.

Let's square both sides of the equation:

and after reduction we get:

.

2) Let now a pair of numbers (x, y) satisfy equation (1) and let M(x, y) be the corresponding point on the coordinate plane Oxy.

Then we substitute equality (1) into the expression for the focal radius of point M:

, from which, by the definition of a parabola, it follows that the point M(x, y) lies on the parabola.

Here we took advantage of the fact that from equality (1) it follows that
and therefore
.

The theorem is proven.

Definition. Equation (1) is called the canonical parabola equation.

Definition. The origin of the canonical coordinate system for a parabola is called the vertex of the parabola.

clause 3. Properties of a parabola.

Theorem. (Properties of a parabola.)

1. In the coordinate system canonical for a parabola, in the strip

no parabola points.

2. In the coordinate system canonical for a parabola, the vertex of the parabola O(0; 0) lies on the parabola.

3. A parabola is a curve that is symmetrical about the focal axis.

Proof. 1, 2) Immediately follows from the canonical parabola equation.

3) Let M(x, y) be an arbitrary point of the parabola. Then its coordinates satisfy equation (1). But then the coordinates of the point
also satisfy equation (1), and, therefore, this point is also a point of a parabola, from which the statement of the theorem follows.

The theorem is proven.

clause 4. Construction of a parabola.

Due to symmetry, it is enough to construct a parabola in the first quarter, where it is the graph of the function

,

and then display the resulting graph symmetrically about the x-axis.

We build a graph of this function, taking into account that this function is increasing on the interval
.

clause 5. The focal parameter of a hyperbola.

Theorem. The focal parameter of a parabola is equal to the length of the perpendicular to its axis of symmetry, restored at the focus of the parabola before intersecting with the parabola.

Proof. Since the point
is the intersection point of the parabola
with perpendicular
(see Fig. 3), then its coordinates satisfy the parabola equation:

.

From here we find
, from which the statement of the theorem follows.

The theorem is proven.

clause 6. Unified definition of ellipse, hyperbola and parabola.

Using the proven properties of the ellipse and hyperbola, and the definition of a parabola, we can give a single definition for all three curves.

Definition. HMT planes for which the ratio of the distance to one fixed point of the plane, called the focus, to the distance to one fixed straight line, called the directrix, is a constant value is called:

a) an ellipse if this constant value is less than 1;

b) a hyperbola if this constant value is greater than 1;

c) a parabola if this constant value is equal to 1.

This constant value referred to in the definition is called eccentricity and is denoted , the distance from a given point to the focus is its focal radius r, the distance from a given point to the directrix is ​​denoted by d.

From the definition it follows that those points of the plane for which the relation there is a constant quantity that forms an ellipse, hyperbola or parabola, depending on the magnitude of this ratio.

If
, then we get an ellipse if
, then we get a hyperbola if
, then we get a parabola.

clause 7. Tangent to a parabola.

Theorem. Let
– arbitrary point of the parabola

.

Then the equation of the tangent to this parabola is

at the point
has the form:

. (2)

Proof. It is enough to consider the case when the point of contact lies in the first quarter. Then the equation of the parabola looks like:

and it can be considered as a graph of the function
.

Let's use the tangent equation to the graph of the function
at the point
:

Where
– the value of the derivative of a given function at a point
.

Let's find the derivative of the function
and its value at the point of contact:

,
.

Here we took advantage of the fact that the tangent point
is a point of a parabola and therefore its coordinates satisfy the equation of the parabola, i.e.

.

We substitute the found derivative value into the tangent equation:

,

where we get:

.

Since the point
belongs to a parabola, then its coordinates satisfy its equation, i.e.
, where we get

or
.

this implies

.

The theorem is proven.

clause 8. Mirror property of a parabola.

Theorem. The tangent to a parabola forms equal angles with its axis of symmetry and with the focal radius of the point of tangency.

Proof. Let
– point of contact, – its focal radius. Let us denote by N the point of intersection of the tangent with the abscissa axis. The ordinate of point N is equal to zero and point N lies on the tangent, therefore, its coordinates satisfy the tangent equation. Substituting the coordinates of point N into the tangent equation, we get:

,

whence the abscissa of the point N is equal to
.

Consider a triangle
. Let us prove that it is isosceles.

Really,
. Here we used the equality obtained when deriving the canonical parabola equation:

.

In an isosceles triangle, the base angles are equal. From here

, etc.

The theorem is proven.

Comment. The proven theorem can be formulated in the form of the mirror property of a parabola.

A ray of light released from the focus of a parabola, after reflection from the mirror of the parabola, goes parallel to the axis of symmetry of the parabola.

Indeed, since the angle of incidence of the ray on the tangent is equal to the angle of reflection from it, the angle between the tangent and the reflected ray is equal to the angle between the tangent and the abscissa axis, which means that the reflected ray is parallel to the abscissa axis.

Comment. This property of a parabola has been widely used in technology. If a parabola is rotated around its axis of symmetry, we obtain a surface called a paraboloid of revolution. If you make a reflective surface in the shape of a paraboloid of revolution and place a light source at the focus, then the reflected rays go parallel to the symmetry axis of the paraboloid. This is how spotlights are designed and car lights. If a device receiving electromagnetic oscillations (waves) is placed at the focus, then they are reflected from the surface of the paraboloid and enter this receiving device. Satellite dishes work on this principle.

There is a legend that in ancient times one commander lined up his warriors along the shore, giving their formation a parabola shape. Sunlight, reflecting from the warriors' shields polished to a shine, was collected in a beam (at the focus of the constructed parabola). In this way the enemy ships were burned. Some sources attribute this to Archimedes. One way or another, the Arabs called the paraboloid of rotation an “incendiary mirror.”

By the way, the word “focus” is Latin and means fire, hearth. Using the “burning mirror” you can light a fire and boil water on a sunny day. So the origin of this term becomes clear.

The word "trick" also means some trick or trick. Previously, the circus was called a booth. So, farcical artists also used the mirror property of the ellipse and, by lighting a light in one focus of the ellipse, they ignited something flammable placed in its other focus. This spectacle also came to be called a magic trick. (Read the wonderful book by N.Ya. Vilenkin “Behind the Pages of a Mathematics Textbook”)

clause 9. Polar equation of ellipse, hyperbola and parabola.

Let a point F be given on the plane, which we will call the focus, and a line D, which we will call the directrix. Let us draw a line perpendicular to the directrix (focal axis) through the focus and introduce a polar coordinate system. Let us place the pole at the focus, and as the polar ray we take that part of the line that does not intersect the directrix (see Fig. 5).

Let point M lie on an ellipse, hyperbola or parabola. In what follows we will simply call a hyperbola or parabola a curve.

Theorem. Let
– polar coordinates of a point on a curve (ellipse, hyperbola or parabola). Then

, (3)

where p is the focal parameter of the curve, – eccentricity of the curve (for a parabola we assume
).

Proof. Let Q be the projection of point M onto the focal axis of the curve, B – onto the directrix of the curve. Let the polar angle point M is obtuse, as in Figure 5. Then

,

where by construction,
– distance from point M to the directrix, and

. (4)

On the other hand, according to the common definition of an ellipse, hyperbola and parabola, the ratio

(5)

equal to the eccentricity of the corresponding curve for any point M on a given curve. Let the point
– the point of intersection of the curve with the perpendicular to the focal axis, restored to focus F and A – its projection onto the directrix. Then

, where
. But
, where

and, substituting into equality (4), we get

or, taking into account equality (5),

whence the equality (3) being proved follows.

Note that equality (4) remains true in the case when the polar angle point M is sharp, because in this case, point Q is to the right of focus F and

The theorem is proven.

Definition. Equation (3) is called the polar equation of the ellipse, hyperbola and parabola.

A function of the form where is called quadratic function.

Graph of a quadratic function – parabola.

Let's consider the cases:

I CASE, CLASSICAL PARABOLA

That is , ,

To construct, fill out the table by substituting the x values ​​into the formula:

Mark the points (0;0); (1;1); (-1;1), etc. on the coordinate plane (the smaller the step we take the x values ​​(in this case, step 1), and the more x values ​​we take, the smoother the curve will be), we get a parabola:

It is easy to see that if we take the case , , , that is, then we get a parabola that is symmetrical about the axis (oh). It’s easy to verify this by filling out a similar table:

II CASE, “a” IS DIFFERENT FROM UNIT

What will happen if we take , , ? How will the behavior of the parabola change? With title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;"> парабола изменит форму, она “похудеет” по сравнению с параболой (не верите – заполните соответствующую таблицу – и убедитесь сами):!}

In the first picture (see above) it is clearly visible that the points from the table for the parabola (1;1), (-1;1) were transformed into points (1;4), (1;-4), that is, with the same values, the ordinate of each point is multiplied by 4. This will happen to all key points of the original table. We reason similarly in the cases of pictures 2 and 3.

And when the parabola “becomes wider” than the parabola:

Let's summarize:

1)The sign of the coefficient determines the direction of the branches. With title="Rendered by QuickLaTeX.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}

2) Absolute value coefficient (modulus) is responsible for the “expansion” and “compression” of the parabola. The larger , the narrower the parabola; the smaller |a|, the wider the parabola.

III CASE, “C” APPEARS

Now let's introduce into the game (that is, consider the case when), we will consider parabolas of the form . It is not difficult to guess (you can always refer to the table) that the parabola will shift up or down along the axis depending on the sign:

IV CASE, “b” APPEARS

When will the parabola “break away” from the axis and finally “walk” along the entire coordinate plane? When will it stop being equal?

Here to construct a parabola we need formula for calculating the vertex: , .

So at this point (as at point (0;0) new system coordinates) we will build a parabola, which we can already do. If we are dealing with the case, then from the vertex we put one unit segment to the right, one up, - the resulting point is ours (similarly, a step to the left, a step up is our point); if we are dealing with, for example, then from the vertex we put one unit segment to the right, two - upward, etc.

For example, the vertex of a parabola:

Now the main thing to understand is that at this vertex we will build a parabola according to the parabola pattern, because in our case.

When constructing a parabola after finding the coordinates of the vertex veryIt is convenient to consider the following points:

1) parabola will definitely pass through the point . Indeed, substituting x=0 into the formula, we obtain that . That is, the ordinate of the point of intersection of the parabola with the axis (oy) is . In our example (above), the parabola intersects the ordinate at point , since .

2) axis of symmetry parabolas is a straight line, so all points of the parabola will be symmetrical about it. In our example, we immediately take the point (0; -2) and build it symmetrical relative to the symmetry axis of the parabola, we get the point (4; -2) through which the parabola will pass.

3) Equating to , we find out the points of intersection of the parabola with the axis (oh). To do this, we solve the equation. Depending on the discriminant, we will get one (, ), two ( title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, our root of the discriminant is not an integer; when constructing, it doesn’t make much sense for us to find the roots, but we clearly see that we will have two points of intersection with the axis (oh) (since title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}

So let's work it out

Algorithm for constructing a parabola if it is given in the form

1) determine the direction of the branches (a>0 – up, a<0 – вниз)

2) we find the coordinates of the vertex of the parabola using the formula , .

3) we find the point of intersection of the parabola with the axis (oy) using the free term, construct a point symmetrical to this point with respect to the symmetry axis of the parabola (it should be noted that it happens that it is unprofitable to mark this point, for example, because the value is large... we skip this point...)

4) At the found point - the vertex of the parabola (as at the point (0;0) of the new coordinate system) we construct a parabola. If title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;">, то парабола становится у’же по сравнению с , если , то парабола расширяется по сравнению с !}

5) We find the points of intersection of the parabola with the axis (oy) (if they have not yet “surfaced”) by solving the equation

Example 1

Example 2

Note 1. If the parabola is initially given to us in the form , where are some numbers (for example, ), then it will be even easier to construct it, because we have already been given the coordinates of the vertex . Why?

Let's take a quadratic trinomial and isolate the complete square in it: Look, we got that , . You and I previously called the vertex of a parabola, that is, now,.

For example, . We mark the vertex of the parabola on the plane, we understand that the branches are directed downward, the parabola is expanded (relative to ). That is, we carry out points 1; 3; 4; 5 from the algorithm for constructing a parabola (see above).

Note 2. If the parabola is given in a form similar to this (that is, presented as a product of two linear factors), then we immediately see the points of intersection of the parabola with the axis (ox). In this case – (0;0) and (4;0). For the rest, we act according to the algorithm, opening the brackets.

I suggest that the rest of the readers significantly expand their school knowledge about parabolas and hyperbolas. Hyperbola and parabola - are they simple? ...Can't wait =)

Hyperbola and its canonical equation

General structure the presentation of the material will resemble the previous paragraph. Let's start with general concept hyperbolas and problems for its construction.

The canonical equation of a hyperbola has the form , where are positive real numbers. Please note that, unlike ellipse, the condition is not imposed here, that is, the value of “a” may be less than the value of “be”.

I must say, quite unexpectedly... the equation of the “school” hyperbola does not even closely resemble the canonical notation. But this mystery will still have to wait for us, but for now let’s scratch our heads and remember what characteristic features does the curve in question have? Let's spread it on the screen of our imagination graph of a function ….

A hyperbola has two symmetrical branches.

Not bad progress! Any hyperbole has these properties, and now we will look with genuine admiration at the neckline of this line:

Example 4

Construct the hyperbola given by the equation

Solution: in the first step, we bring this equation to canonical form. Please remember the standard procedure. On the right you need to get “one”, so we divide both sides of the original equation by 20:

Here you can reduce both fractions, but it is more optimal to do each of them three-story:

And only after that carry out the reduction:

Select the squares in the denominators:

Why is it better to carry out transformations this way? After all, the fractions on the left side can be immediately reduced and obtained. The fact is that in the example under consideration we were a little lucky: the number 20 is divisible by both 4 and 5. In the general case, such a number does not work. Consider, for example, the equation . Here with divisibility everything is sadder and without three-story fractions no longer possible:

So, let's use the fruit of our labors - the canonical equation:

How to construct a hyperbola?

There are two approaches to constructing a hyperbola - geometric and algebraic.
From a practical point of view, drawing with a compass... I would even say utopian, so it is much more profitable to once again use simple calculations to help.

It is advisable to adhere to the following algorithm, first finished drawing, then comments:

In practice, a combination of rotation by an arbitrary angle and parallel translation of the hyperbola is often encountered. This situation discussed in class Reducing the 2nd order line equation to canonical form.

Parabola and its canonical equation

It's finished! She's the one. Ready to reveal many secrets. The canonical equation of a parabola has the form , where – real number. It is easy to notice that in its standard position the parabola “lies on its side” and its vertex is at the origin. In this case, the function specifies the upper branch of this line, and the function – the lower branch. It is obvious that the parabola is symmetrical about the axis. Actually, why bother:

Example 6

Construct a parabola

Solution: the top is known, let's find it additional points. The equation determines the upper arc of the parabola, the equation determines the lower arc.

In order to shorten the recording of the calculations, we will carry out the calculations “with one brush”:

For compact recording, the results could be summarized in a table.

Before performing an elementary point-by-point drawing, let’s formulate a strict

definition of parabola:

A parabola is the set of all points in the plane that are equidistant from a given point and a given line that does not pass through the point.

The point is called focus parabolas, straight line - headmistress (spelled with one "es") parabolas. The constant "pe" of the canonical equation is called focal parameter, which is equal to the distance from the focus to the directrix. In this case . In this case, the focus has coordinates , and the directrix is ​​given by the equation .
In our example:

The definition of a parabola is even simpler to understand than the definitions of an ellipse and a hyperbola. For any point on a parabola, the length of the segment (the distance from the focus to the point) is equal to the length of the perpendicular (the distance from the point to the directrix):

Congratulations! Many of you have made a real discovery today. It turns out that a hyperbola and a parabola are not graphs of “ordinary” functions at all, but have a pronounced geometric origin.

Obviously, as the focal parameter increases, the branches of the graph will “raise” up and down, approaching infinitely close to the axis. As the “pe” value decreases, they will begin to compress and stretch along the axis

The eccentricity of any parabola is equal to unity:

Rotation and parallel translation of a parabola

The parabola is one of the most common lines in mathematics, and you will have to build it really often. Therefore, please pay special attention to the final paragraph of the lesson, where I will discuss typical options for the location of this curve.

! Note : as in the cases with previous curves, it is more correct to talk about rotation and parallel translation coordinate axes, but the author will limit himself to a simplified version of the presentation so that the reader has a basic understanding of these transformations.

Level III

3.1. Hyperbole touches lines 5 x – 6y – 16 = 0, 13x – 10y– – 48 = 0. Write down the equation of the hyperbola provided that its axes coincide with the coordinate axes.

3.2. Write equations for tangents to a hyperbola

1) passing through a point A(4, 1), B(5, 2) and C(5, 6);

2) parallel to straight line 10 x – 3y + 9 = 0;

3) perpendicular to straight line 10 x – 3y + 9 = 0.

Parabola is the geometric locus of points in the plane whose coordinates satisfy the equation

Parabola parameters:

Dot F(p/2, 0) is called focus parabolas, magnitude p – parameter , dot ABOUT(0, 0) – top . In this case, the straight line OF, about which the parabola is symmetrical, defines the axis of this curve.

Magnitude Where M(x, y) – an arbitrary point of a parabola, called focal radius , straight D: x = –p/2 – headmistress (it does not intersect the interior region of the parabola). Magnitude is called the eccentricity of the parabola.

The main characteristic property of a parabola: all points of the parabola are equidistant from the directrix and focus (Fig. 24).

There are other forms of the canonical parabola equation that determine other directions of its branches in the coordinate system (Fig. 25):

For parametric definition of a parabola as a parameter t the ordinate value of the parabola point can be taken:

Where t is an arbitrary real number.

Example 1. Determine the parameters and shape of a parabola using its canonical equation:

Solution. 1. Equation y 2 = –8x defines a parabola with vertex at point ABOUT Oh. Its branches are directed to the left. Comparing this equation with the equation y 2 = –2px, we find: 2 p = 8, p = 4, p/2 = 2. Therefore, the focus is at the point F(–2; 0), directrix equation D: x= 2 (Fig. 26).

2. Equation x 2 = –4y defines a parabola with vertex at point O(0; 0), symmetrical about the axis Oy. Its branches are directed downwards. Comparing this equation with the equation x 2 = –2py, we find: 2 p = 4, p = 2, p/2 = 1. Therefore, the focus is at the point F(0; –1), directrix equation D: y= 1 (Fig. 27).

Example 2. Determine parameters and type of curve x 2 + 8x – 16y– 32 = 0. Make a drawing.

Solution. Let's transform the left side of the equation using the complete square extraction method:

x 2 + 8x– 16y – 32 =0;

(x + 4) 2 – 16 – 16y – 32 =0;

(x + 4) 2 – 16y – 48 =0;

(x + 4) 2 – 16(y + 3).

As a result we get

(x + 4) 2 = 16(y + 3).

This is the canonical equation of a parabola with the vertex at the point (–4, –3), the parameter p= 8, branches pointing upward (), axis x= –4. Focus is on point F(–4; –3 + p/2), i.e. F(–4; 1) Headmistress D given by the equation y = –3 – p/2 or y= –7 (Fig. 28).

Example 4. Write an equation for a parabola with its vertex at the point V(3; –2) and focus at the point F(1; –2).

Solution. The vertex and focus of a given parabola lie on a straight line parallel to the axis Ox(same ordinates), the branches of the parabola are directed to the left (the abscissa of the focus is less than the abscissa of the vertex), the distance from the focus to the vertex is p/2 = 3 – 1 = 2, p= 4. Hence, the required equation

(y+ 2) 2 = –2 4( x– 3) or ( y + 2) 2 = = –8(x – 3).

Tasks for independent decision

I level

1.1. Determine the parameters of the parabola and construct it:

1) y 2 = 2x; 2) y 2 = –3x;

3) x 2 = 6y; 4) x 2 = –y.

1.2. Write the equation of a parabola with its vertex at the origin if you know that:

1) the parabola is located in the left half-plane symmetrically relative to the axis Ox And p = 4;

2) the parabola is located symmetrically relative to the axis Oy and passes through the point M(4; –2).

3) the directrix is ​​given by equation 3 y + 4 = 0.

1.3. Write an equation for a curve all points of which are equidistant from the point (2; 0) and the straight line x = –2.

Level II

2.1. Determine the type and parameters of the curve.