Hyperbola and its canonical equation. Hyperbola and its canonical equation Form and characteristics of hyperbola
Curves of the second order. An algebraic curve of the second order is a curve Г, the equation of which in a Cartesian coordinate system has the form:
If the curve Γ is non-degenerate, then for it there is such a Cartesian rectangular coordinate system in which the equation of this curve takes one of the following three forms (the canonical equation):
Hyperbola,
px is a parabola.
Ellipse- a geometric set of points of the plane, the sum of the distances from which to two points and, called foci, is a constant value 2a, greater than the distance between the foci 2c:
Ellipse given by the canonical equation: symmetrical about the coordinate axes. The parameters a and b are called the semi-axes of the ellipse (large and small, respectively), the points are called its vertices. If a>b, then the foci are on the x-axis at a distance from the center of the ellipse O.
is called the eccentricity of the ellipse and is a measure of its "oblateness" (when an ellipse is a circle, and when it degenerates into a segment of length). If a Hyperbola- a geometric set of points in the plane, the modulus of the difference in distances from which to two points and, called foci, is a constant value 2a, less than the distance between the foci 2c: symmetrical about the coordinate axes. It intersects the OX axis at the points and - vertices of the hyperbola, and does not intersect the OY axis. Parameter a is called the real semiaxis, b is the imaginary semiaxis. Number is called the eccentricity of the hyperbola. Direct are called asymptotes of the hyperbola. Hyperbola given by the canonical equation: is called conjugate (has the same asymptotes). Its foci are located on the OY axis. It intersects the OY axis at the points and - the vertices of the hyperbola, and does not intersect the OX axis. In this case, the parameter b is called the real semiaxis, a is the imaginary semiaxis. The eccentricity is calculated by the formula: Parabola- a set of points of the plane equidistant from a given point F, called the focus, and a given straight line, called the directrix: . The parabola given by the indicated canonical equation is symmetric about the OX axis. The equation defines a parabola symmetrical about the OY axis. Parabola has focus and directrix Parabola has focus and directrix If p>0, then in both cases the branches of the parabola are turned to the positive side of the corresponding axis, and if p<0 - в отрицательную сторону. Examples of problem solving. 1
.Write the canonical equation of the hyperbola, knowing that: a) distance between foci 2c=30, and between vertices 2a=20; b) the real semiaxis is 5, the eccentricity. Solution: a) by condition; ; ; ; from ratios. Answer: . b) by condition; , . 2
. Write the parabola equation knowing that: a) the parabola passes through the points (0,0); (3.6) and is symmetrical with respect to the OX axis, b) the parabola passes through the points (0,0); (4.2) and is symmetric with respect to the OY axis. Decision: a) The point (3.6) lies on the parabola, therefore, is the directrix equation. - parabola equation b) The point (4,2) lies on the parabola, therefore, the directrix equation, Parabola equation. Reduction to the canonical form of the general equation of a curve of the second order Let us consider the general second-order equation in the Cartesian rectangular coordinate system Oxy: Ax 2 + 2Bxy + Cy 2 + 2Dx + 2Ey + F = 0, where not all coefficients A, B and C are equal to zero at the same time. It defines a second order curve. Our goal is to change the coordinate system in such a way as to simplify this equation as much as possible. To do this, first (if B0) we rotate the original basis (coordinate axes Ox and Oy) by an angle b counterclockwise so that the new axes Ox "and Oy" become parallel to the axes of the curve, and the term 2Bxy disappears: Linear transformation matrix: rotation by angle b counterclockwise. Or vice versa, A(x"cosb - y"sinb) 2 + 2B(x"cosb - y"sinb)(x"sinb + y"cosb)+C(x"sinb + y"cosb) 2 + 2D(x"cosb - y "sinb) + 2E (x" sinb + y "cosb) + F \u003d 0 We choose the angle b so that the coefficient at the product of x "y" vanishes, i.e. so that equality holds: 2Acosаsinb + 2B(cos 2 b - sin 2 b) + 2Csinаcosb = 0 In the new coordinate system Ox"y" (after rotation by angle b), given that the equation will look like A "x" 2 + C "y" 2 + 2D "x" + 2E "y" + F" \u003d 0, where the coefficients A" and C" are not equal to zero at the same time. The next stage of simplification consists in the parallel translation of the axes Ox" and Oy" until they coincide with the axes of the curve, while the origin of coordinates coincides with the center (or vertex, in the case of a parabola) of the curve. The transformation technique at this stage is to select a full square. Thus, we obtain the canonical equations of second-order curves. In total, 9 qualitatively different cases are possible (including cases of degeneration and decay): 1. (ellipse), Curves of the 2nd order with displaced centers (vertices). If in the general equation of the 2nd order curve in particular, B \u003d 0, that is, there is no term with the product of variables, this means that the axes of the curve are parallel to the coordinate ones. Consider the equation: In each of cases 1), 2), 3) there may be degenerate curves, which we will not deal with. In order to understand exactly how the curve is located relative to the coordinate system and what are its parameters, the equation can be transformed by selecting full squares. After that, the equation will take the form of one of the non-degenerate equations of a 2nd order curve with a displaced center: these equations define hyperbolas with center and axes parallel to the coordinate ones; these are parabolas with a vertex and an axis parallel to one of the coordinates. Ellipse, hyperbola and parabola as conic sections. Theorem. The section of any round cone by a plane (not passing through its vertex) defines a curve, which can only be an ellipse, hyperbola, or parabola. Moreover, if the plane intersects only one cavity of the cone and along a closed curve, then this curve is an ellipse; if the cutting plane intersects only one cavity of the cone and along an open curve, then this curve is a parabola; if the plane intersects both cavities of the cone, then a hyperbola is formed in the section. The validity of this theorem can be established on the basis of the general proposition that the intersection of a second-order surface by a plane is a second-order line. It can be seen from the figure that by rotating the cutting plane around the straight line PQ, we change the section curve. Being, for example, originally an ellipse, it becomes a parabola for a moment, and then turns into a hyperbola. This curve will be a parabola when the secant plane is parallel to the tangent plane of the cone. Thus, ellipses, hyperbolas and parabolas are called conic sections. Hyperbola is a plane curve, for each point of which the modulus of the difference in distances to two given points ( tricks of hyperbole
) is constant. The distance between the foci of a hyperbola is called focal length
and is denoted by \(2c\). The middle of the segment connecting the foci is called center. A hyperbola has two axes of symmetry: a focal or real axis passing through the foci, and an imaginary axis perpendicular to it passing through the center. The real axis intersects the branches of the hyperbola at points called peaks. The segment connecting the center of the hyperbola with the vertex is called real semiaxis
and is denoted by \(a\). Imaginary axle
denoted by \(b\). Canonical equation of a hyperbola
is written in the form The modulus of the difference in distances from any point of the hyperbola to its foci is a constant value: Hyperbola asymptote equations
Relationship between semi-axes of a hyperbola and focal length Eccentricity
hyperbole Hyperbola Directrix Equations
Equation of the right branch of a hyperbola in parametric form
General equation of a hyperbola
General equation of a hyperbola whose semiaxes are parallel to the coordinate axes Isosceles hyperbola
parabola is called a plane curve, at each point of which the following property holds: the distance to a given point ( focus of the parabola
) is equal to the distance to the given straight line ( parabola directrixes
). The distance from the focus to the directrix is called parabola parameter
and is denoted by \(p\). A parabola has a single axis of symmetry that intersects the parabola at its top
. Canonical parabola equation
has the form Directrix equation
Focus coordinates
Vertex coordinates
General equation of a parabola
The equation of a parabola whose symmetry axis is parallel to the \(Oy\) axis
Directrix equation
Focus coordinates
Vertex coordinates
Equation of a parabola with a vertex at the origin and an axis of symmetry parallel to the \(Oy\) axis
Directrix equation
Focus coordinates
Vertex coordinates
A hyperbola is a set of points in a plane whose distance difference from two given points, foci, is a constant and equal to . Similarly to the ellipse, we place the foci at the points , (see Fig. 1). Rice. 1 It can be seen from the figure that there may be cases and title="(!LANG:Rendered by QuickLaTeX.com" height="15" width="57" style="vertical-align: -4px;">, тогда согласно определению !} It is known that in a triangle the difference of two sides is less than the third side, therefore, for example, with we get: We bring both parts to the square and after further transformations we find: where . The hyperbolic equation (1) is canonical equation of a hyperbola. The hyperbola is symmetrical about the coordinate axes, therefore, as for the ellipse, it is enough to plot its graph in the first quarter, where: Range of value for the first quarter. When we have one of the vertices of the hyperbola . Second peak. If , then from (1) - there are no real roots. We say that and are imaginary vertices of the hyperbola. From the ratio it turns out that for sufficiently large values there is a place for the nearest equality title="(!LANG:Rendered by QuickLaTeX.com" height="24" width="264" style="vertical-align: -6px;">. Поэтому прямая есть линией, расстояние между которой и соответствующей точкой гиперболы направляется к нулю при .!} We investigate the equation (1) the shape and location of the 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 a point in the first quarter, i.e. . In this case , , then the asymptote has the form: , where So the line is the asymptote of the function . Therefore, due to symmetry, the asymptotes of the hyperbola are straight lines. Based on the established characteristics, we 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 hyperbole, the asymptotes, which are the bisectors of the coordinate angles. Example 1 A task Find the axes, vertices, foci, eccentricity and equations of the asymptotes of the hyperbola. Construct a hyperbola and its asymptotes. Decision We reduce the equation of the hyperbola to the canonical form: Comparing this equation with the canonical equation (1), we find , , . Vertices, foci and A task The foci of the hyperbola and its asymptote are given. Write the hyperbola equation: Decision Having written the equation of the asymptote in the form, we find the ratio of the semiaxes of the hyperbola . By the condition 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 . Therefore, the hyperbola has the following equation: Answer Hyperbola and its canonical equation updated: November 22, 2019 by: Scientific Articles.Ru III level 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 the equations of the tangents to the hyperbola 1) passing through a point A(4, 1), B(5, 2) and C(5, 6); 2) parallel to a straight line 10 x – 3y + 9 = 0; 3) perpendicular to the straight line 10 x – 3y + 9 = 0. parabola is the 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 O(0, 0) – summit
. At the same time, the direct OF, about which the parabola is symmetrical, defines the axis of this curve. Value 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 value of the ordinate of the point of the parabola can be taken: where t is an arbitrary real number. Example 1 Determine the parameters and shape of the parabola from its canonical equation: Decision. 1. Equation y 2 = –8x defines a parabola with vertex at a point O Ox. 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 a 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 Define parameters and type of curve x 2 + 8x – 16y– 32 = 0. Make a drawing. Decision. We transform the left side of the equation using the full square 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 vertex at the point (–4; –3), the parameter p= 8, branches pointing up (), axis x= -4. Focus is on point F(–4; –3 + p/2), i.e. F(–4; 1) Headmistress D is given by the equation y = –3 – p/2 or y= -7 (Fig. 28). Example 4 Compose the equation of a parabola with a vertex at a point V(3; –2) and focus at the point F(1; –2). Decision. The vertex and focus of this parabola lie on a straight line parallel to the axis Ox(the 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 desired equation (y+ 2) 2 = –2 4( x– 3) or ( y + 2) 2 = = –8(x – 3). Tasks for independent solution 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 vertex at the origin if you know that: 1) the parabola is located in the left half-plane symmetrically about the axis Ox and p = 4; 2) the parabola is located symmetrically about the axis Oy and passes through the point M(4; –2). 3) 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. II level 2.1. Define the type and parameters of the curve. Define hyperbola and parabola. Write the canonical equations of the hyperbola and parabola, explain the meaning of the quantities included in these equations. Write the equations of directrixes, asymptotes of the hyperbola, show on the drawing their location relative to the hyperbola. What is the eccentricity of the parabola? Show on the drawing the location of the directrix relative to the parabola. Write the equation of a hyperbola whose foci are located on the x-axis, symmetrical about the origin, knowing, in addition, that: a) distance between foci 2 s = 6 and eccentricity ; b) axis 2 a = 16
and eccentricity c) equation of asymptotes d) the distance between directrixes is tricks 2 s = 26. 5.
Determine the points of the hyperbola 6.
Establish that each of the following equations defines a hyperbola and find the coordinates of its center FROM, semiaxes, eccentricity, equations asymptote and directrix: 7.
Write the equation of a parabola whose vertex is at the beginning coordinates, knowing that: a) the parabola is located in the right half-plane symmetrically about the axis Oh, and its parameter p = 3; b) the parabola is located in the left half-plane symmetrically about the axis Oh, and its parameter p = 0,5; c) the parabola is located in the upper half-plane symmetrically about the axis OU, and its parameter p =; d) the parabola is located in the lower half-plane symmetrically about the axis OU, and its parameter p = 3. 8.
Find focus F and the parabola directrix equation 9.
On a parabola 10.
Write the equation of a parabola given its focus F
(7; 2)
and headmistress 11.
Determine points of intersection of a line 12.
In the following cases, determine how the given line is located with respect to a given parabola - whether it intersects, touches or passes outside it: a) b)
in) 1.
a) 2.
headmistress: 7.
a) What is a parallel translation of a coordinate system? Give the formulas for the connection of "old" and "new" coordinates. Give formulas for the connection of "old" and "new" coordinates when rotating the coordinate system without changing its origin. Explain the technique for reducing the general equation of a second-order curve to a canonical form, using successively the rotation of the coordinate system and the parallel translation of the coordinate system. What result is achieved at each of these stages of coordinate system transformation? Tasks 1.
Find out the geometric meaning of the equations: a) d) , e) 2.
By rotating the coordinate axes, convert the equations to the canonical form and plot the curves: a) b) 3.
Convert equations to canonical form and draw: Answers 1.
a) two lines d) point (3; 4), e) two lines X= 0, 2.
a) b) ACTIVITY 3.8. POLAR COORDINATE SYSTEM test questions What are polar coordinates of points? Indicate their connection with the Cartesian coordinates of this point. How to go from Cartesian coordinates of a point to polar coordinates and vice versa? How to write the equation of a line in polar coordinates if its equation in Cartesian coordinates is known and vice versa? Tasks 1.
In polar coordinate system 2.
build a line 3.
Build lines: a) b) 4.
Build lines: a) 5.
Write in polar coordinates the equation of the line that cuts off from polar axis segment " a and perpendicular to it. Write in polar coordinates the equation of a circle centered at a point FROM(0; a) and a radius equal to " a ". b) e) Convert the equations of lines to Cartesian coordinates and construct these lines: a) 9.
Write the canonical equations of curves of the second order: a) Answers 5.
G) 
\(\large\frac(((x^2)))(((a^2)))\normalsize - \large\frac(((y^2)))(((b^2)))\ normalsize = 1\).
\(\left| ((r_1) - (r_2)) \right| = 2a\),
where \((r_1)\), \((r_2)\) are the distances from an arbitrary point \(P\left((x,y) \right)\) of the hyperbola to the foci \((F_1)\) and \( (F_2)\), \(a\) is the real semiaxis of the hyperbola.
\(y = \pm \large\frac(b)(a)\normalsize x\)
\((c^2) = (a^2) + (b^2)\),
where \(c\) is half the focal length, \(a\) is the real semiaxis of the hyperbola, \(b\) is the imaginary semiaxis.
\(e = \large\frac(c)(a)\normalsize > 1\)
The directrix of a hyperbola is a line perpendicular to its real axis and intersecting it at a distance \(\large\frac(a)(e)\normalsize\) from the center. A hyperbola has two directrixes located on opposite sides of the center. The directrix equations have the form
\(x = \pm \large\frac(a)(e)\normalsize = \pm \large\frac(((a^2)))(c)\normalsize\).
\(\left\( \begin(aligned) x &= a \cosh t \\ y &= b \sinh t \end(aligned) \right., \;\;0 \le t \le 2\pi\ ),
where \(a\), \(b\) are the semiaxes of the hyperbola, \(t\) is a parameter.
where \(B^2 - 4AC > 0\).
\(A(x^2) + C(y^2) + Dx + Ey + F = 0\),
where \(AC
The hyperbole is called isosceles
, if its semiaxes are the same: \(a = b\). The asymptotes of such a hyperbola are mutually perpendicular. If the asymptotes are the horizontal and vertical coordinate axes (respectively, \(y = 0\) and \(x = 0\)), then the equation of an isosceles hyperbola has the form
\(xy = \large\frac(((e^2)))(4)\normalsize\) or \(y = \large\frac(k)(x)\normalsize\), where \(k = \ large\frac(e^2)(4)\normalsize .\)
\(y = 2px\).
\(x = - \large\frac(p)(2)\normalsize\),
\(F \left((\large\frac(p)(2)\normalsize, 0) \right)\)
\(M \left((0,0) \right)\)
\(A(x^2) + Bxy + C(y^2) + Dx + Ey + F = 0\),
where \(B^2 - 4AC = 0\).
\(A(x^2) + Dx + Ey + F = 0\;\left((A \ne 0, E \ne 0) \right) \),
or in the equivalent form
\(y = a(x^2) + bx + c,\;\;p = \large\frac(1)(2a)\normalsize\)
\(y = (y_0) - \large\frac(p)(2)\normalsize\),
where \(p\) is the parameter of the parabola.
\(F\left(((x_0),(y_0) + \large\frac(p)(2)\normalsize) \right)\)
\((x_0) = - \large\frac(b)((2a))\normalsize,\;\;(y_0) = ax_0^2 + b(x_0) + c = \large\frac((4ac - ( b^2)))((4a))\normalsize\)
\(y = a(x^2),\;\;p = \large\frac(1)((2a))\normalsize\)
\(y = - \large\frac(p)(2)\normalsize\),
where \(p\) is the parameter of the parabola.
\(F \left((0, \large\frac(p)(2)\normalsize) \right)\)
\(M \left((0,0) \right)\)
The shape and characteristics of the hyperbola
Asymptotes of a hyperbola
Examples of tasks for building a hyperbola
where M(x, y) is an arbitrary point of the parabola, is called focal radius
, straight D: x = –p/2 – headmistress
(it does not intersect the interior of the parabola). Value 
is called the eccentricity of the parabola.
;
and distance between foci 2 s = 20;
and the distance between
, whose distance to the right focus is 4.5.
.
find points whose focal radius is 13.
.
and parabolas 
.
,
;
,
;
,
.
, b) 
, in) 
, G) 
;
,X - 10 =
0;3.
;4.
10;
and 
, asymptote equations: 
;
b) FROM(- 5; 1),a= 8,
b= 6,
, directrix equations: 
and 
, asymptote equations: 
, b) 
, in) 
, G) 
;8.
F
(6; 0),
;9.
(9; 12), (9; - 12);10.
;11.
(- 4; 6) - the line touches the parabola; 12.
a) touches the parabola, b) intersects the parabola at two points, c) passes outside the parabola.Lesson 3.7. Reduction of the equations of curves of the second order to the canonical form Test questions
, b) 
, in) 
,
, e) 
.
,
.
, b) point (0; 0), c) imaginary circle,
, f) two straight lines 
;
, b) 
;3.
a) 
,
, in) 
, d) two lines 
.
build points 
,
,
,
,
,
,
,
,
.
(construction is carried out using a table of values r for 
).
(Spiral of Archimedes),
(cardioid).
, b) 
, in) 
.
, in) y= 3d) y = x, e) 
,
.
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;6.
;7.
a) 
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;8.
a) x = a, b) 
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;9.
a) 
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