IDZ Ryabushko 4.1 Option 6

No. 1 Drawing up canonical equations for curves:

a) ellipse: The equation of the ellipse has the following form: (x-h)²/a² + (y-k)²/b² = 1, where (h,k) are the coordinates of the center of the ellipse, a is the length of the major semi-axis, b is the length of the minor semi-axis. The foci are located at a distance c = √(a²-b²) from the center, the eccentricity is ε = c/a.

b) hyperbolas: The equation of a hyperbola has the following form: (x-h)²/a² - (y-k)²/b² = 1, where (h,k) are the coordinates of the center of the hyperbola, a is the distance from the center to the vertices of the hyperbola, b is the distance from center to asymptotes. The foci are located at a distance c = √(a²+b²) from the center, the eccentricity is ε = c/a. Equations of asymptotes: y = ±(b/a)(x-h) + k.

c) parabolas: The equation of a parabola has the following form: y² = 2px, where (0,p) are the coordinates of the vertex of the parabola, p is the focal length, D is the directrix of the parabola, which is located at a distance p from the vertex.

To solve problems, it is necessary to use known data: the coordinates of points A and B, the coordinates of the focus F, the length of the semi-major axis a, the length of the minor semi-axis b, eccentricity ε, the angle of inclination of the symmetry axis of the parabola, the coordinate of point M and the distance from it to point A and straight line x =8, as well as the angle φ in the polar coordinate system.

No. 2 The equation of a circle passing through point A(0;-3) and having a center at point A can be written as (x−a)²+(y−b)²=r². If the center of the circle is at point A, then the coordinates of the center are (a,b). It is also known that the circle passes through point A, so its equation will be written as (x−a)²+(y−b+3)²=r². It remains to find the radius r. To do this, you can use the coordinates of the left focus of the hyperbola, which has the equation 3x²-4y²=12. The left focus is at a distance c=√(a²+b²) from the center of the hyperbola, where a=√3/2, b=√2. From the equation c²=a²+b² we find c=√7/2. Then the distance from the center of the hyperbola to its vertex is a=√3/2. Obviously, the left focus is located on the segment between the vertices of the hyperbola, so the coordinates of the left focus can be found as (a-c,0). Substituting this point into the equation of the circle, we get (a-c)²+(b+3)²=r². Now it remains to solve a system of two equations with two unknowns a and b to find the coordinates of the center of the circle and its radius r.

No. 3 The equation of a line, each point M of which satisfies the given conditions, can be written as an equation of a circle with a center at point A(1;0) and radius r=1/5 from the distance from point M to straight line x=8. Thus, the equation of a circle will have the form (x-1)²+y²=(1/5d)², where d is the distance from point M to straight line x=8. The distance from point M to point A is 1, so d=5/

No. 1 Drawing up canonical equations for curves:

a) ellipse: The equation of the ellipse has the following form: (x-h)²/a² + (y-k)²/b² = 1, where (h,k) are the coordinates of the center of the ellipse, a is the length of the major semi-axis, b is the length of the minor semi-axis. The foci are located at a distance c = √(a²-b²) from the center, the eccentricity is ε = c/a.

b) hyperbolas: The equation of a hyperbola has the following form: (x-h)²/a² - (y-k)²/b² = 1, where (h,k) are the coordinates of the center of the hyperbola, a is the distance from the center to the vertices of the hyperbola, b is the distance from center to asymptotes. The foci are located at a distance c = √(a²+b²) from the center, the eccentricity is ε = c/a. Equations of asymptotes: y = ±(b/a)(x-h) + k.

c) parabolas: The equation of a parabola has the following form: y² = 2px, where (0,p) are the coordinates of the vertex of the parabola, p is the focal length, D is the directrix of the parabola, which is located at a distance p from the vertex.

To solve problems, it is necessary to use known data: the coordinates of points A and B, the coordinates of the focus F, the length of the semi-major axis a, the length of the minor semi-axis b, eccentricity ε, the angle of inclination of the symmetry axis of the parabola, the coordinate of point M and the distance from it to point A and straight line x =8, as well as the angle φ in the polar coordinate system.

No. 2 The equation of a circle passing through point A(0;-3) and having a center at point A can be written as (x−a)²+(y−b)²=r². If the center of the circle is at point A, then the coordinates of the center are (a,b). It is also known that the circle passes through point A, so its equation will be written as (x−a)²+(y−b+3)²=r². It remains to find the radius r. To do this, you can use the coordinates of the left focus of the hyperbola, which has the equation 3x²-4y²=12. The left focus is at a distance c=√(a²+b²) from the center of the hyperbola, where a=√3/2, b=√2. From the equation c²=a²+b² we find c=√7/2. Then the distance from the center of the hyperbola to its vertex is a=√3/2. Obviously, the left focus is located on the segment between the vertices of the hyperbola, so the coordinates of the left focus can be found as (a-c,0). Substituting this point into the equation of the circle, we get (a-c)²+(b+3)²=r². Now it remains to solve a system of two equations with two unknowns a and b to find the coordinates of the center of the circle and its radius r.

No. 3 The equation of a line, each point M of which satisfies the given conditions, can be written as an equation of a circle with a center at point A(1;0) and radius r=1/5 from the distance from point M to straight line x=8. Thus, the equation of a circle will have the form (x-1)²+y²=(1/5d)², where d is the distance from point M to straight line x=8. The distance from point M to point A is 1, so d=5/6. Substituting this value into the equation of the circle, we get (x-1)²+y²=1/36. Thus, the equation of the desired line is x²+y²-2x=1/36.

No. 4 The curve, defined in the polar coordinate system as ρ=3(1+sinφ), represents a rose petal. To construct it in a Cartesian coordinate system, it is necessary to convert polar coordinates to Cartesian ones. The conversion formulas are x=ρcosφ, y=ρsinφ. Substituting the expression for ρ into them, we obtain x=3cosφ+3cos²φsinφ, y=3sinφ+3sin²φcosφ. Thus, the equation of the desired curve has the form x²+y²=3(3+2sinφ+sin²φ).

No. 5 The curve defined by the parametric equations x=cos(t), y=sin(t) is a circle of unit radius with its center at the origin. To plot its graph on a plane, it is necessary to plot the values ​​of the x and y coordinates for each value of the parameter t from 0 to 2π. The visualization of the curve will be a circle passing through all points with coordinates (cos(t),sin(t)).

IDZ Ryabushko 4.1 Option 6 is a set of problems in mathematics, which includes tasks on composing canonical equations for curves (ellipses, hyperbolas and parabolas), finding the equation of a circle passing through a given point and having a given center, as well as composing an equation lines in the form of an equation of a circle with a given center and radius. To solve problems, it is necessary to use known data, such as coordinates of points, foci, lengths of semi-axes and distances, as well as angles and coordinates of points in different coordinate systems.

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IDZ Ryabushko 4.1 Option 6 is a task that includes five different problems from different areas of mathematics:

1. Create a canonical equation for an ellipse, hyperbola and parabola passing through given points and having given parameters, such as semi-major and minor axes, eccentricity, equations of asymptotes and directrixes, focal length, etc.

2. Write down the equation of a circle with a center at a given point A, passing through another given point and satisfying the condition.

3. Write an equation of a straight line, all points of which are at a certain distance from a given point and from a certain straight line.

4. Construct a curve specified in a polar coordinate system.

5. Construct a curve defined by parametric equations for parameter values ​​from 0 to 2π.

Each problem requires the application of specific mathematical knowledge and skills, such as analytical geometry, trigonometry, algebra, and differential equations. Solving each problem may require different solution methods, depending on its conditions.

IDZ Ryabushko 4.1 Option 6 is an educational task for schoolchildren, published by the publishing house "Ryabushko". This version of the IDL is intended for 4th grade students and contains tasks in mathematics, the Russian language, the environment and other subjects.

IDZ Ryabushko 4.1 Option 6 is one of the options for released tasks and may differ from other options in the number and complexity of tasks. The IDZ usually comes with an explanatory note for parents or teachers, which helps them understand the assignments and properly organize the student’s work.

Typically, IDL is issued on topics studied in the current academic year and is intended to consolidate the knowledge and skills acquired in the lessons. IDZ Ryabushko 4.1 Option 6 can be used as additional material for student independent work at home or as a test in class.

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