IDZ Ryabushko 4.1 Option 5

No. 1. Below are the canonical equations for the ellipse, hyperbola and parabola:

• ?lipse: (x - x₀)² / a² + (y - y₀)² / b² = 1, where (x₀, y₀) are the coordinates of the center, a and b are the semi-major and minor axes, respectively, a > b.
• Hyperbola: (x - x₀)² / a² - (y - y₀)² / b² = 1, where (x₀, y₀) are the coordinates of the center, a and b are the distance from the center to the vertices and the distance from the center to the asymptotes, respectively.
• Parabola: y = a(x - x₀)² + y₀, where (x₀, y₀) are the coordinates of the vertex, a is the parameter of the parabola.

No. 2. The equation of a circle with center at point A(x₀, y₀) and radius r has the form: (x - x₀)² + (y - y₀)² = r². To write the equation of a circle passing through points A and B with its center at point A, we first need to find the radius. To do this, you can find the distance between points A and B, and then divide it in half, since the center of the circle is in the middle of the segment AB. Thus, the radius r = AB / 2. Substitute this value into the equation of the circle and get: (x - x₀)² + (y - y₀)² = (AB / 2)².

No. 3. The condition specified in the problem means that point M is located on the perpendicular bisector of the segment AB. In order to create an equation for this bisector, we need to find its coordinates. This can be done using the formula for the distance between a point and a line. The distance from a point M to a line AB can be found using the formula for the distance from a point to a line in coordinate form. We will then have two equations corresponding to the distances from point M to points A and B, and we can write their sum and set it equal to 28. This will give us the equation of the bisector, which will be the equation of the line we are looking for.

No. 4. To plot a curve in polar coordinates, you need to plot it using the angle and radius values. The equation ρ = 2 / (1 + cosφ) describes a curve that is symmetrical about the x-axis and passes through the origin. To construct a graph, you can plot several points using different values ​​of the angle φ and radius ρ, and then connect them with a line. You can also use a charting program.

No. 5. The curve defined by the parametric equations x = f(t) and y = g(t) is described by points (x, y), which depend on the parameter t. To construct a curve, it is necessary to plot its graph using values ​​of the t parameter in the range from 0 to 2π. To do this, you can plot several points using different t values ​​and then connect them with a line. For example, if we have the parametric equations x = cos(t) and y = sin(t), then we can graph a circle with radius 1 and center at the origin. To do this, you can select several values ​​of t, for example, t = 0, π/4, π/2, 3π/4, π, etc., calculate the corresponding values ​​of x and y and plot points with these coordinates on the coordinate plane. These points can then be connected with a line to create a circle graph.

IDZ Ryabushko 4.1 Option 5

a) The canonical equation of an ellipse has the form: (x - x₀)² / a² + (y - y₀)² / b² = 1, where (x₀, y₀) are the coordinates of the center, a and b are the semi-major and minor axes, respectively, a > b.

For a given ellipse, it is known that 2a = 22, which means a = 11. The eccentricity ε = √57/11 is also known. The semiminor axis b can be found by the formula b = a * √(1 - ε²), that is, b = 2√2.

The coordinates of the foci can be found using the formula c = a * ε. This means c = √57. The focal coordinates will be (x₀ + c, y₀) and (x₀ - c, y₀), where x₀ and y₀ are the coordinates of the center of the ellipse.

b) The canonical equation of a hyperbola has the form: (x - x₀)² / a² - (y - y₀)² / b² = 1, where (x₀, y₀) are the coordinates of the center, a and b are the distance from the center to the vertices and the distance from the center to the asymptotes, respectively.

For a given hyperbola, it is known that 2c - focal length = 10√13, that is, c = 5√13. It is also known that the equation of hyperbola asymptotes has the form y = ± kx, where k = 2/3.

The distance from the center to the vertices a can be found using the formula a² = c² + b². This means a = √(c² + b²) = √(194/3).

c) The canonical equation of a parabola has the form: y = a(x - x₀)² + y₀, where (x₀, y₀) are the coordinates of the vertex, a is the parameter of the parabola.

For a given parabola, the axis of symmetry Ox and the coordinate of the vertex A(27;9) are known, which means the equation will look like: y = a(x - 27)² + 9.

The equation of a circle with center at point A(x₀, y₀) and radius r has the form: (x - x₀)² + (y - y₀)² = r².

The foci of the ellipse 9x² + 25y² = 1 have coordinates (0, ±2/5). The center of the circle passes through the middle of the segment between A(0,6) and (0,-2/5), that is, the point (0, 59/50). The radius of the circle is equal to half the distance between A and (0.59/50), that is, r = √(6.25 + (59/50)² - 6) / 2.

Thus, the equation of a circle will be: (x-0)² + (y-59/50)² = ((√(6.25 + (59/50)² - 6)) / 2)².

The condition means that point M is located on the perpendicular bisector of the segment AB. The distance from point M to line AB can be found using the formula for the distance from a point to a line in the coordinate system:

d = |(y₂ - y₁)x + (x₁ - x₂)y + x₂y₁ - x₁y₂| / √((y₂ - y₁)² + (x₁ - x₂)²),

where (x₁, y₁) and (x₂, y₂) are the coordinates of points A and B, respectively.

Knowing the coordinates of points A(4, 2) and B(-2, 6), you can find the equation of straight line AB: y = -x/2 + 5.

Since point M lies on the perpendicular bisector of segment AB, angle AMB is equal to 90 degrees, which means point M is located on the perpendicular bisector of segment AB. This means that the coordinates of point M will be equal to:

x = (x₁ + x₂) / 2 = (4 - 2) / 2 = 1

y = (y₁ + y₂) / 2 = (2 + 6) / 2 = 4

Thus, the coordinates of point M are (1, 4).

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IDZ Ryabushko 4.1 Option 5 is a set of problems in mathematics, which includes tasks on composing canonical equations and equations of lines, constructing curves in polar and parametric coordinate systems, as well as a task on finding the equation of a circle.

No. 1. This problem requires you to construct canonical equations for an ellipse, a hyperbola, and a parabola, defined in various ways. To do this, you need to use known formulas and data provided in the problem statement.

No. 2. In this problem, you need to write down the equation of a circle with a specified center and passing through specified points. To do this, you can use the standard formula for the equation of a circle, which connects the coordinates of the center and the radius of the circle with the coordinates of an arbitrary point on the circle.

No. 3. In this problem, you need to create an equation for a line that satisfies given conditions. To do this, you can use well-known formulas for the distance from a point to a line and apply the methods of algebra and geometry to find the equation of the line.

No. 4. In this problem, you need to construct a curve defined in a polar coordinate system. To do this, you can use well-known formulas for converting coordinates from a polar to a Cartesian coordinate system and construct a graph of a function specified in Cartesian coordinates.

No. 5. In this problem you need to construct a curve given by parametric equations. To do this, you can use the methods of analytical geometry and construct a graph of a function specified in parametric form.

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