IDZ 4.1 – Option 12. Solutions Ryabushko A.P.

1. Drawing up canonical equations for various curves in the coordinate plane: a) for an ellipse: ((x - A)^2)/a^2 + ((y - B)^2)/b^2 = 1, where A and B - coordinates of the center of the ellipse, a and b are the lengths of the major and minor semi-axes, respectively. b) for a hyperbola: ((x - A)^2)/a^2 - ((y - B)^2)/b^2 = 1, where A and B are the coordinates of the center of the hyperbola, a and b are the lengths of the large and semi-minor axes, respectively. c) for a parabola: y^2 = 2px, where p is the distance between the vertex of the parabola and its focus.

For certain values ​​of the problem parameters, we obtain the following equations: a) ellipse: ((x + 5)^2)/13 + ((y - 15)^2)/4 = 1, where F(-5, 15), a = √13, b = 2, ε = 5√29/29. b) hyperbolas: ((x - 5)^2)/25 - ((y + 12)/13)^2 = 1, where F(5, -12), a = 5, b = 13, ε = √ (a^2 + b^2)/a = 2. c) parabolas: y^2 = 8(x + 5).

1. The equation of a circle with a center at point A and passing through given points will have the form: (x - A_x)^2 + (y - A_y)^2 = r^2, where A(x_A, y_A) are the coordinates of the center of the circle, and r - radius of the circle. The value of r can be found using the distance between the center of the circle and one of the given points.

2. The equation of a line that satisfies given conditions can be found using the formula for the distance between a point and a line. For this problem, the equation of the line will be x = -5, and the distance from point A(2, 1) to this line will be equal to 3 times the distance from point A to the intersection point of a line parallel to x = -5 and passing through point A Having solved this equation, we obtain the equation of the desired line.

3. To construct a curve defined by an equation in a polar coordinate system, you can plot it using the values ​​of ρ and φ, which define the radius vector of a point on the curve and the angle between the initial ray and the radius vector, respectively. For this problem, the curve will have the form of a cardioid.

4. To construct a curve given by parametric equations, you can plot it using the values ​​of x and y calculated for various values ​​of the parameter t in the interval [0, 2π]. For this problem, the curve will look like an ellipse.

This product is a digital product, which represents solutions to problems in mathematical analysis for IDZ 4.1, version 12, made by A.P. Ryabushko. It contains detailed and understandable solutions to problems using various methods and formulas. This digital product is ideal for students and teachers who are studying calculus and want to improve their knowledge and skills. Beautiful design in html format makes using this product even more convenient and attractive. You can easily download this digital product and start using it right away!

IDZ 4.1 – Option 12. Solutions Ryabushko A.P. is a digital product containing detailed solutions to problems in mathematical analysis for IDZ 4.1, version 12. It contains canonical equations for the ellipse, hyperbola and parabola, as well as equations of a circle and line, graphs of curves in polar and parametric coordinate systems.

For an ellipse centered at point F(-5, 15), semi-major axis a = √13 and semi-minor axis b = 2, the canonical equation has the form ((x + 5)^2)/13 + ((y - 15)^2 )/4 = 1, and eccentricity ε = 5√29/29.

For a hyperbola with center at point F(5, -12), semi-major axis a = 5 and semi-minor axis b = 13, the canonical equation has the form ((x - 5)^2)/25 - ((y + 12)/13) ^2 = 1, and the equations of the asymptotes of the hyperbola y = ±kx have the form y = ±12/13x. The directrix of the curve has the equation x = 5 - (25/12) = -5/12, and the focal length 2c = √(a^2 + b^2) = √(5^2 + 13^2) ≈ 14.04.

For a parabola with an axis of symmetry Ox and a vertex at point A(-5, 15), the equation has the form y^2 = 8(x + 5).

The equation of a circle passing through points A and B and having a center at point C (x_C, y_C) can be written as (x - x_C)^2 + (y - y_C)^2 = r^2, where r is the radius of the circle . The value of r can be found using the distance between the center of the circle and one of the given points.

The left focus of the hyperbola is 3x^2 - 5y^2 = 30 with coordinates (c, 0), where c = √(a^2 + b^2) = √(30/3) = 3. Thus, the left focus of the hyperbola has coordinates (3, 0).

The equation of a line, each point of which is located at a distance three times greater from the point A(2, 1) than from the straight line x = -5, has the form y = 7.

The curve defined by the equation in the polar coordinate system ρ = 1/(2 - sinφ) has the form of a cardioid.

The curve defined by the parametric equations x = 4cos^3(t), y = 5sin^3(t) for 0 ≤ t ≤ 2π is an ellipse.

The solutions are prepared in Microsoft Word 2003 using the formula editor, which makes using the product convenient and attractive for students and teachers studying mathematical analysis.

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IDZ 4.1 – Option 12. Solutions Ryabushko A.P.

a) Canonical equation of the ellipse: ((x - A_x)^2) / (a^2) + ((y - A_y)^2) / (b^2) = 1

where A(x,y) are the coordinates of the center of the ellipse, a and b are the lengths of the major and minor semi-axes, respectively.

For a given ellipse: A(-1, 2), a = sqrt(13), b = sqrt(10), F1(0,2), F2(-2,2)

The focal coordinates are defined as F1(x,y) = (A_x + c, A_y), F2(x,y) = (A_x - c, A_y), where c = sqrt(a^2 - b^2)

For a given ellipse: c = sqrt(3), ε = c/a = sqrt(3/13)

b) The canonical equation of a hyperbola: ((x - A_x)^2) / (a^2) - ((y - A_y)^2) / (b^2) = 1

where A(x,y) are the coordinates of the center of the hyperbola, a and b are the lengths of the major and minor semi-axes, respectively.

For a given hyperbola: A(3,0), a = sqrt(17), b = sqrt(8), F1(4,0), F2(2,0)

The focal coordinates are defined as F1(x,y) = (A_x + c, A_y), F2(x,y) = (A_x - c, A_y), where c = sqrt(a^2 + b^2)

For a given hyperbola: c = sqrt(25) = 5, ε = c/a = sqrt(25/17)

y = ±(b/a)x - A_y - (b^2/a^2)(A_x - F_x)

For a given hyperbola: y = ±(2.8284)x - 0 - (8/17)(3 - 4)

c) Canonical equation of a parabola: y = a(x - A_x)^2 + A_y

where A(x,y) is the vertex of the parabola, a is the parameter of the parabola.

For this parabola: A(-5.0), axis of symmetry Ox, a = 1/3

1. The equation of a circle with center at point A(x,y) and passing through points B(x_1,y_1) and C(x_2,y_2) has the form:

(x - x_0)^2 + (y - y_0)^2 = R^2

where R is the radius of the circle, x_0 = A_x, y_0 = A_y.

For this problem: A(-4, 6), B(-2, 8), C(-6, 4)

Let's find the radius of the circle: R = sqrt((x_1 - x_0)^2 + (y_1 - y_0)^2) = sqrt(20)

Equation of a circle: (x + 4)^2 + (y - 6)^2 = 20

1. The equation of a line passing through point A(x_0, y_0) and located at a distance d from the straight line x = k has the form:

|y_0 - k| = d

For this problem: A(2, 1), k = -5, d = 3 * |k - x_0| = 21

Line equation: |y - (-5)| = 21

1. The graph of a curve defined in polar coordinates by the equation ρ = f(φ) is constructed as follows:

• on the interval φ_1 ≤ φ ≤ φ_2 the function f(φ) is constructed and the points (ρ,φ) corresponding to these values ​​are found;
• these points define the curve in polar coordinates.

For this problem: ρ = 1/(2 - sinφ)

1. The graph of a curve defined parametrically by the equations x = f(t), y = g(t) is constructed as follows:

• on the interval t_1 ≤ t ≤ t_2 the values ​​of the functions f(t) and g(t) are found for each value of the parameter t;
• these values ​​correspond to the coordinates of the points (x,y) on the curve graph.

For this problem: x = 4cos(3t), y = 5sin(3t), where 0 ≤ t ≤ 2π.

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