IDZ 4.1 – Option 16. Solutions Ryabushko A.P.

Drawing up canonical equations of curves: a) ellipse: To draw up the equation of an ellipse, you need to know the coordinates of its foci and the lengths of the major and minor semi-axes. The canonical equation of the ellipse has the form: ((x-x0)^2)/a^2 + ((y-y0)^2)/b^2 = 1, where (x0, y0) are the coordinates of the center of the ellipse, a and b - the lengths of the major and minor semi-axes, respectively. The eccentricity value is calculated using the formula ε = √(1 - (b^2/a^2)).

b) hyperbolas: To compose the equation of a hyperbola, you need to know the coordinates of its foci and the distance between them (2c). The canonical equation of a hyperbola has the form: ((x-x0)^2/a^2) - ((y-y0)^2/b^2) = 1, where (x0, y0) are the coordinates of the center of the hyperbola, a and b - the lengths of the major and minor semi-axes, respectively. The eccentricity value is calculated using the formula ε = √(1 + (b^2/a^2)).

c) parabolas: To compose the equation of a parabola, you need to know the coordinates of its vertex and the parabola parameter p (the distance from the vertex to the directrix). The canonical equation of a parabola has the form: y^2 = 2px, where p is the parameter of the parabola.

1.16 a) For an ellipse with eccentricity ε = 3/5 and point A(0, 8), the canonical equation has the form: ((x-0)^2)/(a^2) + ((y-8)^2) /(b^2) = 1, where a = 8/√34, b = 8/√10. b) For a hyperbola with points A(√6, 0), B(−2√2, 1) and focus F(3, 0), the canonical equation has the form: ((x-3)^2)/16 - (( y-0)^2)/2 = 1. c) For a parabola with directrix D: y = 9 and vertex A(0, 9), the canonical equation has the form: y^2 = 36x.

Equation of a circle: The equation of a circle in general form is: (x-a)^2 + (y-b)^2 = r^2, where (a, b) are the coordinates of the center of the circle, r is the radius of the circle. To find the equation of a circle passing through two given points and having a center at point A, it is necessary to find the midpoint of the segment connecting these points and the radius of the circle equal to the distance from the center to any of these points. Thus, the equation of a circle has the form: (x-x0)^2 + (y-y0)^2 = r^2, where (x0, y0) are the coordinates of point A, r is the radius of the circle.

2.16 To construct a circle passing through the point B(1, 4) and having a center at the vertex of the parabola y^2 = (x-4)/3, it is necessary to find the radius of the circle and its center. The radius is equal to the distance from the center of the circle to point B, that is, √((1-4)^2 + (4-4/3)^2) = √(17/3). The center of the circle is located in the middle of the segment between point B and the vertex of the parabola, that is, at the point ((1+4)/2, (4+4/3)/2) = (5/2, 16/3). Thus, the equation of a circle is: (x-5/2)^2 + (y-16/3)^2 = 17/3.

Equation of a straight line: The equation of a straight line in general form is: y = kx + b, where k is the slope coefficient of the straight line, b is the free term. To find the equation of a line passing through point M and satisfying the condition of the ratio of the distances from point M to points A and B, it is necessary to find the coordinates of the point of intersection of this line with the line passing through points A and B. The slope coefficient of the line passing through points A and B , is equal to (5+4)/(-3-2) = -3/7, and its free term is equal to (25-43)/(-3-2) = -2/5. The distance from point M to point A is √((x-2)^2 + (y+4)^2), and the distance from point M to point B is √((x-3)^2 + (y-5 )^2). Therefore, the distance ratio condition can be written as: √((x-2)^2 + (y+4)^2) / √((x-3)^2 + (y-5)^2) = 2/ 3. Solving this equation for y, we get: y = (2x+8)/5 - 4/5√((x-2)^2 + (y+4)^2) / √((x-3)^2 + (y-5)^2).

3.16 The equation of the straight line that satisfies the conditions of the problem has the form: y = (2x+8)/5 - 4/5√((x-2)^2 + (y+4)^2) / √((x-3) ^2 + (y-5)^2).

Equation of a curve in polar coordinates: The equation of a curve in polar coordinates has the form: ρ = f(φ), where ρ is the distance from the origin to a point on the curve, φ is the angle between the radius vector and the positive direction of the x-axis, f(φ) - a function that determines the shape of the curve.

4.16 The equation of the curve in polar coordinates has the form: ρ = 2cos(6φ).

Equation of a curve in parametric form: The equation of a curve in parametric form has the form: x = f(t), y = g(t), where x and y are the coordinates of a point on the curve, t is a parameter, f(t) and g(t ) - functions that determine the coordinates of points on a curve depending on the parameter.

5.16 Equation

IDZ 4.1 – Option 16. Solutions Ryabushko A.P. is a digital product that represents solutions to tasks from a mathematics textbook. In this version, the solutions were prepared by A.P. Ryabushko. The product is designed in a beautiful html format, which allows you to conveniently view and study the material on any device, both stationary and mobile. In addition, this product can be purchased in the online digital goods store, which makes the purchasing process easier and faster. Solutions to assignments are presented in a clear and understandable manner, which will help students easily understand the material and successfully complete assignments. This digital product will be useful for schoolchildren and students studying mathematics and wishing to improve their knowledge and skills in this area.

IDZ 4.1 - Option 16 is a mathematics problem book containing tasks on composing canonical equations of ellipses, hyperbolas and parabolas, constructing circles, equations of straight lines, solving problems of finding an equation of a straight line that satisfies a certain condition, as well as equations of curves in polar coordinates and parametric form . Solutions to the problems in this version were compiled by Ryabushko A.P. The problem book is suitable for students and schoolchildren studying mathematics at the high school level.

***

IDZ 4.1 – Option 16. Solutions Ryabushko A.P. is a collection of solutions to problems in mathematics performed by the author Ryabushko A.P. The collection presents solutions to problems of varying complexity and different branches of mathematics, including analytical geometry, function theory, differential equations and others.

In particular, the collection contains solutions to the following problems:

For three different curves (ellipse, hyperbola and parabola), construct canonical equations given by various points and parameters.
Write down the equation of a circle passing through two points and having a center at a given point.
Write an equation of a straight line, each point of which satisfies the given conditions.
Construct the curve in the polar coordinate system given by the equation.
Construct a curve given by parametric equations.

All solutions are prepared in Microsoft Word 2003 using the formula editor and contain a detailed description of the process of solving the problem.

***

I am very pleased with the acquisition of IDZ 4.1 - Option 16. Solutions by Ryabushko A.P. is a great digital product for exam preparation.
Decisions Ryabushko A.P. in IDZ 4.1 – Option 16 helped me better understand the material and successfully pass the exam.
Quality of materials IDZ 4.1 – Option 16 Solutions Ryabushko A.P. Top notch, I'm very happy with my purchase.
IDZ 4.1 – Option 16 Solutions Ryabushko A.P. is an excellent choice for those who want to get high scores in the exam.
I recommend IDZ 4.1 – Option 16 Solutions by A.P. Ryabushko for anyone looking for a quality digital exam preparation product.
IDZ 4.1 – Option 16 Solutions Ryabushko A.P. very well structured and clear, making exam preparation easy and efficient.
Using IDZ 4.1 – Option 16 Solutions Ryabushko A.P. I was able to improve my knowledge and skills in the subject, which played a big role in the exam.
IDZ 4.1 – Option 16 is an excellent digital product for preparing for the computer science exam.
Decisions Ryabushko A.P. help to understand the material and successfully complete tasks in ILD 4.1 – Option 16.
This digital product contains useful and relevant tasks that will help improve your knowledge in computer science.
IDZ 4.1 – Option 16 is an excellent tool for self-preparation for the exam.
Decisions Ryabushko A.P. in IDZ 4.1 – Option 16 are very clear and easy to understand.
This digital product allows you to effectively use your time to prepare for your computer science exam.
IDZ 4.1 – Option 16 contains many useful tips and recommendations for completing tasks.
Decisions Ryabushko A.P. help improve problem solving skills in computer science.
With this digital product you can quickly and effectively prepare for the computer science exam.
IDZ 4.1 – Option 16 is highly recommended for anyone who wants to successfully pass the computer science exam.