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You will need

• Knowledge of analytical geometry. Basic knowledge of algebra.

Instructions

1

The equation of a straight line is given by the coordinates of two pointsOn the plane through which this line must pass. Let us form the ratio of the coordinates of these points. Let the first point have the coordinates (x1, y1), and the second (x2, y2), then the equation of the line will be written as follows: (x-x1) / (x2-x1) = (y-y1) (y2-y1).

2

We transform the resulting equation to a straight line and express y explicitly in terms of x. After this operation, the equation of the straight line will take the final form: y = (x-x1) / ((x2-x1) * (y2-y1)) + y1.

Instructions

1

In the figure, we chose points A and B. It is convenient to choose the points of intersection with the axes. Two points are sufficient to pinpoint the line.

2

Find the coordinates of the selected points. To do this, lower the perpendiculars from the points on the coordinate axes and write down the numbers from the scale. So for point B from our example, the coordinate x is -2, and the coordinate y is 0. Similarly, for point A, the coordinates are (2; 3).

3

It is known that the equation straight has the form y = kx + b. We substitute in the equation in the general form, the coordinates of the selected points, then for point A we obtain the equation: 3 = 2k + b. For point B we get another the equation: 0 = -2k + b. Obviously, we have a system of two equations with two unknowns: k and b.

4

Next, we solve the system in any convenient way. In our case, we can add the equations of the system, since the unknown k occurs in both equations with coefficients that are the same in absolute value but opposite in sign. Then we get 3 + 0 = 2k - 2k + b + b, or, which is the same: 3 = 2b. Thus, b = 3/2. We substitute the found value of b into any of the equations to find k. Then 0 = -2k + 3/2, k = 3/4.

5

We substitute the found k and b in the equation general form and obtain the desired the equation straight: y = 3x / 4 + 3/2.

Instructions

1

Parabola is a curve thatIts shape resembles an arc and is a graph of the power function. Regardless of what characteristics the parabola has, this function is even. An even function is called an even function whose value does not change for all values ​​of the argument from the domain of the definition when the sign of the argument changes: f (-x) = f (x) Start with the most simple function: y = x ^ 2. From its form, we can conclude that it increases both for positive and negative values ​​of the argument x. The point at which x = 0, and in this case, y = 0 is considered the minimum point of the function.

2

Below are all the main options for buildingthis function and its equation. As a first example, we consider a function of the form: f (x) = x ^ 2 + a, where a is an integer. In order to plot the graph of a given function, it is necessary to shift the graph of the function f (x) by a units. An example is the function y = x ^ 2 + 3, where along the y-axis the function is shifted upwards by two units. If a function with the opposite sign is given, for example, y = x ^ 2-3, then its graph is shifted down along the y axis.

3

Another kind of function that can be specifiedthe parabola is f (x) = (x + a) ^ 2. In such cases, the graph, on the contrary, shifts along the abscissa (x-axis) by a units. For example, we can consider the functions: y = (x +4) ^ 2 and y = (x-4) ^ 2. In the first case, where there is a function with a plus sign, the graph is shifted along the x axis to the left, and in the second case to the right. All these cases are shown in the figure.

4

There are also parabolic relations of the form y = x ^ 4. In such cases, x = const, and y sharply increases. However, this only applies to even functions. parabolas are often present in physical problems, for example, the flight of a body describes a line that is similar to a parabola. Also view parabolas has a longitudinal section of the reflector of the headlamp, the lantern. Unlike a sinusoid, this graph is non-periodic and increasing.

Instructions

1

The basis of any constructions in geometry is the conceptdistance between two points in space. A straight line is a line parallel to this distance, and this line is infinite. Through two points, you can draw only one straight line.

2

Graphically, the line is represented as a line with unlimited ends. Direct can not be depicted as a whole. However, this accepted schematic image implies care straight to infinity in both directions. Straight is indicated on the graph in lowercase Latin letters, for example, a or c.

3

Analytic line in the plane is given the equationm of the first degree, in space - a system of equations. There are general, normal, parametric, vector-parametric, tangential, canonical equations straight through the Cartesian coordinate system.

4

The canonical the equation straight follows from the system of parametric equations. The parametric equations straight are written in the following form: X = x_0 + a * t; y = y_0 + b * t.

5

In this system the following notations are accepted: - x_0 and y_0 are the coordinates of some point N_0 belonging to straight; - a and b are the coordinates of the directing vector straight (owned or parallel to it); - x and y are the coordinates of an arbitrary point N on straight, where the vector N_0N is collinear with the direction vector straight; - t is a parameter whose valueis proportional to the distance from the initial point N_0 to the point N (the physical meaning of this parameter is the time of rectilinear motion of the point N along the direction vector, that is, for t = 0 the point N coincides with the point N_0).

6

Thus, the canonical the equation straight is obtained from the parametric by dividing one equation into another by eliminating the parameter t: (x - x_0) / (y - y_0) = a / b. From: (x - x_0) / a = (y - y_0) / b.

7

The canonical the equation straight in space is given by three coordinates, therefore: (x - x_0) / a = (y - y_0) / b = (z - z_0) / c, where c is the applied vector of the directing vector. Moreover, a ^ 2 + b ^ 2 + c ^ 2? 0.

Instructions

1

Let two lines in the space be given by the canonical equations: (x-x1) / q1 = (y-y1) / w1 = (z-z1) / e1; (x-x2) / q2 = (y-y2) / w2 = ( z-z2) / e2.

2

The numbers q, w and e, represented in the denominators, are the coordinates of the directing vectors to these lines. A non-zero vector is referred to as a guide, which lies on this straight or is parallel to it.

3

The cosine of the angle between the lines has the formula: cosλ = ± (q1 · q2 + w1 · w2 + e1 · e2) / √ [(q1) ² + (w1) ² + (e1) ²] · [(q2) ² + (w2 ) ² + (e2) ²].

4

Direct, given by canonical equations,are mutually perpendicular if and only if their directing vectors are orthogonal. That is, the angle between the straight lines (which is the angle between the directing vectors) is 90 °. The cosine of the angle in this case is zero. Since the cosine is expressed by a fraction, its equality to zero is equivalent to the zero denominator. In coordinates this will be written as q1 · q2 + w1 · w2 + e1 · e2 = 0.

5

For straight lines in the plane, the chain of reasoning looks similar, but the perpendicularity condition will be slightly more simplified: q1 · q2 + w1 · w2 = 0, since the third coordinate is absent.

6

Now let the lines be given by the general equations: J1 · x + K1 · y + L1 · z = 0; J2 · x + K2 · y + L2 · z = 0.

7

Here, the coefficients J, K, L are the coordinates of the normal vectors. The normal is the unit vector perpendicular to straight.

8

The cosine of the angle between the lines is now written in the following form: cosλ = (J1 · J2 + K1 · K2 + L1 · L2) / √ [(J1) ² + (K1) ² + (L1) ²] · [(J2) ² + (K2) ² + (L2) ²].

9

The lines are mutually perpendicular in the case when the normal vectors are orthogonal. In vector form, respectively, this condition looks like this: J1 · J2 + K1 · K2 + L1 · L2 = 0.

10

The straight lines in the plane given by the general equations are perpendicular when J1 · J2 + K1 · K2 = 0.

You will need

• A sheet of paper, a ballpoint pen.

Instructions

1

Set two fixed points F1 and F2 on the plane. The distance between the points will be equal to some fixed value F1F2 = 2c.

2

Draw a straight line on the sheet of paper that iscoordinate axis of abscissa, and draw points F2 and F1. These points represent the foci of an ellipse. The distance from each focus point to the origin must be equal to the same value equal to c.

3

Draw the ordinate axis, thus forming a Cartesian coordinate system, and write the basic equation defining the ellipse: F1M + F2M = 2a. The point M denotes the current point of the ellipse.

4

Determine the value of the segments F1M and F2M usingthe Pythagorean theorem. Keep in mind that the point M has the current coordinates (x, y) relative to the origin, and with respect to, say, the point F1, the point M has the coordinates (x + c, y), that is, the "X" coordinate acquires a shift. Thus, in the expression for the Pythagorean theorem, one of the terms must be equal to the square of the quantity (x + c), or the value of (x-c).

5

Substitute the expressions for the modules of the vectors F1M andF2M in the basic ratio of the ellipse and set both sides of the equation square, first moving one of the square roots to the right side of the equation and opening the brackets. After reducing the same terms, divide the resulting ratio by 4a and re-raise it to the second power.

6

Give such terms and collect the terms with the same factor of the square of the "ix" variable. Put the square of the "Ix" variable outside the bracket.

7

Label a square of a certain value (say,b) the difference between the squares of a and c, and divide the expression obtained by the square of this new value. Thus, you have obtained the canonical equation of the ellipse, in the left part of which the sum of the squares of coordinates divided by the values ​​of the axes, and in the left - one.