Practical lesson 1. Ellipse
So, the first task is on the canonical ellipse equation. The ellipse is given by this equation: 4x2 + 25y2 = 100. Note that this equation itself is not canonical. We need to bring it to the canonical form, and find the main parameters a, b, c, and determine the foci and eccentricity.
Let’s construct the specified ellipse. In order for this equation to result in a canonical form, we divide both sides of the equation by 100, in this case, it’s 1 on the right, but on the left, there is an expression we need: x2/25 + y2/4 = 1. We got the equation which is already canonical, the parameters a and b are in the denominator of the resulting expression, namely, 25 is а2, 4 is b2. The general equation of the ellipse in canonical form looks like this: x2/a2+ y2/b2 = 1, so the parameter a = 5, the parameter b = 2, while a and b are positive numbers, we remember that a is the major semi-axis, b is the minor semi-axis, so a = 5, b = 2. Since we know a and b, we can easily find the inter-focus distance c, more precisely, the inter-focus distance = 2c, and c is the distance from the origin to each focus, while the parameters c, a, b are related by the ratio c2 = a2 – b2, so c2 = 25 – 4 = 21, i.e. c is √21. So, in the canonical coordinate system, we note the origin and two foci – F1 and F2, the distance from the origin to each focus is the parameter c, the inter-focus distance is 2 times greater, so the focus F1 is located at the point (- √21; 0), the second focus F2 has the coordinates (√21; 0). Approximately, of course, you can estimate what the value of c is, it is approximate because the root is not extracted exactly, it turns out somewhere 4.5, it turns out 20.25 squared.
So, the foci are found, the parameter c is found, and the eccentricity is required. Let me remind you that the parameter called the eccentricity allows us to estimate the compression ratio of the ellipse. That is, for different values of this number, an ellipse turns out to have a different compression ratio, and for the case of an ellipse e is always less than 1. If e = 0, the ellipse turns into a circle. In order to find this value, we need to divide the parameter c by the parameter a: e = c/a, in our case, we get √21 divided by 5, and this will be the value of the eccentricity e. That is, the approximate value of e = 0.9, but we have found the exact value, it is √21/5. So, the main parameters are known, there is a canonical equation, let’s give a figure. Let’s fix the equation, all the parameters and look at the figure (see the video). So, the foci F1 and F2 have the values approximately equal to -4.5 and 4.5, the major semi-axis = 5, the minor semi-axis = 2. The ellipse in the canonical coordinate system looks like this (see the video), the eccentricity is approximately 0.9.
Let’s look at another problem. In the first case, we were given the equation of the ellipse, now the problem is reverse, and some conditions are known. We need to make the canonical equation of the ellipse, i.e. to find the parameters a and b, and write the equation of the ellipse in the desired form.
What is known to us? Its eccentricity is known, i.e. the ratio of parameter c to parameter a, and it is known at which points the foci are located. So, the inter-focus distance = 12, i.e. 2c = 12, the parameter c = 6, secondly, the eccentricity e = c/a =2/3, c is known, we substitute c in this ratio, 6/a = 2/3, from here we can easily find the parameter a: 6∙3/2 = 18/2 = 9. Note that knowing the ratio c/a, we cannot say unvalently what c and a are equal to, it is wrong to assume that if the two is in the numerator, it is equal to c, i.e. it would be wrong to conclude that c = 2, and a = 3, the ratio is 2/3, but c is three times more than 2, so a is three times more than 3, that is, a = 9. So, a = 9, let’s find b. We have c2 = a2 – b2, b has a similar ratio: b2 = a2 – c2, since we know a and c, we can easily find b: b2 = 81 - 36 = 45, which means b is √45, we can take out the nine and write it as 3√5, but we need b2 in the equation but not b, and instead of a, we write а2. So, the equation looks like this: x2/81 + y2/45 = 1. So, the canonical ellipse equation is found, and the problem is solved.
Let’s take another problem. It has an equation for a certain curve. We need to make sure that this equation defines an ellipse, and we need to construct this curve. We’ve considered the canonical equation of an ellipse before, i.e. this curve was defined in the canonical coordinate system, but here the coordinate system is such that the equation has a different form. If we remember the lecture, we understand that if necessary we can generalize some determinants, and understand what kind of curve our equation sets. On the other hand, after we find it out, the question arises: how to construct an ellipse? To do this, since we do not have terms with the product of x by y, we can transform our coordinate system in this way, or more precisely, as a transformation, we can transfer the axes so that the new coordinate system already becomes canonical. Let’s write this equation in full form, taking into account the coefficient before x and y, which is zero, and put the free term to the left side: 4x2 + 0xy + 9y2 – 16x + 18y – 11 = 0. So, if we calculate the determinant made up of the first coefficients a, b, c, this is ((4 0) (0 9)), we get 36, which is greater than 0. This means that our figure is either an ellipse, or degenerates into a point (such a degenerate ellipse turns out), or has an empty set of solutions. That is, in general, we can say that we get an ellipse that degenerates into a point, or we get a so-called imaginary ellipse, but one way or another, the figure can be called an ellipse. We suggest that we make sure that the figure is non-degenerate by creating a determinant using all the coefficients. This method is noted in the lecture.
And now we will try to transform this equation in such a way that it is clear how to construct this ellipse. To do this, since we do not have a term with a product of variables, let’s group the terms that contain only the variable x, and the terms that contain only the variable y, and use the method of selecting the full square. The coefficient 4 is taken out of the parentheses, x2–4x is obtained in parentheses, for the other two terms, we make a similar transformation, 9 is taken out, y2 + 2y remains and everything is equal to 11: 4(x2 - 4x)+ 9(y2 + 2y) = 11. Next, to get the full square, we add the necessary terms, namely + 4. In this case, we get the full square, which we will write in this form later (see the video): but note that we added + 4 in parentheses, taking into account the multiplier 4, we added 16, so that nothing changes, we need to subtract 16. A similar operation is performed with the second term, we also select the full square, they say, we add to the full square, + 1, given the coefficient 9, we add 9, so we subtract 9, so that nothing changes, and we get the square of the expression (y + 1): 4(x2 - 4x + 4) – 16 + 9(y2 + 2y + 1) – 9 =1. We move the free terms to the right side: 4(x – 2)2 + 9(y + 1)2 = 36. 36. Now, we have the familiar form, we divide by 36, we get (x – 2)2/9 + (y + 1)2/4 = 1. So, if we now reassign (x-2) by x’, (y + 1) by y’, we get exactly the canonical equation of the ellipse: x’2/9 + y’2/4 =1. The canonical coordinate system has a different origin, let it be Q, and other coordinate axes, but they are derived from these axes by parallel translation. In order to construct our ellipse, we can construct such an ellipse, and make the necessary translation. See what happens in this case. So, let’s see what parameters we have here for our ellipse, parameter a does not change, it is equal to 3, parameter b is equal to 2, the major semi-axis is 3, the minor semi-axis is 2, and the inter-focus distance is calculated as follows: c2 = a2 – b2 = 5 Þ c = √5. Here’s what we should get (see the video). Let me write down on this slide the equation that we had in the Oxy coordinate system: (x – 2)2/9 + (y + 1)2/4 = 1. Note that if we move the Оx axis down by one unit, and the Оy axis to the right by 2 units, in the new coordinate system, we just get the canonical ellipse equation that was written on the previous slide. So, the major semi-axis is 3, the minor semi-axis is 2, and here is the ellipse itself (see the video). So, in the Qx`y` system, it looks like this (see the video). In the original coordinate system, it is obtained by transferring to the vector with the coordinates (2; -1). The problem is solved.