Analytical Geometry, study of the Ellipse - Class 1

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Hello friends greetings, noting that in maths I am very good I want to start a proposal today, I will publish a series of content related to this science of mathematics, which I will name for each publication as classes, in this occasion I will publish classes 1, which is referring to the analytical geometry, specifically to the study of the ellipse, if this geometrical figure which is very important for science since it is much more than that, you knew friend that the figure of the ellipse is taken as reference in the engineer to adapt to the figure more similar to the earth, because if, in my professional area, you take the ellipse as the geometrical figure more adapted to the surface, which allows us to determine certain mathematical parameters, this geometric figure describes in a perfect way the semi-axes that make it up, we use them to determine measurements related to the radius of the earth, it sounds very interesting All this truth, because I invite you to observe this great publication which is based on practical exercises applying the figure of the ellipse.

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The ellipse is the locus of the points - for example point D - whose sum of the distances to two fixed points AB of the plane called foci, is a constant quantity equal to the major axis of the ellipse -C and its symmetric with respect to the Axis y.
If the sum of the distances to those two fixed points called foci is equal to the distance between them (DA + DB = AB), the ellipse becomes a segment of vertices in the foci.

The DA DB segments are called focal rays and by agreement this distance - equal to the major axis - is called 2a. consequently, for every point D of the ellipse DA + DB = 2a.
The distance between the two bulbs AB is called 2c.
The minor axis is named with the letter 2b.

As the sum of the two radii DA + DB is equal to the major axis 2a

DA + DB = 2a,

and knowing that the bulbs AB are symmetric about the axis and the origin of coordinates, with coordinates (-c, zero) and (+ c, zero), we can substitute the formula of the distance for both radii:
DA will be equal to the square root of (x -.- c) squared plus (y-0) raised square.
DB will be equal to the square root of the same elements but positive c, then we will have that is equal to the square root of (x-c) squared plus and squared.
If we develop this equation to obtain a simpler one, we can obtain the canonical equation of the ellipse:

x to the square divided by the semimajor axis to the square plus and to the square divided by the semiaxis less than the square is equal to one.

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In the image we can see an ellipse whose center D has coordinates by x 5 units while by coordinate in y has 3 units. D = (5.3)
As indicated by the blue and yellow arrow we have to place these coordinates on the numerator behind the x and behind the and respectively, so that x minus the first quantity that is 5 all squared (x-5), we divide it by 9 which is the semimajor axis to the square -in green-, then we add (y-3), everything to the square, knowing that 3 was the distance from the center of the ellipse to the horizontal axis, that is, the corresponding measure of the center to the x-axis, we divide all this by 5, which is the semiaxis less than the square (that is, 2.24 squared, measured in red).

Once we have that complete sum we match it to one and in this way we have the equation of the ellipse.

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in the yellow ellipse we have the formula or equation of an ellipse from which we are going to calculate its vertices, foci and graph, as well as the center of it.
Since we have that the canonical equation is the one that appears in the green rectangle, being a squared equal to nine and b squared equal to five, we must divide the given expression between the second member, whose value is 45, in this way we have to the second member will have unity, as it should appear in the canonical equation. Dividing all the terms by 45, as it appears in the upper part of the drawing, we obtain after simplifying, the canonical equation of the green rectangle. Applying the square root of the denominators, we obtain the major and minor semiaxis, whose value is 9 and 5, respectively.

As we have that the sum of the two major axes or also the length of the major axis is equal to the sum of the distances between a point and both bulbs, taking as a particular case the point E, we have the distance from this point to the focus is equal to OV, since EF + EF 'is equal to VV'.
Therefore in the right triangle we have the value of the semiaxis HE and the value of the other semiaxis EF = HV, in this way we can obtain HF by applying the Pythagorean theorem: hypotenuse squared is equal to one leg squared plus the other leg squared , in this way we obtain the length of c, which is the distance from the center of the ellipse to the foci.

As in the ellipse we have that its center passes through the origin of coordinates, the coordinates of the foci are respectively (2,0) and (-2,0).
Since the semimajor axis is worth three, we have that the vertex has coordinates (3,0) and (-3,0).
Since the minor semiaxis is 2.24, which is the root of five, we have the coordinates of the vertices of the minor axis are (zero, 2.24) and (zero, -2.24).

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As we operated in the previous exercise, given the expression of the ellipse in the green rectangle, we divide both members by the coefficient of the second member whose value is 112, in this way we obtain the equation of the canonical ellipse with the second member whose value is the unit, such that it appears in the red rectangle.
Applying the square root of both denominators we have that the semimajor axis b is 2.65 and the semimajor axis is worth 4. In this way we have the coordinates of both vertices: (4.0) and (-4.0) for the semiaxis greater, while (0, 2.65) and (0, -265) we have that they are the vertices of the minor axis.
To calculate the foci, we apply the Pythagorean theorem in which we have the value of the semimajor axis and the semimajor axis, which are the hypotenuse and a leg respectively, thus replacing the data we have the value of the other leg c, which is three .

In the image appears in the lower left corner another method to calculate the foci, as we have that the hypotenuse of the right triangle is the radius of the circumference that cuts the x axis in the focus F, we can calculate the intersection of the circumference with that axis -resolving the system of equations of the circumference and the axis-, thus obtaining the coordinates of both foci: the equation of the circumference is given by the expression x minus the coordinate in x of the center D, whose value is zero, plus and minus the y-coordinate of the center, whose value is 2.65, all squared and equal, (the sum of both terms), the radius squared, which is the hypotenuse of value four.

The intersection of this circle with the line y = 0, which is the equation of the x axis, defines the value of the foci, whose coordinate in x is 3 and -3 (the coordinate in y is zero because it is a center ellipse in the origin of coordinates).
To calculate the straight side, which is the length of segment AB of a vertical line that passes through the foci until it intercepts the elliptical curve, we multiply the constant two by the semimajor axis less than the square and we divide all this between the semimajor axis, substituting the data we obtain the value 3,5, as it appears in the lower right edge of the drawing.

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We have that in the coordinates of the center are -1 and -5, so in the numerator of the equation we put x minus -1, e and minus -5, obtaining x +1 and also y +5.
We take the size of the semimajor axis, which is four units and we raise it to the square, therefore it will remain 16 in the denominator, below x +1, all this squared, we do the same with the minor semiaxis, we raise it to the square obtaining 12 and we put it below and +5, all this squared.
We match all the elements that we have just named corresponding to the first member to one, and we obtain the equation of the ellipse, which is what appears in violet.

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The yellow ellipse we can see the coordinates of point B, what are 3 and 4, that is what appears in the numerator of the first member of the equation, (x-3) squared, and divided by the semimajor axis to the square, then is the sum of (y-4), (four is the other coordinate of point B), and divided by the semiaxis less than the square, 9. The sum of both is equal to unity, and this is the equation of the ellipse.

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In the image we can see that in the yellow rectangle the equation of an ellipse, we add a unit to each member to obtain the binomial (y-1) squared, then we divide everything by 4 to obtain a 1 in the second member. In this way we already have in the blue ellipse the equation of the ellipse with its main elements and that we draw next. What remains in the denominator of each term are the semi-axes squared, while what is subtracted from y and what is subtracted from x are the coordinates of the center of the ellipse, (0,1), not forgetting that both the binomial (y-1) as the x are squared.