When a right circular cone is cut by a plane that passes through the vertex can be generated the following curves:
Ellipse (e): the cutting plane is not parallel to any generating and cuts all the generators of the cone. Is a closed curve without points at infinity.
Parable (p): the cutting plane is parallel to a generator and has a point at infinity.
Hyperbola (h): the cutting plane is parallel to 2 generative and has two points at infinity.
Circumference: the plane is orthogonal to the axis of revolution and is a particular case of the ellipse in which the two axes are equal.
Generation
projective conics.
As homology across all sets of points correspond to other series and straight beams correspond to other beam lines, a conic becomes another conical homology. In the figure we see several properties of homologies on conical points aligned counterparts SS'están the center of projection O, the homologous lines aa 'cut into the shaft, the axis points are double or counterparts of themselves , which means that if a figure cut to stop homology in the points, his counterpart goes through these two points, resulting in these unchanged.
The tangents to a conic are also to its counterpart. In two peer reviews on a pole and polar have their counterparts in other pole and polar of the other.
Any conic can be transformed into a circle, so we get points on the circumference counterparts thus obtaining new points of the conic.
Given two sets of points on a conic SAB EDC, if we join each of them with the other set, except that faces it, we at the intersection of all three lines MNO points are aligned. As we have seen this procedure used to obtain new points of the conic, defined by five points to get the sixth, seventh, and so on.
As in geometry points and lines are correlative terms, can be exchanged for each other, so a conic defined by five points can be transformed into another set of five straight. Any of these five elements can be transformed with the result that we can build any cone with five elements between points and lines.
abcd Given four lines and a point M incident in one of them d determine the conic through the point is tangent to given lines. Make a circle with a diameter either (in yellow) tangent to the incident point on the line d M is going to be the focus. Taking one of the lines d as the axis of homology, the counterpart of the first line c is the tangent to the circle from the intersection of the line d axis c. The intersection of each line c and its counterpart c 'with the other line by its counterpart b' adjacent respectively, determine a new line that is part of a radiation (green lines) in which vertex is the center of homology. If we get new points of the ellipse, we take any two homologous points, eg RR '. Using any straight line passing through R 'we see that intersects the circle at P', we need to align the center of homology with P 'until it intersects the axis of the point V, and V with R we align the intersection with OP "a point of the ellipse P.
Another example similar to above but with the outer edge of the cone. We have several points on the conic and a tangent ABCE m on a point of it A. Making a yellow circle tangent to the given line m and passing through point A, and aligning the center of homology to the points of the ellipse ECB, we at the intersection with the circumference of their counterparts B'C'E '. Using any straight line passing through 2 points of the circle C 'D' intersects the axis at a point L, which together with the known homologous point of the ellipse C have at the intersection of this line CL with the double line AD 'a new point on the ellipse D.
Correlate the previous year, as data have four points and one incident on a tangent to the conic. Trying to determine the conic that passes through the points BCDE and tangent to the line to point B. If we take the point B as the center of homology and make a circle (in yellow) to the line tangent to this point B, we build double lines (lines through the center of homology B and given points) passing through points CDE. These lines intersect the circle at points C'D'E ', respectively. EC counterparts
C'E'se Lines intersect at a point of Y axis, another pair of lines CD C'D 'intersect at another point on the axis S, two points determine YS shaft position.
To construct the tangents to the conic from any point T, we f tangent from that point on the circumference. Align the point of tangency S 'with the center of homology B until it intersects the axis at a point. We calculate the counterpart of this item and get the point of tangency S with the tangent cone and tangent homologous fu f 'to the circle.
In previous cases, the elements, fixed points or are tangent lines were always incidents (where there was a tangent, there was a point of the cone that was about it) and therefore there was a single cone passing for points and was tangent to the lines. In the event that the items are no incidents, the solution is not just a single cone but can be, for example, if we have three points and two tangents to the conic, as is the statement of this drawing, exercise has four solutions. In the exercise free But one solution appears resolved, one in which the yellow circle is tangent to two given lines and passing through two points of the conic.
If you align the center point of homology E intersects the circle O at point E '. At this point you align with a point T 'of the circle and intersects the axis at a point determined together with E T at the intersection of the line OT. " T is a conical point and if we join his counterpart T 'to the point of tangency G' we have an intersection point on the axis have joined the T point in prolonging the G-spot at the intersection with the tangent b.
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