General 4Given five tangent lines, construct the conic section and the five points of tangency.  Here, in order to avoid cluttering the image, the tangent lines are given as the produced sides of pentagon (or anti‑pentagon) ABCDE. This problem is solved in Principia (Book I, Proposition 27, Problem 19). Newton’s proof is provided here, somewhat expanded, because Newton could be rather terse. It refers back to two other sections of Principia, Book I, those being Lemmas 24 and 25. They are presented separately as General 4 Support. This page will now dive straight into the construction. In the Sketchpad document the given tangent lines are arranged to form a convex pentagon as a matter of practicality. Clearly they must circumscribe an ellipse. The same procedure applies to hyperbola cases, but that curve can be difficult to manage with parts flying off toward infinity. When the ellipse is completed correctly, it should be possible to drag the tangent lines and create other forms.  It begins with construction of two diameters, which will intersect at the center. Let tangent line DE be disregarded for now, and let CD and AE intersect at F. Now the sides of quadrilateral ABCF are tangent lines. Construct the diagonals AC and BF, and their midpoints P and Q. By Lemma 25, line PQ is a diameter.  Now do it again. This time AB is disregarded, and let BC and EA meet at G, forming quadrilateral GCDE, all four sides being tangent lines. Points R and S are midpoints of diagonals GD and CE. Line RS is another diameter. The two diameters meet at O, the center of the conic section.  Most of those construction objects can now be hidden, but we will need to keep center O. A circumscribed trapezoid will be needed. Let AB be one base, and let the adjacent sides, BC and AE be the legs. Dilate AB with respect to O using scale factor ‑1. Its image is that green line, intersecting the legs at T and U, so ABTU is the trapezoid, and all four of its sides are tangents. Construct the diagonals AT and BU, and let them intersect at V. By Lemma 24, points V, O, and the points of tangency of the two bases are collinear. Construct line VO and let it intersect AB at a point of tangency, shown here in red. Line VO also intersects TU at a point of tangency, but that is not used here.  That last procedure is repeated with the other four tangent lines, and the sketch now has five points of tangency, the require conditions for Newton’s 5‑point conic, or Pascal’s. There also are plenty of tangent lines, useful for construction of the conic parts. The Web Sketch below has the completed construction. Drag the purple intersection points to see different forms, including hyperbolas. This particular sketch is so cluttered with parts that I have decided to present the section and points of tangency only. The other parts can be shown with the action buttons. |