[Figure 14]

FIG. 14

74.  Conic through five points. Pascal’s theorem furnishes an elegant solution of the problem of drawing a conic through five given points.  To construct a sixth point on the conic, draw through the point numbered 1 an arbitrary line (Fig. 14), and let the desired point 6 be the second point of intersection of this line with the conic.  The point L = 12-45 is obtainable at once; also the point N = 34-61.  But L and N determine Pascal’s line, and the intersection of 23 with 56 must be on this line.  Intersect, then, the line LN with 23 and obtain the point M.  Join M to 5 and intersect with 61 for the desired point 6.

[Figure 15]

FIG. 15

75.  Tangent to a conic. If two points of Pascal’s hexagon approach coincidence, then the line joining them approaches as a limiting position the tangent line at that point.  Pascal’s theorem thus affords a ready method of drawing the tangent line to a conic at a given point.  If the conic is determined by the points 1, 2, 3, 4, 5 (Fig. 15), and it is desired to draw the tangent at the point 1, we may call that point 1, 6.  The points L and M are obtained as usual, and the intersection of 34 with LM gives N.  Join N to the point 1 for the desired tangent at that point.

76.  Inscribed quadrangle. Two pairs of vertices may coalesce, giving an inscribed quadrangle.  Pascal’s theorem gives for this case the very important theorem

Two pairs of opposite sides of any quadrangle inscribed in a conic meet on a straight line, upon which line also intersect the two pairs of tangents at the opposite vertices.

[Figure 16]

FIG. 16

[Figure 17]

FIG. 17

For let the vertices be A, B, C, and D, and call the vertex A the point 1, 6; B, the point 2; C, the point 3, 4; and D, the point 5 (Fig. 16).  Pascal’s theorem then indicates that L = AB-CD, M = AD-BC, and N, which is the intersection of the tangents at A and C, are all on a straight line u.  But if we were to call A the point 2, B the point 6, 1, C the point 5, and D the point 4, 3, then the intersection P of the tangents at B and D are also on this same line u.  Thus L, M, N, and P are four points on a straight line.  The consequences of this theorem are so numerous and important that we shall devote a separate chapter to them.

77.  Inscribed triangle. Finally, three of the vertices of the hexagon may coalesce, giving a triangle inscribed in a conic.  Pascal’s theorem then reads as follows (Fig. 17) for this case: 

The three tangents at the vertices of a triangle inscribed in a conic meet the opposite sides in three points on a straight line.