159. THEOREM. If a line be drawn to the focus from the intersection of a secant with the directrix, it will bisect the exterior angle of the triangle formed by the chord and the focal radii. P'M', and P"M" being parallel, M M R F P" But and Therefore (§ 69, Prob. 7) the line FR bisects the exterior angle of the ▲ P'FP". COROLLARY. If P' and P" should coincide at P, PR would be a tangent and FR would be perpendicular to the focal chord through F and P. 160. THEOREM. A tangent to a parabola bisects the angle formed by PF and PM. The analysis suggests that we establish the equality of the AMPR and FPR NOTE. One of the important uses to which this property is put is the construction of parabolic reflectors. And since the angle of incidence equals the angle of reflection, any ray of light emanating from F will be reflected to a line parallel to the axis. The parabolic reflector is made by revolving a parabola on its axis. As a geometric figure it is called a paraboloid of revolution. If rays of light should enter the parabolic reflector parallel to the axis, they would all be reflected to the focus. Exercises. 1. Show that the distance from the focus to the point of tangency equals the distance from the focus to the foot of the tangent. 2. Show that if FM be drawn, it will be perpendicular to the tangent, and will be bisected by the tangent. 3. Show that the locus of the foot of a perpendicular from the focus to the tangent will be the tangent at the vertex of the curve. 4. Show that the projection on the axis of the segment of the tangent is bisected at the vertex. (This projection is called the sub-tangent.) 161. Definitions. A normal is a line perpendicular to a tangent and passing through the point of tangency. A sub-normal is the projection on the axis of the segment of the normal included between the point of M P tangency and the axis. Thus QN is the sub-normal. THEOREM. The sub-normal of a parabola is constant, and equals the distance of the focus from the directrix. ΔΡΟΝ = Δ ΜDF. .. QN DF. = Exercise. 1. Show how, in three ways, to draw a tangent to a parabola at a given point of the curve. 162. THEOREM. If from the point of intersection of two tangents a straight line be drawn parallel to the axis of a parabola, it will bisect the chord joining the points of tangency. Р. Also since IP, is the perpendicular bisector of FM2, 2 COROLLARY. If at the point of intersection of IB with the curve a tangent be drawn, it will be parallel to the chord P1P, and the segment IB, which bisects the chord P¡P2, is itself bisected by the curve. P II, represent the tangent at C, intersecting IP, and IP By the theorem, a straight line through I, parallel to the axis, bisects CP IH, is parallel to IC (each being parallel to the axis). .. I is the middle point of IP1. In the same manner it is shown that I is the middle point of IP2. II, bisecting two sides of the AIP,P2, is parallel to the side PP, and bisects IB. Q. E. D. PROBLEM. Show how to construct an arc of a parabola having given a pair of tangents and the points of tangency. P E CD B H FIG. 355. The analysis suggests the following: Join PP2; bisect it at B. Join BI; bisect it at A. Join AP; bisect it at C. Draw CE parallel to BI; bisect it at D. Join AP2; bisect it at II. Draw IIJ parallel to BI; bisect it at K. The points P, D, A, K, and P2 are five points on the required curve. As many points as are desired may be found in the same way, and the curve traced through them will be the arc of the parabola required. NOTE.The method described in the above problem is one of the processes frequently used for constructing parabolic railroad curves. 163. PROBLEM. Show that the area included between an arc and a chord of a parabola is two-thirds the area of the triangle formed by the chord and the tangents at its extremities. Let PP represent the chord, PCP, the arc, IP1 one tangent, and IP, the other. B is the middle point of P1P1⁄2. Join IB. tangent at C and draw the chords CP1 and CP2. Draw a The area of the APCP, is double the area of the twice the base II and If at K and at H tangents be drawn to the curve, and chords be drawn from K to P1 and to C, and also from II to C and P, we shall have I I FIG. 356. an inscribed convex polygon, and an escribed re-entrant polygon. The additions to the area of the inscribed triangle (which have formed the inscribed polygon of 5 sides) are double the additions to the area of the escribed triangle (which have formed the escribed polygon of 5 sides). For these reasons the inscribed polygon of 5 sides has twice the area of the escribed polygon. If at the points of intersection of the new tangents with those previously constructed, lines be drawn parallel |