332 INCIDENT AND REFLECTED RAYS. With O as centre and any radius less than OA, describe a circle cutting the circle on OA in a and a,, and the perpendicular through O in b and b,. 1 Draw Aa meeting parallels to OB through and b1 in P and P1, and draw da, meeting the same parallels in Q and Q1. P, Q, P, and Q, will be points on the required locus, for the triangles ObQ, Oa,Q, e.g. are equal in all respects. 1 Similarly any additional number of points can be determined as shewn. The curve extends to an infinite distance on both sides of O, and has an asymptote parallel to OB on the opposite side to A and at the same distance from OB as A; or if AN be drawn perpendicular to OB and NX on it be made equal to AN, the asymptote passes through X. The internal and external bisectors of the angle AOB are tangents at O to the two branches of the curve passing through that point. The tangent at A is inclined to ОA at an angle OAT = angle AOB, and parallels to OB at distances from it=0A are tangents to the curve. The points of contact L and M of these last are determined by drawing LAM perpendicular to OA. At some point beyond M the curve becomes convex to the asymptote. This problem is a solution of the question:-to find the point on a spherical mirror, on which a ray from any point A must impinge in order that it may be reflected parallel to a given direction. For if O be the centre of the mirror, the circular arc representing the section of the mirror by the plane passing through A, O, and the line OB through O parallel to the given direction, will of course cut the curve in points such that the incident and reflected rays make equal angles with the normals at those points. In other words the problem is to find the point P on a given circle at which the lines AP, PB, A being a given point and PB being parallel to a given line make equal angles with the normal at P. The whole curve in such a case need not be drawn, since it is easy to find points on the curve in the neighbourhood of the part of the mirror required and to draw an arc of the curve through them. PROBLEM 167. Given three points A, B, C, to determine the locus of a point P moving so that the angles which PC makes with PA and PB are always equal (Fig. 167). Let AC be greater than BC. On AC and BC as diameters describe circles, and with centre C and any radius not greater than and the Q BC describe an arc cutting the circle on AC in a and circle on BC in b and b1. The lines Aa, Aa, will intersect both the lines Bb and Bb, in points on the required locus. Only three of the intersections are shewn in the figure, viz. the points P, and R, the fourth not falling within the limits of the paper. Similarly any additional number of points can be determined as shewn. The curve extends to an infinite distance on both sides of the line AB, and has an asymptote parallel to the line joining C to the centre point of AB, and which cuts AB between A and D the foot of the perpendicular from C on AB at a distance DE from D, which may be thus determined. 334 Let INCIDENT AND REFLECTED RAYS. BC=a, AC=b, AD=m, BD =n and CD=h. It can be shewn analytically that the length On DA make DF = DB, therefore AF=m ·N. Draw FG perpendicular to AB meeting BC in G, so that FG2DC=2h; 2 2 .. AG2 = AF¦2 + FG|2 = m−n|2 + 2h|2. Draw GK perpendicular to AG meeting AB in K, so that by similar triangles ᎪF : AG :: ᎪᏀ : ᎪK ; .. AG2=AF.AK. In the figure K is beyond the limits of the paper, but if AG is bisected in g, and gk is drawn perpendicular to AG meeting AB in k, Ak=AK and therefore AG 22. AF. Ak. The above expression for DE therefore becomes Draw CL perpendicular to AC and make CL = CB so that On AB make Al= AL, and through 7 draw IM parallel to KL meeting AL in M. (In the figure AL is bisected in L, so that kД, is parallel to KL.) By similar triangles 1 1 i.e. AM will be the required length DE. The asymptote can then be drawn through E parallel to the line joining C to the middle point of AB. The internal and external bisectors of the angle ACB are tangents at C to the two branches of the curve passing through that point. The tangents AT, BT, at A and B make angles CAT, CBT, with CA and CB equal respectively to the angles CAB, CBA. This problem is a solution of the question :—to find the point on a spherical mirror on which a ray from A must impinge in order that it may be reflected to B;—for if C be the centre of the mirror, the circular arc representing the section of the mirror by the plane passing through A, B and C will of course cut the curve in points such that the rays from A and B make equal angles with the normals at the points. In other words the problem is to find the point P on a given circle at which the lines AP, BP, A and B being given points make equal angles with the normal at P. The whole curve in such a case need not be drawn, since it is easy to find points on the curve in the neighbourhood of the point required and to draw an arc of the curve through them. Magnetic curves. The locus of the vertex of a triangle described on a given base and having the sum of the cosines of the base angles constant, is called a magnetic curve. If AB be the given base, and P a point on the locus, we must therefore have cos PAB+ cos PBA =k, and corresponding to different values of k, we get a series of curves passing through A and B. These represent the lines of force in the plane of the paper due to a magnet whose poles are the points A and B. The greatest value of k is 2, since the numerical value of the cosine of an angle is never > 1, 1, and k may have any value between O and 2. PROBLEM 168. To draw a magnetic curve, the base AB and the constant k being given (Fig. 168). On AB as diameter describe a circle ARQB, and on AB take a point M such that AM = k. AB. |