Lemma 23b (Prop 25_ Icon

Proposition 25 Problem 27

Draw a trajectory that shall pass through two given points, and touch three right lines given by position.

Through the concourse of any two of the tangents one with the other, and the concourse of the third tangent with the right line which passes through the two given points, draw an indefinite right line and, taking this line for the first ordinate radius, transform the figure by the preceding

Lemma In this figure those two tangents will become into a new figure. parallel to each other, and the third tangent will be parallel to the right line that passes through the two given points. Suppose hi, kl to be those two parallel tangents, ik the third tangent, and hi a right line parallel thereto, passing through those points a, b, through which the conic section ought to pass in this new figure; and completing the paral- lelogra n BO cut in fiikl, c, d, e, right lines hi, ik, kl be that he may be to the square let the root of the rectangle ahb, ic, to id, and ke to kd. as the sum of the right lines hi and kl is to the sum of the three lines, the first whereof the right line ik, and the other two are the square roots of the rectangles ahb and alb ; and c, d, e, will be the points of contact. For by the properties of the conic sections, he 2 to the rectan is 2 2 2 2 2 gle ahb, and ic to id and ke to kd , and el to the rectangle alb, are all in the same ratio and therefore he to the square root of ahb, ic to id, ke to kdj and el to the square root of alb, are in the subduplicate of that , ; ratio ; and by composition, in the given cedents hi + kl y to the sum ratio of the sum of all the ante of all the consequents ^/ahb -- ik *Jalb, Wherefore from that given ratio we have the points of contact c, d, e, in the new figure. By the inverted operations of the last Lemma, let those : points be transferred into the first figure, and the trajectory will be there described by Prob. XIV. Q.E.F. But according as the points a, b, fall between the points //, /, or without taem, the points c, d, e, must be takenTHE MATHEMATICAL PRINCIPLES 144 Cither between the points, h, a, b, falls i, between the points the Problem is k, or without them. /, h, describe a From I, 18 trajectory that shall pass through a given point, and touch given by position. common the h, impossible. right lines four I.J If one of the points and the other xvithout the points i,

PROPOSITION 26 PROBLEM

any two intersections, of of the tangents to the common intersection of the other two, draw an indefinite right line and taking this line for the first ordinate radius ; x o / s ; transform the figure (by Lem. XXII) into a new figure, and the two pairs of tangents, each of which before concurred in the first ordinate ra- now become parallel. Let hi and kl, Al l
ik and hi, be those pairs of parallels completing the parallelogram hikl. dius, will And let p be the point in this new figure corresponding to the given point the centre of the figure draw pq.= and O? in the first figure. Through will be the other point through which the conic sec to Op, q being equal Let this point be transferred, by the tion must pass in this new figure. O inverse operation of Lem. XXII into the first figure, and there we shall have the two points through which the trajectory is to be described. But through those points that trajectory may be described by Prop. XVII. LEMMA If two right lines, as AC, BD XXIII. given by position, and terminating in given points A, B, are in a given ratio one to the other, and the right are joined is cut in line CD, by which the, indetermined points C, will be placed in a right in a given ratio ; I say, that the point D K K line given by position. BD meet in the right lines AC, as to is to take E, and in to the be given always equal AC, and let will be line and, by construction, For let BG BE BD AE FD EG GD, to EC ; that is, to as EF, AC to BD, and and therefore the therefore in a given ratio be Let will given in kind. triangle %- ; EFC CF be cut in L so as CL may be to CF E
I ,.+++ K cT^"^ in the ratio of CK to CD ; and EFL will be given in kind, and a given ratio, the triangle will be placed in the right line therefore the point given by position. will be similar and because Join LK, and the triangles CLK, because that is EL L CFD is a given line, and LK is to FD in a given ratio, ; LK FD will be also givenOF NATURAL PHILOSOPHY. SEC. V.] To 145 ELKH EH will be always a parallelogram. be taken equal, and is always therefore the point placed in the side (given by po Q.E.D. tiition) of that parallelogram. this let HK K And EFLC COR. Because the figure EF, EL, and EC, that is given in kind, the three right lines GD, HK, and EC, is, will have given ratios to each other.