Proposition 25-26 Problem 27-28

Draw a trajectory that passes through 2 points

by Newton
3 min read 623 words
Table of Contents

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 into a new figure.

In this figure those two tangents will become 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 hl 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 parallelogram hikl, let the right lines hi, ik, kl be so cut in c, d, e, that hc may be to the square 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 is 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, hc² to the rectangle ahb, and ic² to id², and ke² to kd², and el² to the rectangle alb, are all in the same ratio; and therefore hc to the square root of ahb, ic to id, ke to kd, and el to the square root of alb, are in the subduplicate of that ratio; and by composition, in the given ratio of the sum of all the antecedents hi + kl, to the sum of all the consequents …

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 h, l, or without them, the points c, d, e, must be taken either between the points, h, i, k, l, or without them. If one of the points a, b, falls between the points h, i, and the other without the points h, l, the Problem is impossible.

Proposition 26 Problem 28: Describe a trajectory that shall pass through a given point, and touch four right lines given by position.

From the common intersections, of any two 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; 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 radius, will now become parallel.

Let hi and kl, ik and hl, be those pairs of parallels completing the parallelogram hikl. And let p be the point in this new figure corresponding to the given point in the first figure.

Through O the centre of the figure draw pq: and Oq being equal to Op, q will be the other point through which the conic section must pass in this new figure. Let this point be transferred, by the 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. 17.

Send us your comments!