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the angles CAK, DAL equal to the angles FGH, VFH. Let AK, AL meet the right line BD in K and L, and thence draw KM, LN, of which let KM make the angle AKM equal to the angle GHI, and be itself to AK as HI is to GH; and let LN make the angle ALN equal to the angle FHI, and be itself to AL as HI to FH. But AK, KM, AL, LN are to be drawn towards those sides of the lines AD, AK, AL, that the letters CAKMC, ALKA, DALND may be carried round in the same order as the letters FGHIF; and draw MN meeting the right line CE in i. Make the angle iEP equal to the angle IGF, and let PE be to Ei as FG to GI; and through P draw PQf that may with the right line ADE contain an angle PQE equal to the angle FIG, and may meet the right line AB in f, and join fi. But PE and PQ are to be drawn towards those sides of the lines CE, PE, that the circular order of the letters PEiP and PEQP may be the same as of the letters FGHIF; and if upon the line fi, in the same order of letters, and similar to the trapezium FGHI, a trapezium fghi is constructed, and a trajectory given in kind is circumscribed about it, the Problem will be solved.

So far concerning the finding of the orbits. It remains that we determine the motions of bodies in the orbits so found.

SECTION VI.

How the motions are to be found in given orbits.

PROPOSITION XXX. PROBLEM XXII.

To find at any assigned time the place of a body moving in, a given parabolic trajectory.

Let S be the focus, and A the principal vertex of the parabola; and suppose 4AS $\scriptstyle \times$ M equal to the parabolic area to be cut off APS, which either was described by the radius SP, since the body's departure from the vertex, or is to be described thereby before its arrival there. Now the quantity of that area to be cut off is known from the time which is proportional to it. Bisect AS in G, and erect the perpendicular GH equal to 3M, and a circle described about the centre H, with the interval HS, will cut the parabola in the place P required. For letting fall PO perpendicular on the axis, and drawing PH, there will be AG² + GH² (= HP² = $\scriptstyle {\overline {AO-AG|^{2}}}$ + $\scriptstyle {\overline {PO-GH|^{2}}}$) = AO² + PO² - 2GAO + 2 GH + PO +