The composition is as follows: produce EA to N, so that the rectangle EAN may be equal to the square of CE: bisect EN in O, and draw OP at right angles to it, meeting the straight line AL, which is parallel to CE in the point P; and from the centre P, distance PE, describe a circle, and let AP meet its circumference in the points L, M: join EL, EM, and draw AQ at right angles to EL; and between EQ, EL find (13. 6. Elem.) a mean proportional EF, and make EG equal to EF, then, by the 25th proposition of this book, describe an ellipsis, of which FG may be one of the axes, and which may pass through the point A: of this ellipsis, AB, CD are conjugate diameters. For since AQ is perpendicular to the axis FG, and EQ, EF, EL proportionals, AL touches the ellipsis (18. 2.) in the point A; and because CD is parallel to the tangent AL, it is in the same position with the conjugate diameter to AB; and the angle LEM being in a semicircle is a right angle; consequently EM is the other axis : hence the rectangle LAM is equal to the square of the semi-diameter conjugate to AE (21.2:) but the same rectangle-LAM is equal (35.3. Elem.) to the rectangle EAN, that is, by the construction to the square of CE; therefore CE is the semi-conjugate to AE: the ellipsis therefore passes through C; and because ED is equal to EC, and EB equal to EA, it passes likewise through the points D, B. Hence AB, CD are conjugate diameters in the ellipsis described.

PROP. XXVIII. PROB.

The position and magnitude of a diameter of an el lipsis being given, and the position of a straight line, passing through a given point in the ellipsis, and ordinately applied to that diameter being also given; to describe the ellipsis.,

Let AB (fig. 21.) be the given diameter, to which RS, a straight line given in position, is ordinately applied, from a given point R of the ellipsis to be described.

Bisect AB in the point E, and draw through E a straight line parallel to RS, and in that parallel take equal straight lines EC, ED, so that the rectangle ASB may be to the square of RS, as the square of AE to the square of EC or ED. If a mean proportional be found (13. 6. Elem.) between AS and SB; then (22. 6. Elem.) the mean proportional is to RS, as AE to EC, which is therefore found (12. 6. Elem.) then, by the preceding proposition, describe an ellipsis of which AB, CD may be conjugate diameters; this ellipsis will pass through the (2. cor. 15. 2.) point R, and RS will be ordinately applied (4. cor, 14. 2.) to the diameter AB.

PROP. XXIX. THEOR.

If a cone cut by a plane passing through the axis be cut also by another plane, meeting both the sides, of the triangle through the axis, but neither parallel to the base of the cone, nor subcontrarily situated, if that other plane, and the plane in which the base of the cone is situated, meet in the direction of a straight line perpendicular, either to the base of the triangle through the axis, or to that base produced; the line which is the common section of this other plane, and the conical surface, is an

ellipsis, which has for one of its diameters the common section of the triangle through the axis, with this same plane.

Let there be a cone (fig. 22.) the vertex of which is the point A, and the base the circle BC; let it be cut by a plane through the axis, and let the section be the triangle ABC; let it be cut likewise by another plane, meeting both the sides AB, AC, of the triangle through the axis, but neither parallel to the base of the cone, nor subcontrarily situated; let the line DEF be the common section of this other plane with the conical surface ; and let GH, the common section of this plane, with the base of the cone (continued) be perpendicular to BC: then the line DEF is an ellipsis; and DF, the common section of the triangle through the axis, and this same plane, is one of its diameters.

In the section DEF take any point E, and through E to DF draw EK parallel to HG; and through K draw LM parallel to BC: therefore the plane which passes through EK, LM is parallel (15. 11. Elem.) to the plane through BC, GH, that is, to the base of the cone: consequently the plane through EK, LM (23. 1.) is a circle, of which LM is a diameter: but EK is perpendicular (10. 11. Elem.) to LM, because GH is perpendicular to BG; therefore the rectangle LKM is equal (35. 3. Elem.) to the square of EK. In like manner, any other point N being taken in the section DEF; if NO be drawn parallel to EK, or GH, and through O, PQ be drawn parallel to BC; it may be shown, that the rectangle POQ is equal to the square of NO: consequently, the square of EK is to the square of NO, as the rectangle LKM to the rectangle POQ: but (by simil. trian.) LK is to PO, as DK is to DO; and KM is to OQ, as KF is to OF; but the ratios compounded of these ratios are the same to one another; and therefore the rectangle LKM is to the rectangle POQ, as the rectangle DKF is to the rectangle DOF (23.6. Elem :) and therefore (11. 5. Elem.) the square of EK is to the square of NO, as the rectangle DKF to the rectangle DOF. Describe, therefore, an ellipsis (28. 2.) of which DF may be a diameter, and in which EK may be ordinately applied to DF: and because the point E, by construction, is in this ellipsis, the point N is likewise in it (3. cor. 15. 2.) And the same thing may be demonstrated with regard to all the points of the sec tion DEF.

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