To find the point in which the curve meets any other radius P's', draw s'N at right angles to AB; join P'N which is the vertical projection of the slant side passing through s': produce p's to meet A'v' and AB in v' and T; draw V'в, TT" parallel to r'r", and join BT" cutting P'N in E'; draw E'E' parallel to r'r', and E' will be the point in which r's' is intersected by the curve.

In a similar manner we may find any number of points in the required section C'E'D'F'.

When the points v and r become too remote to be conse. quently used in the construction, we may find the required points of the curve by the method used in determining the intersection of the plane by the slant side passing through K.

In this example, in which the cone is divided by the plane into upper and under parts of the conic section, is called an

ellipse; the curve C'E'D'F' is therefore the projection of an ellipse, and consequently, c'E'D'F' is also an ellipse.

PROBLEM IV.

To construct the ellipse of which the curve c'E'D'F' in the preceding problem is the horizontal.

This problem is readily solved by finding by prob. 12, chap. II. the positions of the points of the curve on the horizontal plane by the revolution of the intersecting plane to a coincidence with the horizontal plane.

Or, we may proceed as follows, which is nearly equivalent. Because Ko' is the position on the horizontal plane of the slant side passing through ¤, and KC'A the angle of elevation of the given plane; it is plain, that cu' is the transverse axis

of the ellipse section required. Make c' equal to c'u', and will be the position on the horizontal plane of the vertex determined by u'.

In like manner, take c'r equal to CR' and make the perpendicular equal to P'r', and will be a point in the required section. In a similar manner, we may find any number of points in the circumference of the required ellipse dec'o.

It is evident, from this construction, that the ellipse SEC' is derived from C'E'D'F' by elongating each abscissa from c' as cp' in the constant ratio of c'D' to c'v'; so that cr' is to c'r as c'D' to c'd, while the semiordinate n remains the same as P'H'.

And as the curve dec'p is by the definitions of conic sec. tions an ellipse, it is manifest from the constant ratio of the abscissas cr' and c'r having a common semiordinate r'a' or , that the projection C'E'D'F' is also an ellipse.

PROBLEM V.

To construct the section of a cone by a plane parallel to the axis of the cone.

Let AB be the ground line; p' the centre of the circular base KQ'R' of the cone touching the ground line AB in K: produce the radius P'K to P', and take KP" equal to the axis of the cone which is supposed to be at right angles to the plane of its base, and consequently to the horizontal plane; then "K is the vertical projection of the axis, and p" of the vertex.

Suppose the cutting plane to be parallel to the vertical plane, and to intersect the horizontal plane in the straight line R'YQ', which is therefore parallel to AB, and consequently perpendicular to the diameter CK.

Draw any radius r's'r of the base, meeting R'Q' in s", and the circumference of the base in T. Find by prob. 1, chap. IV, the vertical projection r'v of the slant side passing through T, the corresponding horizontal projection of this slant side being r'T: through s' draw s'ss" at right angles to AB, and meeting vp" in s", and Ns" is the altitude of the conic surface at s', because NS', Ns" are evidently co-ordinates of a point of the slant slide passing through T.

And, since the cutting plane which passes through R'Q' is perpendicular to the horizontal plane, it is evident that Ns', NS" are the co-ordinates of the point in which the slant side

terminating in T penetrates the cutting plane; if therefore we make s'L equal to Ns", it is plain that L will be the position on the horizontal plane of the point denoted by s', s", by the revolution of the cutting plane about the intersection R' Q.

By a similar construction, we may determine any number of points in the curve Q'LXR', which will be the section required.

The curve required may be obtained still more simply by merely finding the perpendiculars NS"H, and describing the curve through L, S, M, &c. without determining the corresponding points in Q'LXR'.

It is evident that the plane meeting the base at right angles in p'yo' must also meet the upper division of the conic surface, and produce another section equal and similar to qʻXR'. The curve determined by this construction is an hyperbola.

PROBLEM VI.

To construct the intersection of a conic surface by a plane pa. rallel to one of the slant sides of the cone.

Let AB be the ground line; p' the centre of the cone's base, which is supposed to be coincident with the horizontal plane ; and let the base EFK touch the ground line in K: in PK produced, take KP" equal to the altitude or axis of the cone, and p" is the vertical projection of the summit of the cone. Let the cutting plane be parallel to the ground line, and meet the

base in the horizontal trace EF, which will consequently be parallel to AB: in KP" produced if necessary, take KY" a fourth