DESCRIPTIVE GEOMETRY. 39 
DD’. Marking off on these horizontal lines, the breadths taken at the 
corresponding parts of the horizontal projection of the parabola, we 
shall obtain the points, 1’, 2’, 3’, 4’, and 5’ E, which determine one half 
of the parabola. The other half is to be drawn symmetrically with 
this. 
A simple method of describing a parabola, when its breadth below and 
its altitude to the vertex are given, is shown in fig. 31. At the middle of 
the line AB, erect the perpendicular, CD=twice the height of the parabola, 
and determine the vertex, C ; through this zy is passed, parallel to AB. At 
A erect the perpendicular, Az, and divide it in f, g,h, &c., into equal 
parts, which are then marked off from D, at a, b, c, &c. From C to &, and 
from A to a, draw straight lines ; their intersection will give one point of the 
parabola. Another point will be obtained by the lines Cz and Ab, &c. 
The second limb of the parabola, being symmetrical with the first, is easily 
constructed. | 
The projection of the third conic section, the hyperbola, is explained by 
means of fig. 30. As this is perpendicular to the base, it can appear only 
as a straight line in-horizontal projection, this line being D’F’. The vertex 
is projected at EK’. Ifthe line E’F’ were also perpendicular to the base line, 
the hyperbola would be projected -on the vertical plane in another straight 
line: this is the case in. fig. 55, with respect to the line CA. As this view 
can give no satisfactory representation, we have in fig. 30 made the plane 
of intersection parallel to the vertical plane. To determine the vertex of 
the hyperbola-in the vertical view, we first draw the axis, CC’, take off 
from C, and parallel to the base, the distance C’E’, and let fall from the point 
thus obtained, a perpendicular to the side of the cone. From the point 
where this meets the side, draw to the axis a. parallel.to the base; this 
determines the point E. on the latter. Place any number of planes of inter- 
section, in the- horizontal -projection; parallel to. the base. these will form 
circles with the common centre C’, and whose vertical projections as straight 
lines may be readily ascertained by. methods already explained. From the 
points 1, 2, 3, &c., of the: ground plane, where the projections of the inter- 
secting surfaces cut those: of ‘the hyperbola, draw perpendiculars to the 
corresponding vertical projections. of the aforesaid planes of intersection ; 
we thus obtain the intersections 1’, 2’, &c., as points of the hyperbola, which 
may then be joined by a line continuous with the vertex E. Another 
method of describing the hyperbola is presented in pl. 4, fig. 29. 
If one body penetrate another, a surface of intersection or penetration 
will be formed. It is one of the problems of projection to determine the out- 
lines of such surfaces of intersection, and. their projections under different 
circumstances. The number of possible cases is infinite, and we can here 
only adduce a few-as examples. In. fig. 32, two cylinders are shown, 
of unequal diameters, and penetrating each other at right angles. The base 
line, or the one in which the surfaces of horizontal and vertical projection 
intersect each other, may be represented by zy. The one cylinder is repre- 
sented in vertical projection as circle KE, and in horizontal projection as 
rectangle AB; the other, being parallel to both planes, is in both cases 
2 
39 
