zontal trace of the given section plane) in W, W will be a point on one asymptote, which will therefore be the line CW. Similarly the second asymptote can be obtained from the point w1.

248

CONIC OF GIVEN ECCENTRICITY.

PROBLEM 131. To cut a conic of given eccentricity from a given cone (Fig. 135).

Let v'a'b' be the elevation of the given

m

eccentricity be given by two lines m and n.

cone, and let the

From o', the foot

of the axis of the cone, set off along the axis a length o'd = n and

o'e=m; through d draw dh parallel to the base of the cone meeting the slant side in h; from e with radius hb' (the distance between h and the foot of the slant side) describe an arc cutting a'b' in g; the required section plane must be inclined to the horizontal plane at the angle ego', and all sections made by planes inclined at this angle will have the same eccentricity.

Proof. Produce ge to meet the slant side of the cone in q, and in the cone inscribe a sphere touching the plane of section geq in the point ƒ and the slant side v'b' in p: through p draw px parallel to the base of the cone meeting geq in x.

but pq =ƒq, the angle pxq = the angle ego', and the angle qpx = the

but ƒ is the focus, q the vertex, and x the trace of the directrix of the section made by the plane geq.

If the conic is to be an hyperbola, i. e. if m>n, there is a limit to the vertical angle of the cone in order that the problem may be

possible. It will be observed that the length eg is

COS a

where a

is the semi-vertical angle of the cone, and eg must evidently be greater than co' or m.

must be greater than the angle whose cosine is or in other

n m

words the ratio of the height of the cone to length of slant side must be less than

m

PROBLEM 132. From a given cone to cut a conic of given eccentricity and having a given distance FX between focus and directrix (Fig. 135).

As in the last problem, draw some one plane of section of the required eccentricity, as geq, and determine its focus f and the trace (x) of its directrix.

Draw v'x, v'f to the vertex of the cone; on xƒ make xf, the given distance FX, and through f, draw fF parallel to xv meeting fo' in F. Through F draw a line parallel to gq meeting the slant sides of the cone in A and A, and xv' in X. This will be the trace of the required plane of section, A and A, being the vertices, Fa focus, and X the trace of one of the directrices.

1

DEF. If a straight line move so as to pass through the circumference of a given circle, and to be perpendicular to the plane of the circle, it traces out a surface called a Right Circular Cylinder. The straight line drawn through the centre of the circle perpendicular to its plane is the axis of the cylinder.

The cylinder may evidently be regarded as a particular case of the cone, the vertex being at an infinite distance from the base so that the generators are ultimately parallel.

As with the right circular cone, it is evident that a section of the surface by any plane perpendicular to the axis is a circle, and that a section by any plane parallel to the axis (i. e. passing through the infinitely distant vertex) consists of two parallel lines.

PROBLEM 133. To determine the section of a right circular cylinder by a plane inclined at any given angle (0) to the axis (Fig. 136).

Let Im be the line of intersection of the given plane of section with the horizontal plane of the base of the cylinder, i.e. the Fig. 136.

horizontal trace of the plane of section. Draw any ground line xly perpendicular to lm, and through I draw ld making the angle

d'ly equal to the complement of the given angle 0. Let o be the plan of the axis of the cone, and through o draw oo'o" perpendicular to xy; o'o" will be the elevation of the axis of the cone on the vertical plane of projection, and lď will be the trace on the same plane of the given section plane.

With centre o and radius equal to that of the cylinder describe a circle ab, and draw aob perpendicular to Im; through a and b draw aa'a", bb'b" perpendicular to xy, meeting it in a' and b', and the rectangle a'"a'b'b" will be the elevation of the cylinder. Let ld' cut a'a" in n' and b′b” in n”.

Imagine a horizontal plane to cut the solid at any height between n' and n", as at p'q'; it will evidently cut the cylinder in a circle of diameter p'q', and which would in plan have o for its centre, and it will cut the section plane in a horizontal line the elevation of which is the point r', in which p'q' cuts lď' and the plan of which is the line rrr' perpendicular to xy. The points of intersection of the circle and line will evidently be points on the desired curve of intersection, and therefore if r, r, are the points in which the plan of the line cuts the circle ab (which is of course the plan of the circle p'q') these will be the plans of the points in which the horizontal plane at the height a'p' above the base of the cylinder cuts the required curve of intersection.

1

Now imagine the plane of section to be rotated round its horizontal trace Im until it coincides with the horizontal plane of projection. In elevation the point r' would travel over the circular arc 'R' struck with 7 as centre, meeting the ground line in R', while on plan the points r and r1 would travel along lines rR, r ̧R ̧ perpendicular to lm reaching the horizontal plane of projection in the points R, R, found by drawing R'RR perpendicular to xy.

1

1

Similarly any number of additional points can be found by drawing a series of planes parallel to p'q', all of which will of course cut the cylinder in circles, the plans of which are the circle ab.