Demonstration of a Problem in Conic Section. 281 
asserted by Dr. Hutton; and the principle which he has 
adopted as general, obtains in one case only. As I have 
not seen this error corrected, nor any other method laid 
down by authors on conic sections for obtaining conjugate 
hyperbolas by means of intersecting four cones situated as 
above, | send you the following, thinking that it may possi- 
bly merit a place among the mathematical articles of your 
Journal. 
Let LAC, HAF, HAL, and : P e 
AF, be four cones, having a: \ 
common vertex A, their axes 
inthe plane of the paper, and 
touching each other in the 
right-lined elements, CAH, 
and LAF 
Ifthe two cones CAL and 
FAH be cut by a plane 
QCBDE, parallel to the line 
PO, it will intersect the 
cones in opposite hyperbo- Ay 
las ; and if we take the plane i 
N 
Al 
i 
perpendicular to that of the 
oe these hyperbolas will 
orthographically project- 
edin the line EQ. If through 
“4 
meets the right-lined ele- , G 
s AC ee * 
plane be passed parallel to R 
the line KAB, this plane will intersect the two cones 
LAH and CAF, in opposite hyperbolas, and these hyper- 
bolas will be conjugates to the former. 
Demonstration. 
Pass any plane as GF perpendicular to the axis of the 
cone GAF, it will intersect its surface in a circle, and the 
plane QE in a right line, which will be a common ordinate 
at the point KE, to the transverse axis of the hyperbola, and 
the diameter GF of the circle. Since the triangles CAB, 
CGE and DEF are similar, CB is to AB as CE to EG, and 
CB is to AB as DE to EFy and by multiplying together 
Vou. VI.—No. 2. 36 
