214 
On Conjugate Hyperbolas. 
[July, 
one being the supplement of the other, and each of the angles RZU, OFW a right 
angle. By this arrangement will be produced two hyperbolas, mutually on 
posite, viz. FGH, KLl\l, and two others, mutually opposite, viz qr S 
PON. Now as the common plane of all these hyperbolas is perpendicular 
to BQSD and ENPA, so it is also perpendicular to the vertical sect! “ 
may be made by cutting the cones ABC, DCE, by a plane perpendicular to 
the paper, passing through the continued straight line XGLY. Hence tiro parts 
of the hyperbola FGH, on each side of this plane, viz. FGX, XGH are simi- 
lar and equal to each other ; and for the same reason, KLY, YLM are also 
mutually similar and equal, and the straight line GL is the transverse axis 
of the hyperbolas FGH, KLM : the question is, therefore, under these circumstances, 
are QRS, PON the conjugate hyperbolas of FGH, KLM, as according to tbe 
common definition they ought to be ? It is very easily seen that they are not ; for 
if they were, then GL should be the conjugate axis of QRS, PON, and its half 
LA the semiconjugate. Now since QRS, PON are perpendicular to the bases of 
their cones, it follows from what was been demonstrated in my former letter, that 
their semiconjugate axis is equal to their distance from the vertical section, parallel 
to them, of the cones BCD, ACE ; that is, to the line UZ in the one cone, and WF 
of the other ; and each of these evidently is equal to the perpendicular from C, the 
point of contact of the apices of the four cones, to the line GL ; that is, the perpendi- 
cular from C to GL must be equal to G A or A L. Now this is impossible, because 
the distance of the hyperbolas QRS, PON from the parallel vertical sections, viz. 
the lines UZ and WF, or in other words, the perpendicular from C to GL may 
easily be conceived to remain the same, while the angles GXT, LYV, and conse- 
quently the magnitude of the line GL, and its half GA, vary. That is, the constant 
perpendicular is equal to a quantity of variable magnitude, which is absurd. 
Let us now suppose the cones placed in the situation represented in the second 
figure (Fig. 4.) of my former letter, and let a plane, perpendicular to the paper, pass 
through the line GL. It is evident this will cut off t*o opposite hyperbolas FGH, 
MLK, of which the semitransverse is G A. By the same notations as in my former 
letter G A = GX=cr /r v Tuv „ S=. Then 
to get the value of that is, the semiconjugate of the hyperbolas FGH, MLK, we 
must take a fourth proportional to /Ft t ^ .w , -r— - , j-rr TyT 
-V (BL x GX) x GX, AG and AX Xh. 
Supposing this done, and 0 to be the line, and let GA be the perpendicular from G 
to t el me AC, then GA is either equal to, less, or greater than 0. First let it be equal, 
t len if a plane perpendicular to the paper be drawn through GA, and produced both 
ways to „ and n, it will cut oft from the cones BCD ACE two hyperbolas, whose 
trai^verse axis will be GA, or the conjugate axis of the two former hyperbolas 
FGH, MLK, and so far the object may be said to be attained : but the reverse of this 
cannot e s own ; namely, that the conjugate axis of the new hyperbolas is equal 
to the transverse axis of the former. Next let GAbe less than 0, then there may be 
two points found in the line AC, which may be joined with G, and a plane drawn 
through them, so as to produce two new pairs of opposite hyperbolas, with the same 
property of their transverse axis as the former pair. If GA be greater than 0, then 
no hyperbola of this kind can be produced by a plane passing through the point G. 
It still, therefore, remains to inquire, how to cut off from the cone ACE a 
hyperbola, whose transverse axis is 0, and its conjugates G A ; or to state the matter 
e geneially, the problem is this: — Having given a cone of infinite magnitu > 
D le at whose apex is known, required to cut from thence a hyperbola, who 
/axes are given. 
