162 
On Conjugate Hyperbolas. ; 
[June, 
withont exception, agreed, that their ideas of the formation of conjugate hyper- 
bolas, by the intersection of the cone, was precisely what Hutton has here laid down. 
And yet, notwithstanding such high authority, it is certain that Hutton’s defini- 
tion is wrong ; and that conjugate hyperbolas will not be produced in the manner 
he describes, unless the four cones are all equal. To prove the truth of this, it is 
necessary to consider two cases. One, Avherein the planes of the hyperbolas 
are perpendicular to the plane of the base of the cone, which is Hutton’s 
supposition ; the other, wherein they are inclined to it. I shall content myself with 
examining the first case for the present, leaving the second to a future opportunity. 
Let us take the two conjugate cones as given by Hutton, PCO, OCN, (Fig 1.) 
and suppose that the plane of the hyperbola SFT, IWH, is perpendicular to the base 
of the cone ; and since the angle between the side and the base of the cone PCO, is 
COP, so that between the side and base of the other cone OCN, is CON, and these 
angles are the complements of each other. Then let the radius of the base of 
PCO, that is, the altitude of OCN, be r, and let the tangent of the angle COP, that 
*s the cotangent of the angle CON, be r, then the altitude of PCO — radius of base 
of OCN = r r; now let the distance of the centre of the base of each cone, re- 
spectively, from the plane of the hyperbolas SFT, IWH, that is, the line joining the 
centre of the base of each cone, and the points A and E, or the lines BA and YE 
which, by Hutton’s hypothesis, are equal, be b, and this is evidently equal 
to the perpendiculars drawn from the points F and W, respectively, to the lines 
joining the apex and centre of the cone PCO, and OCN ; let the point of the inter- 
section of these perpendiculars, and the joining lines, (which cannot well be repre- 
sented in the figures, but may easily be supplied by the imagination,) be X and Z, 
then the A’s FXC, WZC, respectively, are similar to half the vertical A 's of their 
respective cones ; and since FX = b = WZ by hypothesis, so XC, that is, the dist- 
ance of F, the vertex of the hyperbola SFT, from its centre =z b r andZC, the dist- 
ance of W from the centre of IWH = i. that is, the semi-transverse axis 
r 
b 
of SFT, is b r and of IWH is — . The question, therefore, resolves itself into this; 
T 
having given the hyperbola S F T, whose semi-transverse is XC, whose absciss is 
FA, and whose ordinate is the perpendicular AT, to find its semi-conjugate. 
b 
If that be equal to ZC, or — , then is IWH the conjugate hyperbola equal to SFT, 
but not otherwise. 
Now, for this purpose, to avoid confusions in characters, let us express the 
equations to the hyperbola, by Sanscrit letters, thus ^ / 2 
m which, as usual, V is the ordinate, ^ the semi-transverse axis, ? the semi -con- 
Jugate, and ^ the absciss; and then ^ = nowhere as above, 
v 2 x 
^ — b r, V by the nature of the circle = /- — : . , 
\/(r+b) X (r— 6) =v /r*-a 2 and 
TJ =r (r — J) __ j7__ 4 
*J2 b t* (r b) -f—r* (r — b) 2 t^/ 
It follows, therefore, that the true semi-conjugate axis to the hyperbola SFT, « 
; “ ls > the dlstance of the P lane of hyperbola from the vertical triangle of tie 
cone, but the semi-transverse of IWH is 
b 
; that is, the said distance divided by 
T 
