22 REPORT — 1871. 



When A,D, B are finite, solve the equation for if, regarding B as positive. Put 

 ^=D^'-AB, ^^DF-EB, A=F^~CB ; 

 .-. By=+Da;^+F= + V(5'■^''+2^•^+^)• 

 For D=- AB>0, we may assume g=l ; then for the upper sign we get, as Proxi- 

 mate Conic Hyperbola, if D be <\, 



Bj/,'^+Da-'-+F=x'^+A. 

 K D is negative, we have a second Proximate Conic for the lower sign, 



Bi/j=+Da;^+F=-.r^-A. 



Of course the asymptotes are '&y- +'Dx--=+x-, or only By2-{-Da;^=x^ 



If D^— AB=0,' g=^0, and the cui-ve is infinite only when h is positive: then 

 if J) is negative, 



B«/i=+DiB2+ F= + i^{2h)x 



is two Proximate Conic Hyperbolas, and the asymptotes are oblique and parallel in 

 pairs ; they do not pass through the centre, but are equidistant from it. 



Evidently if C = and E, F have opposite signs, the cui-ve crosses itself in the 

 centre; but if C = and E, Fhave the same sign, the centre is a mere Stud. 



An undulation of the curve implies a double tangent. Such double tangents 

 are always parallel to one axis. [I desire a general proof.] There can be only 

 two pairs parallel to one axis. To ascertain whether there is undulation across 

 OY, put x^=Q, and try whether y- is there a 7naximum or a tnitiimtan. 



Making x^ infinitesimal and k positive (which is implied), 



thus for upper sign, 



whence j^2 is a minimum at x^=0 if — D> 0, a maximum if — ^ — D <'0. Yet 



h . . . '^^ ^ 



when — — =D, y^ is a minimum at x'=0 if gk—h^> 0, or a maximum itgk—h^ < 0. 

 y « 



Now gk—h'' = 'BY, and we cannot have V = without degeneracy. Hence this test 

 is final. Also 7; — D s/k is equivalent to BE= - 2DEF+CD2=0. 



If the branch we are investigating is injinite and y^ is a minimum, there is tio un- 

 dulation; but if y'^ is a maximum, it begins to decrease, yet must afterwards in- 

 crease ; hence there is undulation. On the contrary, if the upper branch hejinite 

 and 3/^ be a maximum, there is no undulation ; but there is imdulation if y^ be a 

 minimum. 



In general, for tangents parallel to the axis of x, putting -/ =0 we have the ob- 



dx ' 

 vious solution a? = 0, when there is a vertex on the axis of y. Besides this, we may 

 have a double tangent where 



T>s/{gx'+2hx^+k)=gx-'+h, 

 which yields 



^^ =a/ijE^= V(i7^H2A^=+A)=By=+Da.^+R 



Hence at the points of contact 



gx"-+h==±-D^- (^, gif+h'= + ^{-A\), 



if^'=DE-FA. 



When T=={}?x^—fi?y'^){pV-\-a''y'), the curve has oblique asymptotes, V3?=fi^y^. 

 To^try whether it ever cut its asymptotes, put y=yo ; then at the common point 

 'K^x"=fu^y^, and 2Ex'+2Fj/=+C=b. If the x, y hence deteimined is within the 

 limits of the curve, it does thus cut ; if not, it does not. 



