94 



REPORT — 1856. 



the equation of an equilateral hyperbola whose centre is at V, the vertex of 

 the parabola, and whose transverse axis is the parameter of the parabola. 



The area of this curve, the elements being taken parallel to the axis, or 

 the area between the curve and the vertical axis passing through V, is 

 found by integrating the value of xdy. 

 Now 



iC=2m sec ^, and y=^^m tan 0, 

 therefore 



\xdy^=^'^nr y^ec^ ^d^^=%ri\rti sec ^ tan + mJsec <^ (/^J. 



But it has been shown in IV. that 



n(m . 0) = m sec tan (/> + m \ sec <^ df. 



Hence the hyperbolic area VAQR = 2mll(m. 0). 



(42) 



Therefore as the hyperbolic area is equal to a constant multiplied into the 

 corresponding arc of the parabola, the evaluation of the hyperbolic area 

 depends on the properties of logarithms. 



It also follows, from what 

 has been established in the pre- 

 ceding part of this paper, that 

 hyperbolic areas may be multi- 

 plied and compared according 

 to the laws which regulate pa- 

 rabolic arcs. 



Let (p and 6 be the angles in 

 which the normals to the cor- 

 responding points of the para- 

 bola and the hyperbola cut the 

 axis, then if (p and 6 be these 

 angles, it is easily shown, since 

 VQ = normal at Q, that 



tan 0= sin 0. . . (43) 



This expression will enable 



us to express the hyperbolic 



area in terms of the angle which 



the normal to the hyperbola 



makes with the axis instead of 



the parabolic amplitude ; for as the parabolic amplitude is related to the 



normal angle of the hyperbola 6 by the equation tan 0=sin (p, 



n t , , 2 tan Q x oo 



2 tan <b sec A = r- = tan 20, 



^ ^ l-tan'e 



and 



Now 



sec 0-1- tan 0= v/ sec 20 + tan 20. . . . 



n(jw . 0)=m sec0 tan + m log (sec -j- tan 0), 



or, substituting for the preceding values of 0, 



2n(»i . 0)=m tan 20-f-»J log (sec 20-f tan 20); 



but taking the amplitude 20, 



n (m . 20)= m sec 20 tan 20 + m log (sec 20 + tan 20). 



(44-) 



I 



