for turning Ovals. 3 g 5 
cllipfe thefe tangents will meet each other in the axis ca 
produced. Draw mg perpendicular to tm, and gms will 
be the angle of deviation fought. I fay, the angle mtn, 
between the tangents to correfponding points in the 
ellipfe and circumfcribing circle, is equal to the angle of 
deviation gms. 
For becaufe tnc is a right-angled triangle, and nf 
perpendicular to tc; therefore tnp = ncp = msp, that is 
(in the triangles mtn and gms) the angles tnm and msg 
are equal. In like manner, becaufe tmg is a right-angled 
triangle and mp perpendicular to tg, therefore tmp = 
mgp, and (in the triangle mtn and gms) the angles tmn 
and mgs are equal; therefore in the fame triangles, the 
remaining angles mtn and gms are alfo equal. 
To compute the angle mtn, we have by trigonometry 
tp’' + pmxpn : tp ::mn : tan. mtn, radius being unity. 
Call now c a — cb=t, cp=x, p m ca— cb (or t—c) —d f 
and we ha ve pn = VTT^Tx- alfo cd:cb::pn:pm 
-\/tt~xxx- 0 whence pm x pn = tt-xx x j. Again 
cp : pn :: pn : ft, whence tp='-£z Af. Daftly, cd : bd :: 
pn . mn— \ftt— xx x - ; whence the tangent of mtn the 
angle fought is iagp, and this is a maximum when 
~d’ OT 7+- e = xx -» or w ken ~ =yy, or when cp and pm 
have fuch a proportion that cp’ : pm 1 :: ca 3 : cb 3 . 
Let 
