| Non-homocentric Pencils. 713 
From (1) and (2), 
a 

rcosg _ °F ocn8 0 
di) 7 a aa 
ta cade 
whence 
a(r cos d—d) 
2(cl Or cose eae (3) 
We will now eliminate ? between (1) and (3). Squaring (1) 
we have 

cos 0@= 
ped a eee 
1 (c+ ve tan? 20 
=( Bene! 2 4 cos? (1 — cos? @) 
—\°" eos 0) * (2e0s?O—1)? * 
Substituting for cos @ and simplifying : 
}a?(2 cos 6 —d)?— 2(ed — br cos )*}? 
= cos? fa*(b—c)*{4(cd — br cos db)? a?(r cos $—d)?!. 
This is the polar equation to the shadow, and is of the 
fourth degree in 7. 
Substituting rcos $6=w and 7?=2*? +7 we get the Cartesian 
equation 
f{a?(«—d)? —2(cd—bx)?\?(a? + y?) 
= xa?(b—c)*{4 (ced —bx)?—a*(a—d)??. 
This is of the sixth degree and is met in six points by a 
line through the origin. Two degrees, however, are accounted 
for by a conjugate double point at the origin, introduced in 
the transformation. We are left with four shadow points for 
each point of the object. LHvidently, however, the analysis 
applies to the complete spherical mirror, and one or three of 
these points will be due to rays which would pass virtually 
through the object after reflexion at the missing convex 
surface. The condition for this will be seen on reference to 
fig. 7, which shows a section of the complete mirror and of 
the epicycloid, which is the caustic for reflexion at both 
surfaces, the part to the right of the centre for the convex 
and that to the left for the concave surface. The rays which 
pass through any point are the tangents drawn through it to 
the epicycloid. Of these there are four if the point is outside 
the epicycloid; three from the concave and one from the 
convex surface of the mirror if the point is to the left of the 
centre, and three from the convex and one from the concave 
if it is to the right. If the point is inside the epicycloid 

