296 
HYPERBOLA. 
Section XIV. 
Polar Equation. Centre the Pole. 
1. To prove that of all diameters of an hyperbola the trans 
verse axis is the least. 
The polar equation to the hyperbola, the centre being the pole 
and the transverse semi-axis being the prime radius vector, is 
1 cos' 2 # sin' 2 # 
I/ ’ 
4 - 4 = sin”0 (4 + i 
whence 
The expression for — \ being essentially positive, it is evident 
that 4 is less than 41 or a 1 less than r' 2 , or a than r. a con 
elusion which establishes the proposition. 
De la Hire : Sectiones Coniece, lib. II. prop. 36. 
2. An hyperbola, of which C is the centre, is cut in two 
points P, P', by a given straight line: to find the tangent of 
the sum of the inclinations of CP, CP', to the transverse axis. 
The equations to the hyperbola and to the straight line being 
'cos' 2 # sin' 2 #\ _ 
~d l V~) ~ ’ 
and 8 = rcos(# — X), 
we have, at their intersections, 
'cos' 2 9 
S' 2 
sin' 2 # 
P 
= cos' 2 (# — X), 
and therefore 
P\ S 2 
sin‘ 2 X + ] tan' 2 # + sin2X.tan# + cos‘ 2 X r, = 0, 
b7 or 
and consequently, # ( , # (( , being the inclinations of CP, CP', 
to the transverse axis, 
„ . sin2X 
tan(# + #J 
cos 2X