2.21. Polar Equations For Conic Sections
Setting the fous at the origin, i.e.,
F=0
,
eq(2.22) simplifies to
‖ X ‖ =e
X⋅
N
ˆ
−d

(2.23)
which, in polar coordinates, becomes
r=e
rcosθ−d

(2.24)
where
r= ‖ X ‖
and
X⋅
N
ˆ
=rcosθ
[see Fig.2.13].
If X
is on the same side of the directrix as the focus,
rcosθ−d<0
so that (2.24) becomes
r=e(
d−rcosθ
)
Þ
r=
ed
ecosθ+1
(2.25)
If X
is on the other side of the directrix from the focus,
rcosθ−d>0
and (2.24) becomes
r=e(
rcosθ−d
)
Þ
r=
ed
ecosθ−1
(2.26)
Since, here, q is restricted to
0<θ<
π
2
while
r,e,d>0
,
eq(2.26) cannot be satisfied for any q if
e≤1
. In other word, only hyperbola can have a
branch on the other side of the directrix from the focus.
Consider now the branch indicated by
(2.25).
If
e<1
,
there is no restriction on q so that
0<θ<2π
as befits an ellipse.
For
e>1
,
q is
restricted to
θ≤
cos
−1
(
−
1
e
)
.