## 2.24. Conic Sections Symmetric About The Origin

Consider the cartesian equation

‖ X ‖
2
−2X⋅F+
‖ F ‖
2
=
e
2
(
X⋅
N
ˆ
)
2
−2eaX⋅
N
ˆ
+
a
2
(2.30)

where

a=e(
d+F⋅
N
ˆ
)
(2.30a)

If (2.30) is to be symmetric about the
origin, i.e., invariant under
X→−X
,
all terms in odd powers of *X* must
vanish. Hence

F=ae
N
ˆ
(2.35)

Combining this with (2.30a) gives

F⋅
N
ˆ
=ae
=(
d+F⋅
N
ˆ
)
e
2
=
e
2
d
1−
e
2
if
e≠1

Þ
a=
ed
1−
e
2
(2.36)

Putting these back to (2.30) gives

‖ X ‖
2
+
a
2
e
2
=
e
2
(
X⋅
N
ˆ
)
2
+
a
2

Þ
‖ X ‖
2
−
e
2
(
X⋅
N
ˆ
)
2
=
a
2
(
1−
e
2
)
where
e≠1
(2.37)

which is clearly symmetric about the
origin. Since (2.37) is also invariant
under
N
ˆ
→−
N
ˆ
,
there are 2 pairs of symmetrically placed directrices and foci. According to (2.35), the foci are at points
±ae
N
ˆ
. From exercise 7 in sec 2.28, the directrices
intersect the line joining the foci at points
±
a
e
N
ˆ
.

To find the intersects of the curve (2.37)
on the line joining the foci, we set
X=β
N
ˆ
so that (2.37) becomes

β
2
−
e
2
β
2
=
a
2
(
1−
e
2
)

Þ
β=±a

These points at
X=±a
N
ˆ
are called the **vertices** of the **central
conics** given by (2.37). The segment
that joins them are called the **major
axis** if the conic is an ellipse, the **transverse
axis** if the conic is a hyperbola.

To find the intersects of the curve (2.37)
on the line that passes through the origin and perpendicular to the
major/transverse axis, we set
X=γ
M
ˆ
where
M
ˆ
⋅
N
ˆ
=0
. Eq(2.37) then becomes

γ
2
=
a
2
(
1−
e
2
)
Þ
γ=±a
1−
e
2

Since *X*
and hence *g* must be real, there is no solution for
e>1
. For ellipses with
e<1
,
the segment that joins these points are called the **minor axis**.