2.24. Conic Sections Symmetric About The Origin

Consider the cartesian equation

        X 2 2XF+ 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 XX , 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      e1

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     e1         (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.