## 2.23. Cartesian Equation For A General Conic

Eq(2.22) can be written as

XF     =e| X N ˆ F N ˆ d |   =  | eX N ˆ a |                  (2.29)

where  a=e( FN+d ) .  Taking the square of (2.29) gives

X 2 2XF+ F 2 = e 2 ( X N ˆ ) 2 2eaX N ˆ + a 2          (2.30)

Setting

X=( x,y )           F=( f 1 , f 2 )            N ˆ ( n 1 , n 2 )

where n 1 2 + n 2 2 =1 , eq(2.30) becomes

x 2 + y 2 2( x f 1 +y f 2 )+ f 1 2 + f 2 2 = e 2 ( x n 1 +y n 2 ) 2 2ea( x n 1 +y n 2 )+ a 2

Þ            A x 2 +Bxy+C y 2 +Dx+Ey+G=0                         (2.32)

with

A=1 e 2 n 1 2              B=2 e 2 n 1 n 2             C=1 e 2 n 2 2               (2.31)

D=2( ae n 1 f 1 )                E=2( ae n 2 f 2 )                    G= f 1 2 + f 2 2 a 2

Note that the discriminant

Δ= B 2 4AC= ( 2 e 2 n 1 n 2 ) 2 4( 1 e 2 n 1 2 )( 1 e 2 n 2 2 )

=4 e 4 n 1 2 n 2 2 4( 1 e 2 + e 4 n 1 2 n 2 2 )   =4( e 2 1 )                       (2.33')

so that eq(2.32) represents an ellipse, parabola, or hyperbola if D is negative, zero or positive, respectively.  [Caution: our D is the negative of that given in the text].