equations and standard equations for plane conics
Suppose that C is the conic with focus F, directrix L and eccentricity e.
Probaly the simplest choice of axes is to take L as the yaxis, and the line
through F perpendicular to L as the xaxis. We direct the latter so that the
focus F is (f,0) with f > 0.
P(x,y) is on C if and only if 
PF = ePQ 

i.e. 
PF^{2} = e^{2} PQ^{2} 

i.e. 
(xf)^{2} + y^{2} = e^{2}x^{2} = e^{2} x^{2} 

i.e. 
(1e^{2} )x^{2} + y^{2} 2fx +f^{2} = 0. 

Thus, we have established the
Lemma
A plane conic defined by the focusdirectrix property has an equation
which is quadratic in x and y.
Thus, a plane conics are examples of loci defined by a quadratic equation.
Such loci are either plane conics, or degenerate conics, so that we can take plane conics
as the family of nondegenerate loci withe equations quadratic in x and y.
As indicated earlier, we can also define conics as plane sections of a cone. Such
sections also have equations quadratic in x and y, so are plane or degenerate conics.
The degenerate cases arise by taking a plane through the vertex, so the plane conics
are the sections of a cone by a plane not through the vertex. There is one snag in
this approach  it is not at all clear that every plane conic arises as a conic section.
The elegant way to resolve this problem is to use projective geometry.
To get a simpler form of equation, we need to choose axes on a casebycase basis,
Throughout, we will assume that the distance from F to the directrix is given by f.
e = 1  the parabola
We choose as origin the point on the axis midway between F and the directrix, then
F is the point (a,0) and L is the line x = a, where a = f/2, so a > 0.
P(x,y) is on C if and only if 
PF = ePQ 

i.e. 
PF^{2} = e^{2} PQ^{2} 

i.e. 
(xa)^{2} + y^{2} = e^{2}x+a^{2} = e^{2} (x+a)^{2} 

i.e. 
y^{2} = 4ax 

e ≠ 1  the ellipse and hyperbola
Here, we choose the origin on the axis so that F is (ae,0) and L is the line x = a/e.
Since the distance from the focus to the duirectrix is f, this requires that we have
f = aea/e = ae^{2}1/e. We will choose the positive value of a.
P(x,y) is on C if and only if 
PF = ePQ 

i.e. 
PF^{2} = e^{2} PQ^{2} 

i.e. 
(xae)^{2} + y^{2} = e^{2}xa/e^{2} = e^{2} (xa/e)^{2} 

i.e. 
(1e^{2})x^{2} + y^{2} = a^{2}(1e^{2}). 

i.e. 
x^{2}/a^{2} + y^{2}/a^{2}(1e^{2}) = 1. 


At this point, it is convenient to separate the cases, according as e < 1 or e > 1.
e < 1  the ellipse
Here, (1e^{2}) > 0, so we can choose b > 0 with b^{2} = a^{2}(1e^{2}). The equation becomes
x^{2}/a^{2} + y^{2}/b^{2} = 1.
e > 1  the hyperbola
Now, (e^{2}1) > 0, so we can choose b > 0 with b^{2} = a^{2}(e^{2}1). The equation becomes
x^{2}/a^{2}  y^{2}/b^{2} = 1.
These are the standard forms for the equation of a plane conic. Note that the value
of a is determined by those of e and f in each case, and b, where relevant, by a and e.
Since f is simply the distance from the focus to the directrix, the equation is uniquely
determined by the focus, directrix and eccentricity of the conic.
This reduction to standard form allows us to classify the congruence classes of
conics in euclidean and similarity geometries.
It also allows us to identify the symmetries in euclidean and other geometries.
focusdirectrix page
