equations of conic sections
Given a point V, a line L through V, and αε(0,π/2),
the cone with vertex V, axis L and angle α
all points on lines through V making angle α with L.
These lines are the generators
of the cone.
To simplify the algebra, we consider the intersection of a general cone
in R3 with the xy-plane.
Suppose that the cone C has vertex V =(u,v,w), that d is a unit direction vector for L, the
axis of C, and that C has angle α. Then the point P lies on C if and only if vector VP makes
an angle α or π-α with d. In terms of scalar products, this condition can then be stated as
VP.d = ±|VP|cos(α). Squaring this, and writing P = (x,y,z), and d = (a,b,c), we see that
P is on C if and only if ((x-u)a + (y-v)b+ (z-w)c)2 =
((x-u)2 + (y-v)2 + (z-w)2)(cos(α))2.
Even without multiplying out, this is easily seen to be, at worst, quadratic in x, y and z.
In fact it must have at least one quadratic term since it is not a plane as α < π/2.
We obtain an equation for the intersection with the xy-plane by setting z = 0. The resulting..+a2x2+b2y2+2abxy+..=
This has a non-zero xy-term unless ab = 0. If a = 0, then the x2-term on the left vanishes,
equation will be at most quadratic in x and y. If we concentrate on the terms in x2, y2
and xy, we get
but, as cos(α) ≠ 0, that on the right does not, so the equation has a non-zero x2-term.
Similarly, if b = 0, there is a non-zero y2-term.
For the general case, given a plane Π and a cone C, we may choose Π as the xy-plane.
Then the above analysis yields
Theoremf(x,y) = Ax2+Bxy+Cy2+Fx+Gy+H = 0,
where at least one of A, B and C is non-zero, i.e. f is actually quadratic in x and y.
The curve obtained by intersecting a cone with a plane has an equation of the form