# The general equation of a conic section in the plane of the conic

In my Calculus book it says:

We have shown that planes are represented by first-degree equations and cones by second-degree equations. Therefore, all conics can be represented analytically (in terms of Cartessian coordinates $$x$$ and $$y$$ in the plane of the conic) by a second-degree equation of the general form $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,$$ where $$A, B, ..., F$$ are constants...

How can the conclusion about the general form of the equation of a conic be drawn from the premises in the first sentence?

• Apply a "rigid" transformation that moves the "plane of the conic" to the $xy$-plane itself. The transformed cone will still have a second-degree Cartesian equation in $x$, $y$, $z$; we get the equation of cone's intersection with the now-conveniently-placed plane of the conic by setting $z=0$, leaving the general $xy$-form shown.
– Blue
Commented Jul 5, 2019 at 11:47

In fact, the intersection of a plane with any quadric surface results in a conic. I find it most convenient to derive this using the homogeneous matrix form of these equations.

Letting $$\mathbf X = (X,Y,Z,1)^T$$, consider the quadric surface with the implicit equation $$\mathbf X^TQ\mathbf X=0$$, where $$Q$$ is a symmetric $$4\times4$$ matrix. If we write $$\mathbf x=(x,y,1)^T$$, where $$x$$ and $$y$$ are the coordinates of a point on the plane in its coordinate system, then there is a $$4\times3$$ matrix $$M$$ such that $$\mathbf X=M\mathbf x$$ is the mapping from plane coordinates to space coordinates. In particular, if $$\tilde{\mathbf X}$$ and $$\tilde{\mathbf Y}$$ are the (inhomogeneous) coordinates in $$\mathbb R^3$$ of the unit coordinate vectors of the plane’s coordinate system, and $$\tilde{\mathbf C}$$ the origin of this coordinate system, then $$M = \begin{bmatrix}\tilde{\mathbf X} & \tilde{\mathbf Y}&\tilde{\mathbf C}\\0&0&1\end{bmatrix}$$ is one such matrix. (In fact, every parameterization of the plane has a corresponding matrix $$M$$: $$\mathbf X=M\mathbf x$$ is just a compact way to write this parameterization.)

Substituting into the equation of the quadric, we have $$\mathbf X^TQ\mathbf X = (M\mathbf x)^TQ(M\mathbf x) = \mathbf x^T(M^TQM)\mathbf x = 0. \tag{*}$$ Now, $$M^TQM$$ is itself a symmetric matrix, so if $$M^TQM=\begin{bmatrix} A & \frac B2 & \frac D2 \\ \frac B2 & C & \frac E2 \\ \frac D2 & \frac E2 & F\end{bmatrix}$$ then equation (*) expands into $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$. So, every intersection of a plane with this surface has an equation of this form in the plane’s Cartesian coordinate system.

Every circular cone is an affine image of the cone $$x^2+y^2=z^2$$, so its implicit equation is, in homogeneous matrix form, $$\mathbf X^T(A^T\operatorname{diag}(1,1,-1,0)\,A)\mathbf X=0$$, where $$A$$ is the matrix of an invertible affine transformation. The central matrix in this equation is clearly symmetric, so this is a special case of the above and therefore every conic has a general equation of the required form. Actually, to cover every type of conic, one also has to consider intersections of a plane with a cylinder, which can be considered a degenerate cone, but clearly that’s also a special case of the above derivation.

Well, it's a bit of a stretch. Take a conic, represented by a homogeneous quadratic like $$ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0,$$ which is what I assume was meant by "cones are represented by second-degree equations." Then take a plane, represented by something like $$px + qy + rz + s = 0.$$ Supposing, for a moment, that $$r \ne 0$$, we can write an equivalent representation, namely $$\frac{p}{r}x + \frac{q}{r}y + z + \frac{s}{r} = 0,$$ which I'll rewrite by assigning new names, as $$p'x + q'y + z + s' = 0$$ and then find, that for every point $$(x, y, z)$$ on this plane, we have $$z = -s' -p'x -q'y$$ So if a point lies on this plane, it looks like $$(x, y, z) = (x, y, -s' -p'x -q'y)$$ Furthermore, the numbers $$x$$ and $$y$$ constitute 'coordinates' for that plane, i.e., every point of the plane corresponds to a unique $$xy$$-pair, and vice versa.

Now if that point also lies on the conic I mentioned earlier, then we must have $$ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0,$$ which means (replacing $$z$$ with $$-s' -p'x -q'y$$) that $$ax^2 + by^2 + c(-s' -p'x -q'y)^2 + dxy + ex(-s' -p'x -q'y) + fy(-s' -p'x -q'y) = 0,$$ which we can multiply out, and gather like terms in powers of $$x$$ and $$y$$ to get something that looks like $$(a + p'^2 -ep')x^2 + (b + q'^2-fq')y^2 + dots = 0.$$ Letting $$A = (a + p'^2 -ep'), B = (b + q'^2-fq'), \ldots,$$ we have exactly the form promised in the quoted section of your question.

Of course, this all assumed that $$r \ne 0$$. In the case where $$r = 0$$, we know that one of $$p$$ and $$q$$ must be nonzero, and we simply do the same thing with whichever one is nonzero (or pick one at random, if both are nonzero). Suppose that $$q \ne 0$$. Then we end up with an expression for $$y$$ in terms of $$x$$ and $$z$$, and we discover that $$x$$ and $$z$$ work as coordinates for the plane, and we end up with a quadratic in $$x$$ and $$z$$ instead of in $$x$$ and $$y$$, but that's still fine. Similarly, if $$p \ne 0$$, we can end up with a quadratic in $$y$$ and $$z$$.