The Journal of Online Mathematics and Its Applications, Volume 8 (2008)
The Most Marvelous Theorem in Mathematics, Dan Kalman

Recognizing a Quadratic Equation as a Conic Section

Here we begin with a generic quadratic equation and see that it is always possible to rotate and translate the corresponding curve to something that has a much simpler equation. Let us start with the equation

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

representing a curve Γ. The first step is to show that for an appropriate rotation of Γ, the corresponding curve has no xy term. As shown here, rotating Γ counterclockwise about the origin through an angle α has the effect of replacing x with x cos α + y sin α and replacing y with y cos αx sin α in the equation. Therefore, the equation of the rotated curve will be

A(x cos α + y sin α)2 + B(x cos α + y sin α)(y cos αx sin α) + C(y cos αx sin α)2 + D(x cos α + y sin α) + E(y cos αx sin α) + F = 0

If this is expanded fully, what are the terms containing an xy factor? We will have the cross term from the two perfect squares, as well as the firsts and lasts products from B(x cos α + y sin α)(y cos αx sin α). Combining just those terms shows that the coefficient of xy in the new equation will be

Bnew = 2A cos α sin α − 2C cos α sin α + B( cos2α − sin2α) = 2 cos α sin α(A − C) + ( cos2α − sin2α)B.

We would like this to equal 0. Using trigonometric identities, we can write the desired condition as

Bnew = sin(2α)(AC) + cos(2α)B = 0

or equivalently sin(2α)(CA) = cos(2α)B.

Now we may suppose that B is not zero, since in that case no rotation is necessary to eliminate the xy term from the original equation. Also, by agreeing that we will not rotate through an angle of 0, π/2,  or π, we avoid the possibility that sin(2α) is zero. Accordingly, we can rewrite the preceding equation as

 cos 2alpha/ sin 2alpha = (C − A)/B

or more simply

cot 2 alpha = (C − A)/2 .

This shows that it is possible to rotate the curve in such a way that the new equation has no xy term: we rotate through an angle of alpha = (1/2) cot inv ((C − A)/B)

After rotating the curve, the new equation will have the form

A′ x2+ C′ y2+ D′ x + E′ y + F′ = 0

where the new coefficients will depend on the original coefficients and on α. It is possible that A or C will now be 0. If not, we can combine the quadratic and linear terms in each variable by completing the square. For example, if A is nonzero, we can write the x terms as follows

completing square with x terms

A similar substitution can be made for the y terms if C′ is not zero. After making these changes, each variable appears just once, either in a linear term, or as the square of a linear term. With a final translation, we can replace each square of a linear term with a simple x2 or y2. That will reduce the original equation to one of the following forms

Ax2 + C″ y2 + F″ = 0
Cy2 + Dx + F″ = 0
Ax2 + Ey + F″ = 0.

These are all conic sections, or degenerate forms of conic sections (for example, a pair of crossing lines).