Is the discriminant of a second order equation related to the graph of $ax^2+bxy+cy^2+dx+ey+f=0$?

Question

Is the discriminant of a second order equation related to the graph of $ax^2+bxy+cy^2+dx+ey+f=0$?

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Most people who took precalculus know that $ax^2+bxy+cy^2+dx+ey+f=0$ is the graph of:

• An ellipse if $b^2-4ac\lt0$
• A parabola if $b^2-4ac=0$
• A hyperbola if $b^2-4ac\gt0$

Is this related to the fact that the discriminant of $ax^2+bx+c=0$ is $\Delta=b^2-4ac$, which tell us that the equation has 2 solutions if $b^2-4ac\gt0$, one if $b^2-4ac=0$, and none (in the reals) $b^2-4ac\lt0$?

Answer 1

As far as I know there is no conceptual relation between the two discriminants.

For a conics $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ the key fact is that the discriminant $B^2-4AC$ is invariant for rotations. This means that if the rotation gives an equation $$A'x^2+B'xy+C'y^2+D'x+E'y+F'=0$$
than $B^2-4AC=B'^2-4A'C'$ ( for a proof you can see here).

So we can always transform the equation of a conic in a form where the mixed term is absent ($B'=0$) choosing an angle $\theta=\frac{1}{2}\arctan\left(\frac{B}{A-C}\right)$.

With such a rotation we find a conic that has an axis of symmetry parallel to a coordinate axis and the classification of the type of conic (if it is not degenerate) is simple since we have a parabola if $A'C' =0 \iff B^2-AC=-4A'C'=0$, an ellipse if $A'C'>0 \iff B^2-4AC=-4A'C'<0$ and an hyperbola if $A'C' <0 \iff B^2-AC=-4A'C'>0$. But all this this has no simple relation with the existence of real roots as in the case of the discriminant of a second degree equation.

Answer 2

Both solving a quadratic and diagonalising a quadratic form involve completing the square: $$ 0 = ax^2+bx+c = a\left( x+\frac{b}{2a}\right)^2- \frac{1}{4a}(b^2-4ac) \\ ax^2+bxy+cy^2 = a\left( x+\frac{b}{2a}y\right)^2- \frac{1}{4a}(b^2-4ac)y^2, $$ so in both cases the sign of the quantity $b^2-4ac$ determines the nature of solutions.

Geometrically, $ax^2+bxy+cy^2+dx+ey+f=0$ is the intersection of the conic $z=ax^2+bxy+cy^2$ and the plane $ dx+ey+z+f=0 $. Choosing the plane as $z=0$, we can then ask the question: what does the conic $$ ax^2+bxy+cy^2 = 0 $$ look like? Clearly the point $x=y=0$ is part of it. Does it contain any other points? If $a=0$, it includes the line $y=0$. Now suppose $a \neq 0$. Then there are clearly no points with $y=0$ other than $x=y=0$. $$ a\frac{x^2}{y^2} + b \frac{x}{y}+c=0 $$ is a quadratic in $x/y$, and has real solutions if and only if $b^2-4ac>0$. Therefore the conclusion is:

1. If $a=0$, the conic $ ax^2+bxy+cy^2 = 0 $ consists of the lines $y=0$ and $bx+cy=0$.
2. If $a \neq 0$, the conic $ ax^2+bxy+cy^2 = 0 $ consists of the lines $x-\alpha_i y=0$, where $\alpha_i$ are the roots of the quadratic $az^2+bz+c=0$.

Therefore the conclusion is that if $b^2-4ac>0$, $ ax^2+bxy+cy^2 = 0 $ consists of 2 lines, if $b^2-4ac=0$, it is one line, and if $b^2-4ac<0$, it is only the point $x=y=0$.

Those are the connections, as far as I can see.