The relation between discriminant of conic sections and oblique asymptote

Question

The general form of the conic sections is $Ax^2 + Bxy + Cy^2 + ax + by + c = 0$ . If we divide both sides by $x^2$ then $$A + B(\frac{y}{x}) + C(\frac{y}{x})^2 + a(\frac{1}{x}) + b(\frac{y}{x^2}) + c(\frac{1}{x^2}) = 0 \qquad (1)$$ Also $\lim_{x \to \pm \infty} \frac{y}{x}$ gives us the slope of oblique asymptote . So if we let $x \to \pm \infty$ $$A +Bm + Cm^2 = 0 \qquad (2)$$ The discriminant of $(2)$ is $\Delta = B^2 - 4AC$ which is the same expression for conic sections . How we can explain this connection ?

Answer 1

A short answer is that the slopes of the asymptotes, and indeed, whether or not there are any, are determined entirely by the quadratic part of the equation. What you’ve essentially done in deriving your equation for their slopes is isolate that quadratic part, so it’s not unsurprising that the discriminant should still make an appearance.

We can see how the discriminant enters into the values slopes of the asymptotes by directly constructing their equations. Their slopes are unaffected by translation, so w.l.o.g. we can assume that $a=b=0$. The asymptotes of a hyperbola are the degenerate member of a family of hyperbolas that share these asymptotes; we get the equation of this degenerate conic by setting $c=0$: $$Ax^2+Bxy+Cy^2=0.$$ If $C\ne0$ this equation factors into the lines $$2Cx - \left(B\pm\sqrt{B^2-4AC}\right)y = 0.$$ This is a pair of real lines iff $B^2-4AC\gt0$, which is exactly the condition that the original conic is a hyperbola. The slopes of these lines are $$m = -{2C \over B\pm\sqrt{B^2-4AC}} = \frac1{2C}\left(B\pm\sqrt{B^2-4AC}\right),$$ which are precisely the roots of your equation for the slopes. If $C=0$, then $A\ne0$ and we find similar equations for the asymptotes that also involve the square root of the discriminant.

To get the equations of the original conic’s asymptotes, translate the origin to the hyperbola’s center. The center can be found by setting the partial derivatives to zero, which leads to the system of linear equations $$\begin{align} 2Ax+By+c&=0 \\ Bx+2Cy+b &= 0. \end{align}$$ The discriminant again makes an appearance as the determinant in the denominators of the Cramer’s rule solution to this system: the conic has a well-defined center iff it is nonzero, i.e., the conic is an ellipse or hyperbola.

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To my mind, a natural reason for the conic’s discriminant’s playing such a central role in the above can be found in projective geometry. In the real projective plane, the discriminant of a conic determines how many real intersections it has with the line at infinity: two, one or none. These intersections, if any, correspond to the directions of the asymptotes, which are the tangents to the conic at these intersection points at infinity.

Working in homogeneous coordinates, the matrix of the general conic equation in the question is $$Q=\begin{bmatrix} A&\frac B2&\frac a2 \\ \frac B2&C&\frac b2 \\ \frac a2&\frac b2&c \end{bmatrix}$$ and the line at infinity is $\mathbf l_\infty = [0:0:1]$. The dual conic $$D=\mathcal M_{\mathbf l_\infty}^TQ\mathcal M_{\mathbf l_\infty} = \begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix} \begin{bmatrix} A&\frac B2&\frac a2 \\ \frac B2&C&\frac b2 \\ \frac a2&\frac b2&c \end{bmatrix} \begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix} = \begin{bmatrix} C&-\frac B2&0 \\ -\frac B2&A&0 \\ 0&0&0 \end{bmatrix}$$ is in general a rank-two matrix that represents the intersection points of the conic and line at infinity: the quadratic form vanishes for those lines that pass through at least one of the points. Notice that this matrix depends only on the quadratic part of the conic. This makes sense: translation leaves the quadratic part unchanged and doesn’t change the direction of the asymptotes.

The intersection points $\mathbf p$ and $\mathbf q$ can be extracted by “splitting” the matrix: find a value of $\alpha$ such that $D+\alpha M_{\mathbf l_\infty}$ has rank one. The resulting matrix will be a scalar multiple of either $\mathbf p\mathbf q^T$ or $\mathbf q\mathbf p^T$, from which we can read the two points.

For $D+\alpha M_{\mathbf l_\infty}$ to be a rank-one matrix, all of its $2\times2$ minors must vanish. There’s only one that’s not identically zero, and it leads to the equation $\alpha^2=\frac {B^2}4+AC$ with solutions $\pm\frac12\sqrt{B^2-4AC}$. We can see from this that the discriminant of the conic determines the number of its (real) intersections with $\mathbf l_\infty$. Taking the positive root produces $$\begin{bmatrix} C & -\frac B2-\frac12\sqrt{B^2-4AC} & 0 \\ -\frac B2+\frac12\sqrt{B^2-4AC} & A & 0 \\ 0&0&0 \end{bmatrix}$$ and the intersection point(s) are any row/column pair of a non-zero diagonal entry. If $A\ne0$, we can take the second row and column, and the slopes of the asymptotes are then $$m=-\frac B{2A}\pm{\sqrt{B^2-4AC}\over2A}$$ which are the roots of your equation $Am^2+Bm+C=0$.