On the intuitive explanation for the discriminant property of conics

Question

In this old post detailing the intuition for the classification of conics based on the sign of the discriminant, the author has considered the coordinates $x$ and $y$ of the point on a conic to become very large as you zoom out on it, and used that idea in the initial simplification of the general conic equation. This is apparent when you consider, say, a hyperbola with the coordinate axes as its axes.

However, this simplification seems to run into a problem when you consider a hyperbolas like $xy=c^2$ or even just a parabola of the form $y^2=4ax$. For the hyperbola case, when you zoom out, one coordinate dwindles into zero close to either axis while the other gets large as stated. For the parabola case, $y$ becomes negligible compared to $x$. Thus, the original treatment approximating the conic equation seems to fail here.

I managed to work things out for these specific cases; for $xy=c^2$, the coefficients $A$ and $B$ as in the original post would be zero anyway, and the sole real solution for $\frac{y}{x}$ results (zero, representing the $x$-axis). Similar justification can be provided for the parabola case I brought up.

But that could still be just coincidence; I can't seem to find a general justification of this intuitive proof for conics which pose similar problems as the above examples. For example, hyperbolas obtained by slightly tilting $xy=c^2$ pose the same problem; one variable does not become larger, but remains small compared to the other variable.

Can someone help prove how the original author's initial treatment of the general equation, with the coordinates growing large, still holds for such cases?

Answer 1

I'd restate that intuitive explanation as follows. Retain only the quadratic terms of the equation of the conic and let $f(x,y)=0$ be the resulting equation.

If $f(x,y)$ can be factored into the product of two different linear polynomials (i.e. $\Delta>0$), then you get the equation of two distinct lines and the conic is a hyperbola (two points "at infinity").

If $f(x,y)$ can be factored into the product of two equal linear polynomials ($\Delta=0$), then you get two coincident lines and the conic is a parabola (one point "at infinity").

If $f(x,y)$ cannot be factored over the reals ($\Delta<0$), then you have no points "at infinity" and the conic is an ellipse.

Note also that this argument can be made more rigorous by using homogeneous coordinates.

Answer 2

Apart from these two cases (and the parabola with the y axis as its principal axis), this problem actually never arises. No matter how small the slope of the asymptote (or principal axis, for parabolas) is, the coordinates of the points on it still go to infinity as we zoom out. This simplification is thus justified for all cases.

Update: as noted in the comments, it has to be proven that the parabola can be approximated as its axis far enough along the axis toward infinity. For this, it is sufficient to prove that the slope of the parabola approaches that of the axis at infinite distance along the axis.

Consider a general parabola with axis equation $a x+b y+c=0$ and tangent at vertex $b x-a y+c^{\prime}=0$, and latus rectum $4 \alpha$. So the equation of parabola will n be; $$ \frac{(a x+b y+c)^{2}}{\sqrt{a^{2}+b^{2}}}=4 \alpha\left(b x-a y+c^{\prime}\right)...(1) $$ On differentiating with respect to $x$, $$ \begin{aligned} & 2(a x+b y+c)(a+b m)=4 \alpha \sqrt{a^{2}+b^{2}}(b-a m) \\ \Rightarrow & m=\frac{2 \alpha b \sqrt{a^{2}+b^{2}}-a(a x+b y+c)}{b(a x+b y+c)+2 \alpha a \sqrt{a^{2}+b^{2}}} \end{aligned} .... (2)$$ Substituting $(ax+by+c)$ from $(1)$ in $(2)$, $$ \begin{aligned} m &= \frac{2 \alpha b \sqrt{a^{2}+b^{2}}+2 a \sqrt{\alpha \sqrt{a^{2}+b^{2}(b x-a y+c')}}} {\pm 2 b \sqrt{\alpha \sqrt{a^{2}+b^{2}}\left(b x-a y+c'\right)}+2 \alpha a \sqrt{a^{2}+b^{2}}} \end{aligned} $$ As we move an infinite distance along the axis, the distance of the point on the parabola from the tangent, i. e. , $\frac{b x-a y+c^{\prime}}{\sqrt{a^{2}+b^{2}}}$ and consequently, $b x-a y+c^{\prime}$ will go to infinity (since the equation of the tangent is selected so as to provide positive values for points on the parabola).

Considering $b x-a y+c^{\prime}=\Psi$, $m=\frac{\frac{2 \alpha b \sqrt{a^{2}+b^{2}}}{\sqrt{\psi}} \mp 2 a \sqrt{\alpha \sqrt{a^{2}+b^{2}}}} {\pm 2b \sqrt{\alpha \sqrt{a^{2}+b^{2}}}+\frac{2 \alpha a \sqrt{a^{2}+b^{2}}}{\sqrt{\psi}}}$ And as $\Psi \rightarrow \infty, m \rightarrow \frac{-a}{b}$, which is the slope of the axis of the parabola. Thus, the parabola can be assumed to taper off into a line as we move farther and farther along the axis, which, since it is of the same slope as its axis, seems to coincide with it on scaling it down.