Classifying $P(x, y) = ax^2 + bxy + cy^2 + dx + ey + h$ when $b^2 - 4ac > 0$

Question

Let $P(x, y) = ax^2 + bxy + cy^2 + dx + ey + h$ and suppose $b^2 - 4ac > 0.$

I know that we can re-write $P(x, y)$ as a polynomial of $x:$ $$P(x, y) = ax^2 + (by+d)x + (cy^2 + ey + h).$$ From here, we can get the discriminant $\Delta_x(y)$ in terms of $y:$ $$\Delta_x(y) = (by+d)^2 - 4a(cy^2 + ey + h) = (b^2 - 4ac)y^2 + (2bd - 4ae)y + (d^2 - 4ah).$$

Given the assumptions, I'm supposed to show that one of the following occurs:

1) $\{ y \mid \Delta_x(y) \geq 0\} = \mathbb{R} \text{ and } \Delta_x(y) \neq 0 $

2) $\{ y \mid \Delta_x(y) = 0\} = \{y_0\} \text{ and }\{ y \mid\Delta_x(y) > 0\} = \{ y\mid y \neq y_0\}$

3) there exist real numbers $\alpha$ and $\beta$, $\alpha < \beta$, such that $\{y \mid \Delta_x(y)\geq0\} = \{y \mid y \leq \alpha\} \cup \{y \mid y \geq \beta\}.$

In the first case, we're supposed to get a hyperbola opening left and right. In the second case, we'll get two lines intersecting at a point. In the final case, we'll get a hyperbola opening up and down. However, I can't see how to show these things rigorously. Any insight would be appreciated.

Answer 1

Hint:

Saying that $b^2-4ac$ means you have two real asymptotes, the product of the equations thereof is given by the quadratic part of the equation of the conic: $$ax^2+bxy+cy^2=0.$$

On the other hand,

1. means that for all $y$, there are exactly two roots in $x$, i.e. two points on the conic.
2. means that for $y=y_0$, there is a double root $x_0$, i.e. a double point $(x_0,y_0)$ on the conic, and two points for all other values of $y$.
3. means that there are two points $(x,y)$ on the conic for all $y$, except if $\alpha<y<\beta$.

Answer 2

Here's a more geometric approach. Let $b=2B,d=2D,e=2E$ and write the conic as $$\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}a&B&D\\B&c&E\\D&E&h\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=[0]$$,Rotate every point in the plane through anangle of $\theta$ counterclockwise about the origin. (Notice the points move, not the axes,i.e. we are using the alibi, not the alias approach to geometric transformations.) Then if the point (x,y) is rotated into the point (x',y') we have $$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} $$ and $$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{bmatrix}\begin{bmatrix}x'\\y'\end{bmatrix} $$Thus a point (x',y') is on the rotated conic iff $$\begin{bmatrix}x'&y'&1\end{bmatrix}\begin{bmatrix}\cos \theta &-\sin \theta&0\\\sin \theta&\cos \theta&0\\0&0&1\end{bmatrix}\begin{bmatrix}a&B&D\\B&c&E\\D&E&h\end{bmatrix}\begin{bmatrix}\cos \theta &\sin \theta&0\\-\sin \theta&\cos \theta&0\\0&0&1\end{bmatrix}\begin{bmatrix}x'\\y'\\1\end{bmatrix}=[0]$$ The $x'y'$term in the rotated conic is $(a-c) \sin (2 \theta)+b \cos(2\theta)$ ,which is 0 if $\tan (2\theta)=b/(c-a)$ in the case $a \ne c$ or if $\cos (2 \theta)=0$ in the case $a=c.$(The case $b=0$ is trivial.) Thus by an appropriate choice of $\theta$ we have an equation of the form $$\begin{bmatrix}x'&y'&1\end{bmatrix} \begin{bmatrix}\alpha&0&\zeta\\0&\gamma&\xi\\\zeta&\xi&h\end{bmatrix}\begin{bmatrix}x'\\y'\\1\end{bmatrix}=[0]$$ Looking at the upper left 2x2 bloks of the matrices, we have $$\begin{bmatrix}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{bmatrix}\begin{bmatrix}a&B\\B&c\end{bmatrix}\begin{bmatrix}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{bmatrix}=\begin{bmatrix}\alpha&0\\0&\gamma\end{bmatrix}$$. Taking determinants and remembering that $b=B/2$ we bave $$ac-b^2/4=\alpha \gamma$$ ,so $$b^2-4ac=-4\alpha \gamma$$ Roation does not change the type of the conic, which is a is a hyperbola iff $\alpha$ and $\gamma$ have opposite signs iff the discriminant $b^2-4ac$ is positive.