Show that $P(x,y)=0$ is an ellipse if $b^2-4ac<0$.

Question

I tried the following:

I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $\Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. The discriminant of the $\Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$ is $\Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.

My questions:

i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={\{y \ | \ \Delta_x(y) \ge 0}\} = \emptyset$, $K_2={\{y \ | \ \Delta_x(y) \ge 0}\} = {\{y_0}\}$, or there exist real numbers $\alpha$ and $\beta$ such that $K_3={\{y \ | \ \Delta_x(y) \ge 0}\} = {\{\alpha \le y \le \beta}\}$. How to show that only one of this three cases happens?

ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $\pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).

Answer 1

For i.

Let $Y=\Delta_x(y)$. Then, note that $Y=\Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.

For ii.

for example it can be a part of a hyperbola too [the normal one $\pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation

It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
In the case $K_3$ where there exist real numbers $\alpha,\beta$ such that $\alpha\le y\le \beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.

also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).

I think that in this context, it may be supposed that circles are a special case of ellipses.

Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.

If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have $$b^2-4ac=-4\lt 0,\qquad \Delta_x(y) \ge 0\iff 1-\sqrt 2\le y\le 1+\sqrt 2$$ and $$P(x,y)=0\iff (x-1)^2+(y-1)^2=2$$ which is the equation of a circle.

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Added :

Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $\theta$ :

$$x=X\cos\theta+Y\sin\theta,\qquad y=-X\sin\theta+Y\cos\theta$$

Then, we get $$\begin{align}A&=a\cos^2\theta-b\sin\theta\cos\theta+c\sin^2\theta \\\\B&=(a-c)\sin(2\theta)+b\cos(2\theta) \\\\C&=a\sin^2\theta+b\sin\theta\cos\theta+c\sin^2\theta\end{align}$$

Since $$A+C=a+c,\qquad A-C=(a-c)\cos(2\theta)-b\sin(2\theta)$$ we get $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$

So, when $B^2=0$, we get $$b^2-4ac=-4AC$$

It follows that if $b^2-4ac\lt 0$, i.e. $AC\gt 0$, i.e. either $A\gt 0,C\gt 0$ or $A\lt 0,C\lt 0$, then the equation $$AX^2+0XY+CY^2+\cdots =0$$ represents an ellipse.