Krdytkyu

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).

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 $alphale yle beta$, 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
$$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
and
$$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
which is the equation of a circle.

Added :

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

$$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$

Then, we get
$$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
\\B&=(a-c)sin(2theta)+bcos(2theta)
\\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$

Since
$$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
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 $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.