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

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 $$\alpha\le y\le \beta$$, 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.

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.

• ♥♥ LOVE YOU!! ♥♥ – 72D Nov 27 '18 at 0:25