Classify a conic on a plane

Question

Given a conic C : $$ \left\{ \begin{array}{c} x^2+y^2+z^2-6y = 0 \\ x-y+z=0 \\ \end{array} \right. $$ How can i classify C defined as an orthogonal projection of C on the plane z=0 and how to find the equation ( in canonical form ) of it in this case. Thanks in advance.

Answer 1

Classification is easy: The intersection of a sphere with a plane is a circle or a single point. The circle’s orthogonal projection onto $z=0$ is an ellipse (unless the cutting plane is perpendicular to $z=0$, which it’s not).

As for finding an equation for this ellipse, various methods are available, but in this case simply eliminating $z$ from the system, as suggested by Dr. Sonnhard Graubner, is the simplest. You might also want to take a look at this related question.

Answer 2

The conic $C$, which is the orthogonal projection onto the plane $z=0$, can be obtained from replacing $z$ in the equation

$$ \begin{array}{c} x^2+y^2+z^2-6y = 0 \\ \end{array} $$

with $z= y-x$, which results in

$$x^2+y^2-(x+3)y = 0$$

In order to cast the above conic into its canonical form, the following substitutions can be used,

$$x+3=u+v, \>\>\>\>\> y=u-v$$

which leads to its canonical equation

$$\frac{(u-3)^2}{3}+\frac{(v-1)^2}{1}=1$$

As can be seen explicitly, the conic $C$ is an eclipse.