3
$\begingroup$

It is well known that any (possibly degenerated) conic section in $\mathbb{R}P^2$ is given by, up to a projective transformation, a point, a line, two lines or a circle (given by the equation $x_0^2+x_1^2=x_2^2$ ).

Now, I want to look at these solutions as subsets in $\mathbb{R}^3$, under the usual identification $\mathbb{R}P^2=S^2$ (with antipodal points identified), where I draw lines through the origin and all points on a given conic of $\mathbb{R}P^2$. The result is in most cases a cone, (where I define a cone as the solid obtained by rotating a line about a line through one of its point):

$\bullet$ A circle becomes a regular cone

$\bullet$ A line becomes a plane (which can be seen as a 'flat' cone)

$\bullet$ A point becomes a line (again, a degenerated cone)

$\bullet$ Two lines become two intersecting planes.

This last case doesn't really fit into the picture. My question thus is how we can interpret two intersecting planes in $\mathbb{R}^3$ as some kind of degenerated cone.

$\endgroup$
3
$\begingroup$

Consider the equation $x_0^2 - x_1^2 = rx_2^2$ (visualize this as a cone in any way you choose) and then let $r\to 0$ to see how the cone degenerates into two intersecting planes.

$\endgroup$
  • 1
    $\begingroup$ Nice idea, hadn't thought of, 'stretching out' an ellipse to two parallel lines by seperating the foci. It's hard to think of this as some 'rotating line' but I think this is a fine answer. thanks! $\endgroup$ – ArtW Jun 18 '16 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.