If the line at infinity is a secant line of a conic then the conic is a hyperbola? I'm learning perspective geometry and I have a question about the line at infinity. Is it true that if the conic has the line at infinity as its secant then the conic represents a hyperbola in Euclidean geometry?
If it is, how can I imagine the shape hyperbola using the line at infinity? I think I can lift up the line to infinity but it seems like a parabola than a hyperbola.
I draw a conic going through $A, B, C, D, E$, and I choose AB as the line at infinity. Then I lift it up and it seems like a parabola

P.S: I'm quite new to this field so I may use the wrong jargon. Thank you
 A: I assume you are experimenting in Geogebra to get an intuition for this concept, and I'll answer from that point of view.
What you're currently doing is making $A,B$ approach the same point at infinity.  As a result you are constructing a conic that meets the line at infinity at a single point, and thus has the line at infinity as a tangent, i.e. the conic is a parabola.
Instead, draw two lines $a,b$ intersecting at the origin, and then create a third  line $c$. Let points $A,B$ be the intersections of $c$ with $a,b$.  Then create a conic that goes through the points $A,B,C,D,E$ and move $c$ towards infinity. You'll tend towards a hyperbola. (It's more interesting if you choose $C,D,E$ so that your conic starts off as an ellipse.).  Using this construction you'll ensure that $A,B$ approach different points at infinity, and you'll get a conic for which the line at infinity is a true secant.
A: Imagine three dimensional Euclidean space with the origin at $\,O\,$ and a
plane $\,P\,$ not containing $\,O.\,$ The projective plane associated with
the space is all of the lines that pass through $\,O.\,$ Each of these lines
intersects the plane $\,P\,$ except those that lie in the plane $\,Q\,$ that
passes through the origin $\,O\,$ and is parallel to $\,P.\,$ Thus, the points
in $\,P\,$ represent ordinary projective points. Pick a circle $\,L\,$ with
center $\,O\,$ that lies in $\,Q.\,$ Pairs of antipodal points in $\,L\,$ are
the points in the line at infinity.
Now pick a double cone $\,D\,$ with center at $\,O.\,$ The intersection of
$\,D\,$ with plane $\,P\,$ is a conic section $\,C.\,$ The number of points
at infinity of $\,D\,$ is at most two. The case with two points at infinity
are those where $\,C\,$ is a hyperbola because two lines of $\,D\,$ pass
through $\,O\,$ and also intersect $\,L\,$ in two points at infinity. Note
that a Euclidean hyperbola has two connected open real components because
they omit the two points at infinity that connect them. Similarly a Euclidean
parabola has one open real component and is missing one point at infinity.
A: Suppose you have an infinite horizontal table. Some distance above the table you have a point light. You're holding a metal ring and observing it's shadow on the table.
If the ring is fully below the horizontal plane through the light, you get an ellipse. That horizontal plane is the 3d counterpart to the line at infinity, and you're not intersecting that.
If the ring touches the horizontal plane in a single point, you get a parabola which intersects the line at infinity in a single point (of algebraic multiplicity two) so the line at infinity is a tangent.
If the ring is partially below and partially above the horizontal plane is the light, only the part below casts a shadow on the table. You get one component of a hyperbola, with the two points where the ring intersects the plane of the light as points at infinity, defining the asymptotic directions of the hyperbola. The comparison to projective geometry is suffering a bit here because the physical model uses rays where the mathematical setup uses lines. To better model that you'd take a pair of rings with the light as a center of point reflection. That way you get both arms of the hyperbola.
This explanation may not be clearer than the other answers when you just read it, but it should be possible to actually try this out with a real lamp, (finite) table and wire hoop. So perhaps that experiment can help your intuition.
