Can asymptotes be considered as common tangent to two branches of hyperbola? Today I was solving archive of jee advanced and I found a matrix question of 2008 regarding whether two branches of a hyperbola has a common tangent or not. The answer was no they don't. But then I  thought of asymptotes and it's definition everywhere comprises of the fact that it's a tangent at infinity to the hyperbola. And so now I'm a bit confused regarding the very definition of the hyperbola. Thanks in advance.
 A: An asymptote is a line that approaches a curve arbitrarily closely. Being a tangent at infinity is a mere misconception so we can say that we don't have any common tangent for a hyperbola.
A: In projective geometry, the two asymptotes are two lines tangent to the projective closure of the hyperbola (at two different points) whereas a common tangent is one line tangent to two different curves (at two different points). Projectively the hyperbola is just one curve as the circle is one curve.
To make this precise. In projective geometry we have $x^2-y^2=z^2.$ If we let $z=1$ we get $x^2-y^2=1$ which is a hyperbola in the affine plane.

Now if we look at the line at infinity, $z=0,$ we get $x^2-y^2=0$ in the ${\Bbb P}^1$ or algebraically $k[x,y].$ This only means that in the affine $y=1$, $x=\pm 1$ or the curve $x^2-y^2=z^2$ intersects the line at infinity in two points: $(x:y:z)=(\pm 1:1:0).$ The asymptotes $x^2-y^2=0$ also intersect the line at infinity in these two points. To make things clearer we can look at the other two affines $x=1$ and $y=1$: $1^2-y^2=z^2$ and $x^2-1^2=z^2$ or $y^2+z^2=1$ and $x^2-z^2=1.$ One is the circle and the other is a hyperbola like in the affine $z=1.$ Let's look closer at the circle: $y^2+z^2=1$ and the line $z=0.$ This is now just a line through the origin and intersects the circle in two diametrically opposite points: $(1:\pm 1:0).$ These are the same two points as before, projectively. Now look at the lines $x+y=0$ and $x-y=0$ in the $z=1$ affine. In the $x=1$ affine these become $1+y=0$ and $1-y=0$ or the tangent lines $y=1$ at $(1:1:0)$ and $y=-1$ at $(1:-1:0)$ to $y^2+z^2=1.$

Added: over the reals remember that ${\Bbb P}^2$ is not orientable.
