# If every tangent line of a curve passes through a point, it is a line

1. Sep 28, 2011

### demonelite123

If every tangent line of some curve B(s) which has the unit speed parametrization passes through a fixed point P, then the curve B(s) must be a line.

As a hint, my book says p = B(s) + r(s)B'(s) where r(s) is some function.

so i have that any tangent line L(t) = B(s) + t B'(s) and for some t, L(t) = p = B(s) + r(s)B'(s). i am having trouble continuing with this problem. i cannot think of anything else to do. could someone give me a hint or two on how to proceed? thanks!

2. Sep 28, 2011

### lavinia

Here are two ideas.

- To use your equation recall that the acceleration of a unit speed curve must be perpendicular to the curve.

- Pick two points on the curve and draw the two lines from these two points to the fixed point,p. These lines are tangent to the curve and form a cone. Show that the entire curve must lie inside this cone.

Last edited: Sep 28, 2011
3. Sep 29, 2011

### demonelite123

hi, using your suggestion i differentiated the equation p = B(s) + r(s)B'(s) on both sides to get:
0 = B'(s) + r'(s)B'(s) + r(s)B''(s) and since |B'(s)| = 1, B'(s) * B''(s) = 0 so i take the dot product of B''(s) on both sides to get: 0 = r(s)|B''(s)|2. now if r(s) is equal to 0 for all s, then it is the trivial case of a line p + sv where v = 0. if r(s) is not equal to 0 for any s then i can divide both sides by it and i end up with |B''(s)| = 0 which means B''(s) is the zero vector or that B'(s) is a constant and that B(s) = p + sv where v = B'(s).

the only trouble i'm having is the case when r(s) is 0 for only some values of s. Then p must be some point on the curve itself. I argue that in order for every tangent line to pass through p, they must all be parallel to the tangent line that passes through p itself. however this is more of an intuitive thought and i don't really have a rigorous justification for it. what can be used to prove the statement in this case when r(s) is 0 for only some values of s?

4. Sep 30, 2011

### Bacle

I think you can also use the Frenet-Serret frames to conclude that the curvature of the
curve is identically zero. May be a good idea to translate/shift the curve so that it goes
thru the origin.

EDIT: I know that to use Frenet-Serret frames we need to have non-trivial curvature;
the idea was to get a contradiction by assuming we have a curve with non-trivial torsion
and curvature (a line having both equal to zero), and then arriving at a contradiction, but
I have not been able to get much from this yet; I'll try for a while longer.

Last edited: Sep 30, 2011
5. Oct 5, 2011

### mathwonk

for a counter example in characteristic p algebraic geometry google "strange curves".

6. Oct 7, 2011

### mathwonk

i.e. in characteristic 2, a conic can have concurrent tangents.