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If every tangent line to a curve $\Gamma$ runs through a common point $\textbf{P}_o$, then that curve is a line.

Any help? I don't know how to start.

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    Via the translational change of coordinates, ${\bf X} \to {\bf X} - {\bf P}_o$, we may take ${\bf P}_0 = 0$. Then the condition simply says that ${\bf Q}(t) \parallel {\bf Q}'(t)$, or equivalently that ${\bf Q}(t) = \lambda(t) {\bf Q}'(t)$ for some function $\lambda$. Of course, we may as well reparameterize so that $|{\bf Q}'| \equiv 1$. (Note that if one assumes that $\Gamma$ is at least $C^2$, we can give an alternative proof by differentiating and showing that ${\bf Q}''(t) = 0$, i.e., that ${\bf Q}$ is linear in $t$ and hence traces out a line.) – 2017-02-15
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    Travis: I notice that lately more and more users post perfect answers as comments to the question. What is the reason behind this? – 2017-02-15
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    @ChristianBlatter I am certainly sometimes guilty of this---and I have tried to rid myself of the habit! But in this case I regard my comment as small collection of possibly helpful but unorganized observations and so not quite substantive enough to constitute a proper answer (in fact, I had started my remarks as an answer and was dissatisfied with it, and so I posted it as a comment instead). – 2017-02-16

2 Answers 2

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Let $P_0(x_0,y_0)$ and $y=y(x)$ the equation for the curve $\Gamma$.

$$y'(x)=\frac{y-y_0}{x-x_0}$$

$$\frac{dy}{y-y_0}=\frac{dx}{x-x_0}$$

$$\ln(y-y_0)=\ln(x-x_0)+k$$

$$y-y_0=C(x-x_0)$$

is the equation for a straight line. Being $C$ it's slope.

ADDED

I follow the differential idea because one of the tags the question has attached.

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    Can you explain a little bit? I understand that a derivative gives the slope of the tangent line to a curve, but the reason you can express every tangent line to $\Gamma$ as what you've written in your second line because they all pass through $\textbf{P}_o$, correct? – 2017-02-15
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    I try to answer in a comment. There are in the first equation two points at play: $P_0$ with coordinates $(x_0,y_0)$ and a generic point of $\Gamma$ with coordinates $(x,y)$ and with derivative, at this point, $y'(x)$: Now, we have two points connected by a straight line. Now, the slope of this line is $(y-y_0)/(x-x_0)$ has to be equal to the one of the tangent (because, by hypothesis, this line is the tangent), so is, to the derivative at $(x,y)$ The resulting equation has to be satisfied by $\Gamma$ at any point $(x,y)$ – 2017-02-15
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    Ignore the very last words "at any point $(x,y)$". – 2017-02-15
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    Note that the method in this solution entails a few extra assumptions on $\Gamma$: By writing the curve as the graph of a function $y(x)$, one assumes (1) that the curve is a subset of $\Bbb R^2$ (or at least planar), and (2) that $\Gamma$ *is* a graph in some coordinates. This is always possible at least locally, i.e., for any point $g \in \Gamma$ there is some open set $U \subset \Gamma$ containing $g$ such that $U$ is the graph of a function in some coordinates, but not all curves are graphs. – 2017-02-18
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    Also, by using the expression $\frac{y - y_0}{x - x_0}$, one assumes that the curve $\Gamma$ does not contain the common point $(x_0, y_0)$---which is a problem, as the answer concludes that every such curve $\Gamma$ *does* contain the common point. – 2017-02-18
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    @Travis I think you mean that the equation, if has to be defined for every $(x,y)$ belonging to the curve, isn't defined in the case of $P_0$ belonging to it. You are right, but it can be modified easily considering separately the point $P_0$. We have an removable discontinuity. Ditto, extend the result to $P_0$ and confirm that trivially, the tangent passes on $P_0$ – 2017-02-18
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    I know the issues trying to jump from local to global. Well, we have a graph locally and try to extend it. Now we have a graph, after all. I rekon I have to consider in the future global properties, but I cannot imagine the miriad of problems solved here without a "local approach and later will see". I am sure that the heuristics in maths works too that way in a surprisingly big amount of problems. People "tune" his steps and the heuristics, the interesting part of a proof, is unconfortably hidden many times. – 2017-02-18
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    I agree that the answer here contains the spirit of a full solution. The purpose of my comments is rather only to make more explicit the assumptions that have been added. – 2017-02-18
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    We can resolve both the local-to-global issue and the restriction to planar curves by abandoning explicit coordinates and writing the condition on $\Gamma$ in terms of a regular parameterization ${\bf Q}(t)$ as ${\bf Q}'(t) = \lambda(t) {\bf Q}(t)$ for some $\lambda(t)$; we've translated so that the common point is ${\bf P}_0 = {\bf 0}$. Note that this formulation again requires that $\Gamma$ does not contain ${\bf P}_0$! Integrating gives ${\bf Q}(t) = \left(\exp \int_0^t \lambda(\tau) d\tau\right) {\bf Q}(0)$. Using a unit speed parametrization leads to an explicit formula for $\lambda$. – 2017-02-18
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    Oh, it's challenging and good to learn try to fix the problems :) In fact, you pointed to something I have to improve: to check always the, say, global properties. – 2017-02-18
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I am writing this answer but I may not have been able to frame it properly.

But,you can use my idea.

A tangent to a curve is given by the derivative of the curve's equation at that point.If $f(x)$ represents the curve and it has degree $>1$ then its derivative is never independent of $x$.But,if it a function $f(x)$ is linear (straight line) then it's derivative is a constant independent of $x$.

Hence,the result follows.