Tangent lines to a curve passing through a given point Here is a cool exercise from Shafarevich's book (Ex. 7 in Chapter 1, Section 1):

Given an irreducible (affine) plane curve $C$ over a field of characteristic
  $0$, and point $P$ in the plane, prove that there are only finitely
  many lines through $P$ that are tangent to $C$ at some point.

Remark 1. In terms of the dual curve, this implies that, if $\operatorname{char}(k)=0$, then $C^{*}$ has no line as a component (because a line in a dual projective space exactly corresponds to the set of lines passing through a given point $P$).
Remark 2. The conclusion is false if the characteristic is positive. Indeed, it is not hard to show that for the plane curve $y=x^{p+1}$ where $p=\operatorname{char}(k)>0$, every line through the origin is tangent to this curve (this example is the earlier part of the same exercise).
I am aware that a problem similar to this has been discussed in 
Number of tangent lines to an algebraic curve passing through a given point
However, the solution there uses the language of polar curves and proves something more precise for smooth curves. I think there should be a satisfactory proof that works for irreducible curves, and just shows finiteness of tangent lines through a given point.
Attempt. Let's try the special case when $y=f(x)$ is the equation of the plane algebraic curve. After translating variables, we can assume that $P=(0, 0)$ is the origin. Then every line through the origin has the equation $y=mx$. We want to show that there are only finitely many values of $m$ for which the equation $mx=f(x)$ has a repeated root. This is where we need to use characteristic $0$ assumption. Maybe we can analyze the discriminant? Also, this is just a special case and most curves don't have the form $y=f(x)$. 
Overkill. I am fairly certain that Bertini's theorem can give the solution immediately. Indeed, the lines through a given point $P$ defines a linear system. This induces a linear system of divisors on the curve $C$ (by intersecting the lines with $C$). Since we are in characteristic $0$, Bertini's theorem guarantees that a general member of this linear series will be non-singular, which just means that a general line through $P$ will intersect the curve at distinct points (i.e. transversely).
But this exercise appears in page 22 of the book, so there is gotta be a clean elementary solution! Thanks for your time!
 A: I think you can avoid using the word polar but still get an easy solution. Let's suppose we have a smooth curve in $\mathbb{P}^2$, then we want to count the tangent line to our curve from $P=[r,s,t]$. We can use homogenous coordinates $[x,y,z]$ and our curve is defined by a degree d homogenous polynomial $F(x,y,z)=0$. Imposing that the tangent line through a point of our curve passes through P we get $r\frac{\partial{F}}{\partial{x}}+s\frac{\partial{F}}{\partial{y}}+t\frac{\partial{F}}{\partial{z}}=0$, which is a homogenous polynomial of degree $d-1$, hence a curve. By Bezout the new curve and our original curve intersect in a finite numbers of points ($d(d-1)$) and from our construction we see that the points of intersection of the two curves are exactly the points whose tangent passes through P, hence these points are finite.
If you don't like this approach you can use Bertini and in this case I wouldn't say it is an overkill.
A: I think I have found an elementary solution which does not assume that the curve is smooth. It is quite lengthy, so bear with me.
Let's first deal with affine curves. Without loss of generality, set $P=0$. Let $f(x,y)=0$ be the equation of the irreducible curve. Consider lines of the form $x=at, y=t, a\in \mathbb{k}$. This line is a tangent to the curve if for some $t\in \mathbb{k}\setminus 0$ we have $f(at,t)=0$ and $(f'_y+af'_x)(at,t)=0$. Set $g(a,t)=f(at,t),\ h(a,t)=(f'_y+af'_x)(at,t)$.
The idea (as it was presented concisely by Prism in the comments) is that $g(a,t)=0$ and $h(a,t)=0$ must have finitely many points in common, where we are working inside $\mathbb{A}^2$ where the coordinates are $a$ and $t$. And intuitively, $g$ will be irreducible (except for a factor of $t^r$), and degree of $h$ is smaller than the degree of $g$ (which helps handle most cases).
Say $g(a,t)$ factors as $p(a,t)q(a,t)$. Since $x=at$, $y=t$ this gives us a factorization $$f(x,y)=p(x/y,y)q(x/y,y).$$ Multiplying by $y^N$ for some large enough $N$ gives us a factorization of $f(x,y)$ into polynomials: $$y^Nf(x,y)=\hat{p}(x/y,y)\hat{q}(x/y,y).$$ But $\mathbb{k}[x,y]$ is a UFD and $f(x,y)$ is irreducible, so one of the factors must be divisible by $f(x,y)$, let's agree that it's $\hat{p}(x,y)$. Then $p(x/y,y)=\hat{p}(x,y)/y^n$ for some $n$. This shows that in any factorization of $g(a,t)$ at least one of the factors is just a power of $t$.
Now we write $f(x,y)$ explicitly as $$f(x,y)=\sum_{i,j} c_{i,j}x^iy^j.$$ A direct computation then shows that
$$g(a,t)=\sum_{i,j} c_{i,j}a^it^{i+j},$$
$$h(a,t)=\sum_{i,j} (j+i)c_{i,j}a^it^{i+j-1}.$$
Let's set $g(a,t)=\hat{g}(a,t)t^r$, $h(a,t)=\hat{h}(a,t)t^m$, where $r,m$ are as large as possible. Then $\hat{g}(a,t)$ is irreducible, and we need to show that it does not divide $\hat{h}(a,t)$. Now we have 2 cases:

*

*$r=0$, in which case $m>0$. Thus we only need to show that $\hat{h}(a,t)$ is nonzero. Indeed, it is zero only if we have $j+i=0$ for all nonzero $c_{i,j}$, which is obviously impossible in characteristic 0.

*$r>0$, in which case $m\geqslant r-1$. If $m>r-1$, nothing new happens compared to the previous case. However, if $m=r-1$ we might have $$\text{deg }\hat{g}(a,t)=\text{deg }\hat{h}(a,t).$$ We need to consider the case in which $\hat{g}(a,t)=C\hat{h}(a,t)$ for some $C\in\mathbb{k}$. For this to happen we need to have $i+j=C$, which means that $f(x,y)$ has to be homogeneous. But a homogeneous polynomial of two variables splits into linear factors.

All in all, for affine curves we get finitely many tangents passing through the origin. Write $F(x,y,z)$ for the projectivization of $f(x,y)$. To deal with points at infinity we notice that the line $y=ax$ is a tangent at the point $(a,1,0)$ if and only if $F'_z(a,1,0)=0$. This is a polynomial equation in $a$, so we only have finitely many solutions.
