Tangent lines of conics Let $k$ be algebraically closed. 
Let $P\in k[x,y,z]$ be a homogeneous quadratic polynomial.
Let $C$ be the zero locus of $(P)$ in $\mathbb{P}^2$. Let $Q \in \mathbb{P}^2$.
Is there a tangent line at $Q$ of $C$, i.e. a line that intersects $C$ only in $Q$? Is this line unique? How can one get the polynomial of this line from $P$?
My geometric intuition says YES!, but the only algebraic thing I know about tangents and varieties is the definition of Zariski what a tangent space is, namely
$$T_{X,Q}= \mbox{dual of }\mathfrak{m}/\mathfrak{m^2}\mbox{as $k$ vector space},$$
with $\mathfrak{m}$ the unique maximal ideal of the coordinate ring of $C$ localized at $Q$. Which I cannot tie to the geometric question. 
Any hints or references to solve this algebraically?
 A: The answer is yes and there are indeed two such lines in general. The intuition can come from a picture, but over $\mathbb{R}$ you can sometimes have no solution (choose the point inside a circle..)
To solve it algebraically, you can change coordinates so that your point becomes $p=[1:0:0]$ and the lines through $p$ are of the form $\lambda y+\mu z=0$, where $[\lambda:\mu]\in\mathbb{P}^1$.
If the conic passes through the point, you have of course only one solution, the tangent at the point, when the conic is smooth, and infinitely many solution if the conic is singular at the point.
If the conic does no pass through the point, you write the equation as $$x^2+axy+bxz+cy^2+dyz+ez^2=0,$$ where $a,b,c,d,e\in k$.
Then, let us compute the intersection of a line of the form $\lambda y+\mu z=0$ with your conic. You can parametrise the line via $[u:v]\mapsto [u:-\mu v:\lambda v]$. Then just replace in the equation and you get
$$u^2+uv(b\lambda-a\mu)+v^2(c\mu^2-d\lambda\mu+e\mu^2)=0$$
so for general $\lambda$ and $\mu$ you get two solutions: the intersection of a line with a conic gives two points. The two solutions coincide exactly when the line is tangent. Computing the discriminant you get something of degree $2$ in $\lambda,\mu$ (and homogeneous), so two solutions.
