Let us have elliptical curve $E$ over $\mathbb{R}$ in standard (Weierstrass) form $y^2 = x^2 + ax + b$. What is a geometrical meaning of points of order 2, 3 and 4?
I was able to solve task for points of order 2. Definition says that order $n$ of point $P$ is $n \in \mathbb{N}$ that $nP = O$. Then for order 2: $2P = P \oplus P = O$. Operation $\oplus$ is standard addition over elliptic curves. I could say $P \oplus P = O \rightarrow P = -P$. Lets say $P = (x,y)$, then we looking for points that $(x,y) = (x,-y)$ then the solution is $\{(x,y)\in E |y=-y \}$ .. well, only number $y=-y$ is $0$, so my interpretation of this result is that points of order 2 lays on x-axis $(y=0)$.
With orders 3 and 4 I am stuck. I could find some relations but I am not able to interpret that results geometrically. Results should be: order 3 - inflection points and order 4 - intersection of tangent of point order 4 and $E$ lays on x-axis.
Thank dr. Lahtonen. For the sake of completeness I answer my question with his idea.
Let's say we have $3P = O$, then we want to find a point that $2P = -P$. It means we are looking for a point which tangent intersects elliptic curve $E$ exactly in the point P. Only points which satisfies this are points of inflection.
If we have $4P = O$ it means that $2(2P)=O$. It means we are looking for points which tangent intersects curve in the point of where $2P=O$ which solution is in the question.