A vector approach to orthogonality, rotation, and reflexion on unit conics and an application to physics

ABSTRACT

We present a unified framework to illustrate the similarities between unit circle and unit hyperbola when we consider orthogonality, rotation, and reflection. To illustrate the usefulness of this approach, a simple application to physics is given.

KEYWORDS:

• Unit conics
• orthogonality
• rotation
• reflection
• elastic collision

1. Introduction

Unit circle and unit hyperbola present similarities when we consider orthogonality, rotation, and reflection. Therefore, we present those concepts using a unified approach. Our presentation could be a complement to the standard approach presented in Assem and Bustamante (2017), Glaeser et al. (2016), and Ladegaillerie (2003). An application to physics (Galperin, 2003) concerning a simple problem of elastic collisions is given to illustrate the usefulness of this approach.

2. The mathematical setting

Let $\mathrm{\Delta }=\left(\begin{array}{cc}1& 0\\ 0& \delta \end{array}\right),$ where $\delta =\pm 1$. Let us define the inner product of the two vectors $(x,y)$ and $(\bar{x},\bar{y})$ by

$(x,y){\bullet }_{\mathrm{\Delta }}(\bar{x},\bar{y})=(x,y)\mathrm{\Delta }\left(\begin{array}{c}\bar{x}\\ \bar{y}\end{array}\right)=x\bar{x}+\delta y\bar{y},$

and use the notation ${∥(x,y)∥}_{\mathrm{\Delta }}^2$ for the squared norm (or the energy) of the vector $(x,y)$

${∥(x,y)∥}_{\mathrm{\Delta }}^2=(x,y){\bullet }_{\mathrm{\Delta }}(x,y)=x^2+\delta y^2.$

We observe that

${∥(x,y)∥}_{\mathrm{\Delta }}^2=0ifandonlyif\left\{\begin{array}{c}x=0=yfor\delta =1,\\ \left|x\right|=\left|y\right|for\delta =-1\end{array}\right.$

Cauchy–Bunyakovsky–Schwarz inequality holds

$\delta \left[{∥(x,y)∥}_{\mathrm{\Delta }}^2{∥(\bar{x},\bar{y})∥}_{\mathrm{\Delta }}^2-{\left[(x,y){\bullet }_{\mathrm{\Delta }}(\bar{x},\bar{y})\right]}^2\right]={\left(x\bar{y}-\bar{x}y\right)}^2\ge 0,$

with equality only for proportional vectors, i.e. if equality holds and $(x,y)\ne (0,0)$, then there exits a $\lambda$ such that $(\bar{x},\bar{y})=\lambda (x,y)$.

3. Orthogonality

Two vectors $(u,v)$ and $(\bar{u},\bar{v})$ are said to be $\mathrm{\Delta }$-orthogonal, noted $(u,v){\mathrm{\perp }}_{\mathrm{\Delta }}(\bar{u},\bar{v})$, if and only if

$(u,v){\bullet }_{\mathrm{\Delta }}(\bar{u},\bar{v})=u\bar{u}+\delta v\bar{v}=0.$

For a given $(u,v)$, we can take $(\bar{u},\bar{v})=(-\delta v,u)$ to get a set of two $\mathrm{\Delta }$-orthogonal vectors. Let us remark that

${∥(-\delta v,u)∥}_{\mathrm{\Delta }}^2=\delta (u^2+\delta v^2).$

4. Conic

The unit conic is defined to be the set

$\mathcal{C}=\left\{(x,y)\in {\mathbb{R}}^2|{∥(x,y)∥}_{\mathrm{\Delta }}^2=1\right\}.$

It is the unit circle for $\delta =1$ and the unit hyperbola for $\delta =-1$.

5. Decomposition

Let us take $(u,v)\in \mathcal{C}$ and consider the $\mathrm{\Delta }$-orthogonal basis of ${\mathbb{R}}^2$

$\mathcal{B}=\left\{(u,v),(-\delta v,u)\right\}.$

The decomposition of any $(x,y)\in {\mathbb{R}}^2$ with respect to $\mathcal{B}$ is

$\left(\begin{array}{c}x\\ y\end{array}\right)=p\left(\begin{array}{c}u\\ v\end{array}\right)+q\left(\begin{array}{c}-\delta v\\ u\end{array}\right)=\left(\begin{array}{cc}u& -\delta v\\ v& u\end{array}\right)\left(\begin{array}{c}p\\ q\end{array}\right),$

where

$\left\{\begin{array}{ccc}p& =& (u,v){\bullet }_{\mathrm{\Delta }}(x,y),\\ q& =& \delta (-\delta v,u){\bullet }_{\mathrm{\Delta }}(x,y).\end{array}\right.$

For a given $p$, the line of points $(x(\lambda ),y(\lambda ))$ such that $(u,v){\bullet }_{\mathrm{\Delta }}(x(\lambda ),y(\lambda ))=p$ is given by

$\left(\begin{array}{c}x(\lambda )\\ y(\lambda )\end{array}\right)=p\left(\begin{array}{c}u\\ v\end{array}\right)+\lambda \left(\begin{array}{c}-\delta v\\ u\end{array}\right)\left(\lambda \in \mathbb{R}\right).$

Similarly, for a given $q$, the line of points $(x(\lambda ),y(\lambda ))$ such that $\delta (-\delta v,u){\bullet }_{\mathrm{\Delta }}(x(\lambda ),y(\lambda ))=q$ is

$\left(\begin{array}{c}x(\lambda )\\ y(\lambda )\end{array}\right)=\lambda \left(\begin{array}{c}u\\ v\end{array}\right)+q\left(\begin{array}{c}-\delta v\\ u\end{array}\right)\left(\lambda \in \mathbb{R}\right).$

Those two lines are said to be $\mathrm{\Delta }$-orthogonal.

Let $\sigma =1$ or $-1$. The line ${\mathcal{L}}_{\sigma }$ given by

$\left(\begin{array}{c}x(\lambda )\\ y(\lambda )\end{array}\right)=\sigma \left(\begin{array}{c}u\\ v\end{array}\right)+\lambda \left(\begin{array}{c}-\delta v\\ u\end{array}\right)\left(\lambda \in \mathbb{R}\right),$

intersects $\mathcal{C}$ at only one point $\sigma (u,v)\in \mathcal{C}$, so ${\mathcal{L}}_{\sigma }$ is tangent to $\mathcal{C}$ at the point $\sigma (u,v)$, and the direction of this tangent is $(-\delta v,u)$.

6. Rotation

Let $(u,v)\in \mathcal{C}$. A rotation with respect to the vector $(u,v)$ is a linear application, noted $\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}$, defined by its action on the standard basis $\mathcal{S}=\left\{(1,0),(0,1)\right\}$ of ${\mathbb{R}}^2$. This action is

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}u\\ v\end{array}\right)\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}-\delta v\\ u\end{array}\right),$

so

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}x\\ y\end{array}\right)=x\left(\begin{array}{c}u\\ v\end{array}\right)+y\left(\begin{array}{c}-\delta v\\ u\end{array}\right),$

and hence

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}=\left(\begin{array}{cc}u& -\delta v\\ v& u\end{array}\right).$

Obviously, $\mathrm{r}\mathrm{o}{\mathrm{t}}_{(1,0)}=I$.

From the decomposition of $(x,y)$ with respect to $\mathcal{B}$, we observe that

$\left(\begin{array}{c}x\\ y\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}p\\ q\end{array}\right).$

From the $\mathrm{\Delta }$-orthogonality of the basis, we get

${\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^t\mathrm{\Delta }\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}=\mathrm{\Delta }.$

Hence,

${∥(x,y){\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^t∥}_{\mathrm{\Delta }}^2={∥(x,y)∥}_{\mathrm{\Delta }}^2,$

so, if $(x,y)\in \mathcal{C}$, then $(x,y){\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^t\in \mathcal{C}$. Also,

${∥(x,y)∥}_{\mathrm{\Delta }}^2={∥(p,q){\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^t∥}_{\mathrm{\Delta }}^2={∥(p,q)∥}_{\mathrm{\Delta }}^2,$

so

$x^2+\delta y^2=p^2+\delta q^2,$

and if one of $(x,y)$ or $(p,q)$ is in $\mathcal{C}$, then both are.

The inverse of a rotation is also a rotation because

${\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^{-1}=\left(\begin{array}{cc}u& \delta v\\ -v& u\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}.$

Then, we get

$\left(\begin{array}{c}p\\ q\end{array}\right)={\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^{-1}\left(\begin{array}{c}x\\ y\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}\left(\begin{array}{c}x\\ y\end{array}\right).$

Let us consider the composition of two rotations. Let $(u,v)$ and $(\bar{u},\bar{v})\in \mathcal{C}$. We have

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(\bar{u},\bar{v})}=\left(\begin{array}{cc}u& -\delta v\\ v& u\end{array}\right)\left(\begin{array}{cc}\bar{u}& -\delta \bar{v}\\ \bar{v}& \bar{u}\end{array}\right)=\left(\begin{array}{cc}u\bar{u}-\delta v\bar{v}& -\delta (u\bar{v}+v\bar{u})\\ u\bar{v}+v\bar{u}& u\bar{u}-\delta v\bar{v}\end{array}\right).$

If we set

$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}u\bar{u}-\delta v\bar{v}\\ u\bar{v}+v\bar{u}\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}\bar{u}\\ \bar{v}\end{array}\right)$

we verify directly that ${∥(a,b)∥}_{\mathrm{\Delta }}^2=1.$ Therefore, $(a,b)\in \mathcal{C}$, and

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(\bar{u},\bar{v})}=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(a,b)}=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(\bar{u},\bar{v})}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}.$

which means that the composition of two rotations is a rotation.

It follows that ${\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^n=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u_n,v_n)}$ where the sequence ${\left\{(u_n,v_n)\right\}}_{n=0}^{+\mathrm{\infty }}$ in $\mathcal{C}$ is generated by the recursive process

$\left\{\begin{array}{ccccc}(u_0,v_0)& =& (1,0)& \hspace{0.17em}& \hspace{0.17em}\\ (u_{n+1},v_{n+1}& =& (u{u}_n-\delta v{v}_n,u{v}_n+v{u}_n)& \mathrm{f}\mathrm{o}\mathrm{r}& n=0,1,2,\dots \end{array}\right.$

Moreover, ${\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,-v)}^n={\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^{-n}={\mathrm{r}\mathrm{o}\mathrm{t}}_{(u_n,v_n)}^{-1}=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u_n,-v_n)}$.

7. Reflection

Let $(u,v)\in \mathcal{C}$ and its corresponding $\mathrm{\Delta }$-orthogonal basis $\mathcal{B}$. A reflection with respect to the vector $(u,v)$ is the linear application, noted $\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}$, defined by its action on $\mathcal{B}$ as follows

$\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}u\\ v\end{array}\right)=\left(\begin{array}{c}u\\ v\end{array}\right)\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}-\delta v\\ u\end{array}\right)=-\left(\begin{array}{c}-\delta v\\ u\end{array}\right).$

A direct consequence is that ${\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^2=I$, so ${\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^{-1}=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}$. Moreover $\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

Using the decomposition associated to $\mathcal{B}$, we have

$\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}x\\ y\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left[p\left(\begin{array}{c}u\\ v\end{array}\right)+q\left(\begin{array}{c}-\delta v\\ u\end{array}\right)\right]$

$=p\left(\begin{array}{c}u\\ v\end{array}\right)-q\left(\begin{array}{c}-\delta v\\ u\end{array}\right)$

$=\left(\begin{array}{cc}u& \delta v\\ v& -u\end{array}\right)\left(\begin{array}{c}p\\ q\end{array}\right)$

$=\left(\begin{array}{cc}u& \delta v\\ v& -u\end{array}\right)\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}\left(\begin{array}{c}x\\ y\end{array}\right)$

$=\left(\begin{array}{cc}u& \delta v\\ v& -u\end{array}\right)\left(\begin{array}{cc}u& \delta v\\ -v& u\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right),$

and

$\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}=\left(\begin{array}{cc}{u}^2-\delta v^2& 2\delta uv\\ 2uv& -(u^2-\delta v^2)\end{array}\right).$

We observe that ${\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^t=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,\delta v)}$. Moreover

$\left(\begin{array}{cc}u& \delta v\\ v& -u\end{array}\right)=\left\{\begin{array}{cc}\left(\begin{array}{cc}u& -\delta v\\ v& u\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=& \mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)},\\ \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}u& \delta v\\ -v& u\end{array}\right)=& \mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)},\end{array}\right.$

then

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)},$

or

$\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}.$

We obtain

$\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}=\left\{\begin{array}{c}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^2\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)},\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,-v)}^2.\end{array}\right.$

From these relations, we get

${\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^t\mathrm{\Delta }\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}=\mathrm{\Delta }.$

Therefore, if

$\left(\begin{array}{c}\bar{x}\\ \bar{y}\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}x\\ y\end{array}\right),$

then

${∥(\bar{x},\bar{y})∥}_{\mathrm{\Delta }}^2={∥(x,y){\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^t∥}_{\mathrm{\Delta }}^2={∥(x,y)∥}_{\mathrm{\Delta }}^2,$

or

${\bar{x}}^2+\delta {\bar{y}}^2=x^2+\delta y^2.$

Consequently, if (x,y) $\in \mathcal{C}$ then $(\bar{x},\bar{y})=(x,y){\mathrm{r}\mathrm{e}\mathrm{f}}_{(u,v)}^t\in \mathcal{C}$.

8. Composition of rotations and reflections

From the results of the preceding sections, we can easily prove that

$\left\{\begin{array}{cc}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(\bar{u},\bar{v})}=& \left\{\begin{array}{c}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},\bar{v})}^2\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)},\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},-\bar{v})}^2,\end{array}\right.\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(\bar{u},\bar{v})}\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}=& \left\{\begin{array}{c}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},\bar{v})}^2\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)},\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},-\bar{v})}^2\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)},\end{array}\right.\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(\bar{u},\bar{v})}=& \left\{\begin{array}{c}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,v)}^2{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},-\bar{v})}^2,\\ \mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}{\mathrm{r}\mathrm{o}\mathrm{t}}_{(u,-v)}^2{\mathrm{r}\mathrm{o}\mathrm{t}}_{(\bar{u},\bar{v})}^2\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)},\end{array}\right.\end{array}\right.$

9. A basic problem and its solution

The basic problem presented below is, for $\delta =+1$, a simplified form of the problem of elastic collisions of particles in classical mechanics. It corresponds to the well-known Euclidean situation. The case $\delta =-1$ introduces an hyperbolic setting for which a physical interpretation still remains to be found. It is interesting to see that both problems can easily be solved with the help of the preceding sections.

We consider two unit point masses moving on a straight line and a wall. The velocity vector of the two point masses is $(x,y)$. The kinetic energy $E$ and the momentum $Q$ are the two important quantities to consider. They are given by

$\left\{\begin{array}{cc}E={∥(x,y)∥}_{\mathrm{\Delta }}^2& =x^2+\delta y^2\\ Q=(u,v){\bullet }_{\mathrm{\Delta }}(x,y)& =ux+\delta vy.\end{array}\right.$

where $(u,v)$ is a given vector in $\mathcal{C}$. The problem is in three parts.

9.1. Part I: elastic collisions between two point masses

For an elastic collision between two point masses, the kinetic energy and the momentum remain constant. Therefore, for given values of $E>0$ and $Q>0$, we have to find $(x,y)\in {\mathbb{R}}^2$ such that

$\left\{\begin{array}{c}E=x^2+\delta y^2\\ Q=ux+\delta vy.\end{array}\right.$

It might exist at most two points as solutions of this problem. They are the possible points of intersection of a conic and a straight line. From the Cauchy–Bunyakovsky–Schwarz inequality, existence of at least one solution is related to the condition $\delta (E-Q^2)\ge 0$. In fact, there exists at most two points in ${\mathbb{R}}^2$ that solve the problem under that condition, they are given by

$Q\left(\begin{array}{c}u\\ v\end{array}\right)\pm \sqrt{\delta (E-Q^2)}\left(\begin{array}{c}-\delta v\\ u\end{array}\right).$

Therefore, from the preceding decomposition, if one solution is

$\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)=p\left(\begin{array}{c}u\\ v\end{array}\right)+q\left(\begin{array}{c}-\delta v\\ u\end{array}\right),$

then

$\left(\begin{array}{c}{x}_{+}\\ y_{+}\end{array}\right)=p\left(\begin{array}{c}u\\ v\end{array}\right)-q\left(\begin{array}{c}-\delta v\\ u\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right),$

is the other solution. Therefore, from a mathematical point of view, an elastic collision between two point masses is a reflection.

9.2. Part II: elastic collision of one point mass with a wall

For an elastic collision of a point mass with a wall, the sign of the velocity of the point mass who hits the wall changes. Suppose the $y$-component changes of sign, so we have

$\left(\begin{array}{c}{x}_{+}\\ y_{+}\end{array}\right)=\left(\begin{array}{c}{x}_{-}\\ -y_{-}\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right),$

and again, from a mathematical point of view, we still have a reflection. In this case, the kinetic energy remains constant

${∥(x_{+},y_{+})∥}_{\mathrm{\Delta }}^2={∥(x_{-},y_{-})∥}_{\mathrm{\Delta }}^2=E,$

while the momentum changes

$Q_{+}=(u,v){\bullet }_{\mathrm{\Delta }}(x_{+},y_{+})$

$=(u,v){\bullet }_{\mathrm{\Delta }}(x_{-},-y_{-})$

$=(u,v){\bullet }_{\mathrm{\Delta }}(x_{-},y_{-}-2y_{-})$

$=(u,v){\bullet }_{\mathrm{\Delta }}(x_{-},y_{-})-2(u,v){\bullet }_{\mathrm{\Delta }}(0,y_{-})$

$=Q_{-}-2\delta v{y}_{-}.$

9.3. Part III: two successive collisions

We can combine the preceding kind of collisions. Therefore, let us consider a sequence of two collisions. Firstly, let us consider an elastic collision between the two point masses followed by an elastic collision of the second point mass with a wall. Then, we get

$\left(\begin{array}{c}{x}_{+}\\ y_{+}\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,-v)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right).$

Secondly, let us consider first an elastic collisions of the second point mass with a wall followed by an elastic collision between the two point masses. Then, we get

$\left(\begin{array}{c}{x}_{+}\\ y_{+}\end{array}\right)=\mathrm{r}\mathrm{e}{\mathrm{f}}_{(u,v)}\mathrm{r}\mathrm{e}{\mathrm{f}}_{(1,0)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right)=\mathrm{r}\mathrm{o}{\mathrm{t}}_{(u,v)}\left(\begin{array}{c}{x}_{-}\\ y_{-}\end{array}\right).$

The result is that in both cases, we get a rotation. Also, each rotation is the inverse of the other.

10. Conclusion

We have presented, in a simple way, how to obtain quite elementary and useful results about orthogonality, rotation, and reflection on conics (circle or ellipse and hyperbola) without considering any parametrization of the curves. We have solved an elementary problem in physics which involve reflections and rotations. Physicists like to compare the first two parts of the problem to Heron’s Law of reflection (see Kocik, 1999), and the third part leads to a comparison with Kepler’s Second Law (see Rafat & Dobie, 2020).

Public Interest Statement

The study of conics is an important part of analytic geometry. Also, conics appear very often in applied problems in mathematics and physics. Orthogonality, rotation and reflexion are properties and operations that are very important and helpful to solve problem. In this paper a unified approach to these concepts on conics is presented which use vectors without the introduction of angle. We illustrate the usefulness of the approach to identify that the basic structure of a collision problem in physics is essentially a problem of reflection and rotation on a conic.

Acknowledgements

This work has been financially supported by an individual discovery grant from the Natural Sciences and Engineering Research Council of Canada.

Disclosure statement

No potential conflict of interest was reported by the authors.

Additional information

Funding

This work was supported by the Canadian Network for Research and Innovation in Machining Technology, Natural Sciences and Engineering Research Council of Canada [RGPIN-2016-05572].

Notes on contributors

François Dubeau

François Dubeau received his BScA degree in Engineering Physics in 1971 and his MScA degree in Industrial Engineering in 1973, both from Ecole Polytechnique de Montréal, and his PhD degree in Mathematics in 1981 from University of Montréal. He taught at the Royal Military College of St-Jean from 1982 to 1992 and at University of Sherbrooke from 1992 up to 2015. He retired in 2015, and since that time, he is an associate professor at the Mathematics Department of University of Sherbrooke. His research interests include applied mathematics, operational research, numerical analysis, mathematical modeling, and digital image processing.