I'm trying to prove the following statements, and I'd appreciate a little help.

  • 1) Find the vertex of the parabola of equation $ \ y^2 + 2mx + 2ny + d = 0$, without changing the reference system, $m\neq 0$.

  • 2) In the usual euclidean-affine plane $\mathbb{E_2}$, with $\mathbb{R^2}$ as associated vectorial space. We consider a right triangle PQR, with two equal legs, being P the vertex where the right angle is situated. We'll say the bisector of the triangle is the bisector of the right angle.

    • 2.1) Prove that, for every vectorial line $U$ in $\mathbb{R^2}$, there exists one parabolic conic (degenerate or non-degenerate) going through all of the vertices of the triangle (P,Q,R) and direction of main axis $U$.

    • 2.2) Deduce the bijection between the following sets:

      • The set of vectorial lines of $\mathbb{R^2}$
      • The set of parabolic conics going through the three vertices of the triangle.
  • 3) Say how many of the conics from question 2) are degenerate and identify them in terms of the given triangle.

  • 4) Prove that the only conic from question 2) with its main axis being parallel to the triangle's bisector is non-degenerate, therefore it's a parabola. Find its vertex, main axis, focus, directrix, and focal parameter $p$, all in terms of the given triangle.

The definition of a parabola is:

  • A parabola involves a point (the focus) and a line (the directrix). The focus does not lie on the directrix. The parabola is the locus of points in that plane that are equidistant from both the directrix and the focus.

A parabolic conic is one of the following: A parabola ($\Delta \neq 0$ & $\delta=0$), a double line or two parallel real lines ($\Delta = 0$ & $\delta=0$).

($\Delta$, $\delta$ are explained below) ($\Delta=0$ means it's degenerate, and $\delta=0$ means it's a parabolic conic)

I've solved 1) myself. I put it here for you to see if it can help you. A conic's general formula is $ax^2 + 2bxy + cy^2 + 2ex + 2fy + d = 0$. And it's projective matrix:

$Â = \left( \begin{array}{l c} a & b & e \\ b & c & f \\ e & f & d \\ \end{array} \right) $ and main matrix $A = \left( \begin{array}{l} a & b \\ b & c \\ \end{array} \right) $

Therefore ours is $Â = \left( \begin{array}{l c} 0 & 0 & m \\ 0 & 1 & n \\ n & n & d \\ \end{array} \right) $ and its main matrix is $A = \left( \begin{array}{l} 0 & 0 \\ 0 & 1 \\ \end{array} \right) $

So $\Delta = |Â| = -m^2$, $\delta = |A| = 0$ (so it's a non-degenerate parabolic conic AKA parabola).

Now we derive IMPLICITLY w.r.t. x getting: $2yy' +2m +2ny'=0$ isolating $\ y'$ we get $y'=\frac{-2m}{2y+2n}$; and now we find the direction of the axis (Ker(A)) generated by vector <(0,1)>. Although the expression is 'illegal' this line has a slope of '1/0', so for both expressions to make sense, $2y + 2n$ has to be equal to $0$, therefore $y=-n$. Another condition we have is that the vertex of the parabola belongs to it, so it must fit its equation, meaning $(-n)^2 +2mx-2n^2+d=0$ so $x=\frac{n^2 - d}{2m}$. So the vertex is the point $(\frac{n^2 - d}{2m},-n)$

  • $\begingroup$ What’s a “vectorial line?” $\endgroup$ – amd Nov 8 '18 at 1:30
  • $\begingroup$ A line generated by a vector (a,b) in $\mathbb{R^2}$. I’m Spanish sorry for the inconvenience. I just wanted to make clear that there’s no point involved (the line isn’t affine). $\endgroup$ – M. Navarro Nov 8 '18 at 8:42

Notice first of all that if line $U$ is parallel to one of the sides of triangle $PQR$, then there exists no parabola going through all vertices of $PQR$ and with axis parallel to $U$. In that case, however, there exist two lines, both parallel to $U$ and passing through $PQR$, hence a degenerate parabola.

If $U$ is not parallel to any of the sides of triangle $PQR$, then a parabola through $PQR$ can always be constructed (for ANY triangle): it is Problem 51 (page 79) in Eagles' book Constructive Geometry of Plane Curves.

Here's a sketch of the construction. If $U$ is not parallel to any side of triangle $PQR$, then through one of the vertices (e.g. $P$) we can draw a line parallel to $U$ and intersecting the opposite side ($QR$) at point $D$.

Let $E$ be the midpoint of $QR$: draw a line through $E$ parallel to $U$ intersecting one of the other sides of the triangle (e.g. $PQ$) and thus intersecting line $PR$ at some point $L$. Draw now a line through $L$, parallel to $QR$, to meet line $PD$ at $G$. Line $QG$ will then cut $EL$ in $H$.

Construct now point $U$ on $HE$ such that $QE^2=4HE\cdot HU$, draw through $U$ a line parallel to $QR$ and construct $F$ on that line such that $HF=HU$. Point $F$ is then the focus of the parabola passing through $PQR$ and one can then easily construct its directrix.

For a proof that this construction works you can consult Eagles' book.

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For an analytic solution you could consider a parabola of equation $y=x^2$ and a generic slope $m\ne 0$. You can check that points $$ P=(a,a^2),\quad Q=(m-a,(m-a)^2),\quad R=(-1/m-a,(-1/m-a)^2), $$ form a right isosceles triangle, with side $PQ$ having slope $m$, if $$ a={m^3\pm1\over2m(m\mp1)}. $$

  • $\begingroup$ I'm having some issues understanding the proof in eagle's book, would you mind to explain what $::$ stands for? And the triple point (triangle shape)? Also, what does exactly $QE^2=4HE \cdot HU$ mean? $\endgroup$ – M. Navarro Nov 9 '18 at 20:40
  • 1
    $\begingroup$ Symbol $::$ is the same as = when used in proportions. $\therefore$ means "therefore" and is used to introduce a deduction. $QE^2=4HE\cdot HU$ means that the square with side $QE$ has four times the area of a rectangle with sides $HE$ and $HU$. $\endgroup$ – Aretino Nov 9 '18 at 20:48
  • $\begingroup$ If you prefer an analytic solution, I added a hint which should be of help. $\endgroup$ – Aretino Nov 9 '18 at 22:51
  • $\begingroup$ I'm very thankful for your help Aretino, but being honest, I believe the knowledge required for this assignment actually exceeds my knowledge in geometry and algebra, and I'd need a really detailed solution to fully understand its reasoning. I can't help but wonder the level of understanding on the subject you must have. This task was assigned by my Ampliation of Linear Algebra and Geometry teacher, who happens to be really demanding. Once the deadline (13/11) is passed, I'd love to see your elaboration on the analytic solution. Thanks again for your help and I'll hand in what I've done myself $\endgroup$ – M. Navarro Nov 9 '18 at 23:28
  • $\begingroup$ There is little to elaborate, actually: once you show that for any $m\ne0$ there is an isosceles right triangle with its vertices on the parabola, you have proven problem 2.1. Of course the point of view is reversed, but the result is the same. And of course you must consider separately the case $m=0$, when a leg of the triangle is perpendicular to the axis and the other leg is parallel: in that case you must consider the degenerate conic formed by two parallel lines. $\endgroup$ – Aretino Nov 10 '18 at 11:38

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