4
$\begingroup$

By constructing the parabola I mean

Given two points $P_1$ and $P_2.~~$ Given axis of symmetry $L$ that is not $\overleftrightarrow{P_1P_2}$ nor perpendicular to $\overleftrightarrow{P_1P_2}.~~$ Find the directrix $\Gamma$ and the focal point $F$ such that $d(F,\, P_1) = d(P_1,\, \Gamma)$ and $d(F,\, P_2) = d(P_2,\, \Gamma)$

Algebraically this is easy to solve, and I have translated the solution to a geometric construct in GeoGebra.

My problem is that I've basically just constructed the desired lengths (the vertex coordinates and the focal length) "somewhere else" and then shuffle the lengths to the appropriate place. As can be seen in my GeoGebra worksheet, this doesn't tell me why the focus and the directrix satisfy the "length-matching" criterion for a parabola.

What I am hoping for is an informative compass-and-straightedge construct, where lengths are built around the given two points and the axis of symmetry, showing geometrically why/how the lengths equal $\overline{P_iF} = d(P_i,\, \Gamma)$ for $i =1,2$

Thank you.

P.S.

Below are some details regarding my algebraic formulation that got translated into my GeoGebra worksheet. Feel free to skip it:

The given two points can be on the same side of $L$ or they can be on opposite sides. It doesn't matter since one can obtain the mirror images and end up with 4 points.

Without loss of generality, set the given two points as $(0,0)$ the origin and $(a, b)$ in the first quadrant. Set the axis of symmetry as a vertical line $\Gamma:\, x = x_0$, where the two points are on the same side $x_0 > a > 0$

(one might wonder why I don't set the axis of symmetry as one of the coordinate axes$\ldots$ well, I made a bad choice I guess)

At any rate, the parabola is completely determined with 2 unknowns and 2 equations: the focal length $f$ and vertex height $y_0$: $$ y = \frac{ -(x - x_0)^2 }{4f} + y_0 \\ y_0 = x_0 \frac{b}a \frac{x_0}{ 2x_0 - a} \qquad , \qquad f = \frac{ (2x_0 - a) a}{4b}$$ where I explicitly write $y_0$ as such to show the geometric construct I adopted, with similar triangles doing the scaling. As for $f$ I used the power-of-a-point.

$\endgroup$
3
$\begingroup$

One uses Pascal's theorem for hexagons inscribed in conics. The hexagons do not need to be convex and embedded, but the order of the points (following cyclicity) is important.

You have two points on the parabola $P_1$ and $P_2$ and the axis $L$ of the parabola. Then the point at infinity $P_{\infty}$ of on $L$ is also on the parabola. Reflect points $P_1$ and $P_2$ with respect to $L$ and obtain the points $P'_1$ and $P'_2$ respectively, which also lie on the parabola.

Big Step 1. Construct the tip of the tip of the parabola $P_0$, i.e. the point where the parabola intersects the parabola's axis $L$, together with the line $b$ through $P_0$ orthogonal to $L$. The line $b$ is the tangent to the parabola at point $P_0$.

Given the five points $P_1, \, P_2, \, P'_1, \, P'_2$ and $P_{\infty}$ and the line $L$ you can construct the sixth $P_0$ using Pascal's theorem for the hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$ (again, the order is importan).

  1. Denote by $P_{\infty}P_2$ the line through $P_2$ parallel to $L$. Construct $Q_1 = P_{\infty}P_2 \cap P_1P_1'$;

  2. Construct $Q_0 = L \cap P_1'P_2'$;

  3. Line $Q_0Q_1$ is Pascal's line for hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$.

  4. Construct $Q_2 = Q_0Q_1 \cap P_2P_2'$;

  5. Then construct point $P_0 = P_1Q_2 \cap L$ which is the sought point (Pascal's theorem). Draw line $b$ through $P_0$ orthogonal to axis $L$.

Big Step 2. Construct that tangent $t_2$ to the parabola at point $P_2$. To do that one can use the degenerate version of Pascal's theorem where the hexagon is $P_0P_0P_1P_2P_2P_2'$ where the line defined by the degenerate edge $P_0P_0$ is tangent line $b$ at $P_0$ and the line defined by the degenerate edge $P_2P_2$ is tangent line $t_2$ at $P_2$.

  1. As already constructed, point $Q_2 = P_0P_1 \cap P_2P_2'$;

  2. Construct $\hat{Q} = P_1P_2 \cap P_0P_2'$

  3. Line $\hat{Q}Q_2 $ is the Pascal line for degenerate hexagon $P_0P_0P_1P_2P_2P_2'$;

  4. Construct $M = \hat{Q}Q_2 \cap b$;

  5. Construct line $t_2 = MP_2$ which is the tangent to the parabola at point $P_2$ (Pascal's theorem, degenerate version).

Concluding Big Step 3.

  1. Draw line $m$ passing through point $M$ and orthogonal to line $t_2$.

  2. Construct the point $F = m \cap L$. This is the focus of the parabola.

  3. Reflect point $F$ with respect to line $b$ and obtain the point $S$ on $L$ such that $SP_0 = FP_0$, i.e. $P_0$ is the midpoint of segment $FS$.

  4. Construct the line $\Gamma$ through $S$ orthogonal to axis $L$. This is the directrix.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. If I'm not mistaken, georg's answer was basically the same, and yours is very detailed and complete. Too bad the credit can only go to one response. I have one question though: you stressed that the order (cylicity) is important that the "first" hexagon is $~P_{\infty}P_2P_2'P_1'P_1P_0~$, so the sides of this hexagon actually zigzags across the axis $L$? How do you place the initial two points $P_1$ and $P_2$ in relation to $L$? $\endgroup$ – Lee David Chung Lin Oct 1 '16 at 8:21
  • $\begingroup$ The hexagon has to include the axis $L$ as a line, formed by on of its edges, which means you should traverse $P_0P_{\infty}$ at some point. The rext of the polygon has to include as many edges as possible that you know and the simplest to use. In fact you can also choose a slightly different polygon, as long as $L$ is an edge. I picked this one because it looked fine to me after I had tried different versions. The order is important in order to form the Pascal line. Different polygons with the same vertices give different lines. $\endgroup$ – Futurologist Oct 1 '16 at 12:38
  • $\begingroup$ I just found out about this. I am sooooo sorry. I thought I accepted your answer back then but I must have double clicked it or something. This was all before I posted my own variation of solution. $\endgroup$ – Lee David Chung Lin Oct 12 '16 at 17:28
  • $\begingroup$ @LeeDavidChungLin Don't worry about it. I have totally forgotten about this post too. It is normal, people are busy and it's easy to forget details of this kind. Thanks for the recognition and you have my respect for your integrity! Cheers! $\endgroup$ – Futurologist Oct 12 '16 at 18:25
3
$\begingroup$

Georg and Futurologist have answered my question solidly, and I'm here posting another answer I got from Georg's hints just to add some variety.

I guess I spotted this construction because I had a more shallow understanding of the Pascal's Theorem, ha :P

Before stating the construction, I shall make some observations with two figures.

The hexagon in the Pascal's Theorem can be formed by connecting the points in any order, and just below is a more simple-minded arrangement on an ellipse.

enter image description here

In this figure above, the points in the lower half $~P_1'~$ and $~P_2'~$ are the mirror images of $~P_1~$ and $~P_2~$ so with the symmetry the Pascal line is perpendicular to the axis of symmetry $L$, here also the $x$-axis.

Now we bring the right most vertice $P_{\infty}$ to infinity and have a parabola, with the symmetry and orthogonality intact, as the figure just below:

enter image description here

Here the blue line $\overline{P_{\infty}P_2}$ (as well as the mirror image green $\overline{P_{\infty}P_2'}$ )becomes parallel to the axis $L$, as a crucial thing in Georg and Futurologist's solution.

Here begins the construction (see a figure below):

  1. Construct $\overrightarrow{P_2P_1}$ to intersect with the axis of symmetry $L$ at $Q_0$
  2. Construct a line through $Q_0$ that is perpendicular to $L$ and intersects with $\overline{P_{\infty}P_2}$ at $Q_1$. (Here the blue $\overline{P_{\infty}P_2}$ is constructed as the line through $P_2$ and parallel to $L$; similar goes for the green $\overline{P_{\infty}P_2'}$)
  3. Construct $\overleftrightarrow{Q_1P_1'}$ to intersect with $L$ at $P_0~$. This $P_0$ will be the vertex. (so far up to this step is basically a different version of the same construct as Georg and Futurologist's)

enter image description here

  1. Construct from point $P_0$ a line perpendicular to the green $\overline{P_{\infty}P_2'}$ and intersects it at $Q_2$
  2. The middle point of $\overline{P_0Q_2}$ will be denoted as $M$, such that $\overline{P_0M} = \overline{Q_2M} $
  3. Connect the line segment $\overline{P_2'M}$ so that we have a right triangle $\triangle Q_2MP_2'$ where $\angle MQ_2P_2' = \pi/2$ is the right angle.
  4. Find a point $N$ on the other side of $Q_2$ (opposing $P_2'$) so that we have the similar right triangle, $\angle NMP_2' =\angle MQ_2P_2' = \pi/2~$ and $\angle MNQ_2 = \angle Q_2MP_2'$
  5. The length $\overline{NQ_2}$ gives the focal length. The directirx $\Gamma$ and focus can be done easily. See the figure.

This construction itself shows why it works: there's the isosceles triangle showing $~\overline{NP_2'} = d(\Gamma,\,P_2') = d(F,\,P_2')~$, where point $F$ is the focus (the orange point on the right; not labeled for clarity).

Note that the steps 4 to 8 is based on $P_2'$, but one can do the same construction for any of the points $~P_1,\,P_1',\,P_2$.

I have checked algebraically that this construction is correct. I wonder if there's a way to NOT invoke the Pascal's theorem in the argument for this construction. Basically I would need to prove that the directrix-focus pair constructed based on $P_2'$ through this special point $P_0$ (which we know is the vertex of the parabola) is the same directrix-focus pair based on $P_1'$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

HINT

  1. Construct points $P_1', P_2'$ symmetrically stacked to points $P_1, P_2$ according to the axis of the parabola.

  2. --> We have 5 points of parabola $(P_1, P_2, P_1', P_2')$ and point $U^\infty$ at infinity in the direction of the axis of the parabola.

  3. Use Pascal's theorem to construct the tangent to the parabola at one of the points Pi.

  4. --> We have the tangent parabola with point of contact-we can determine the vertex of parabola, parameter, ...

Example (the tangent line at point P1 on Pascal's theorem)--> vertex V and so on.

enter image description here

Supplement

Pascal's line is given two points - the intersections of the lines

$23-56: \quad P_1P_2 \times P_1'P_2'$

$34-61: \quad P_2U^\infty \times P_1'P_1$

Through of the third point of intersection passes the tangent line:

$45-12: \quad P_2'U^\infty \times P_1P_1 \equiv$ tangent at point $P_1$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see you just edited. I was actually trying to understand how Brianchon applies to this situation per your hint$\ldots$, ha. Either way I need some time to figure it out. $\endgroup$ – Lee David Chung Lin Sep 25 '16 at 19:07
  • $\begingroup$ @LeeDavidChungLin My fault (not Brianchon, but Pacal). I will update the example. $\endgroup$ – georg Sep 25 '16 at 19:14
  • $\begingroup$ @georg I'm not sure how to get the tangent line just yet, but I noticed that: with the symmetry, the pascal's line is perpendicular to the axis of symmetry (x-axis), so once we make the line passing through $P_1$ and $P_2$ and get that intersection on the x-axis, through the Pascal line we can get the other two intersections. The line connecting the "upper" point with $P_1'$ actually passes through the vertex, so does the line connecting the "lower" point with $P_1$. $\endgroup$ – Lee David Chung Lin Sep 25 '16 at 19:40
  • $\begingroup$ @LeeDavidChungLin With my English I can't discuss in detail. Pascal's line is not necessarily perpendicular to the axis. Ještě doplním text Yet I shall add your reply text. $\endgroup$ – georg Sep 25 '16 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.