[Math] Construct Parabola given two points and axis of symmetry

algebra-precalculusconic sectionsgeometric-constructionplane-geometry

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.

Best Answer

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.

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