How to construct the point of intersection of a line and a parabola whose focus and directrix are known?

Question

I found this problem in Polya's "How to solve it". It goes as follows

Using only a straight edge and a compass, construct the point(s) of intersection of a given line and a parabola whose focus and directrix are known.

I think I've managed to do it for the special case when the directrix and the given line are parallel, but I'm stuck at the general case when they're not. I could add my method for the special case in my question but I think it's convoluted and inelegant. I was wondering whether anyone could help me figure it out. I could just look it up in the book, but Polya just uses that problem to illustrate restatement of a problem and does not provide a solution. I'd be glad if someone could give me some pointers.

Answer 1

Using the definition of parabola, each of the intersection points $P_1,P_2$ lies on L and is the center of a (different) circle through point F and tangent to D (e.g. $|P_1F| = |P_1T|$). But this only gives two pieces of information for each circle; three are needed.

See diagram.

But, the missing third point can be obtained by reflecting $F$ about the line $L$ to get $F'$ ($P_1$ and $P_2$ lie along L and so their circles must be symmetrical about L). See diagram.

Then extend $FF'$ until it intersects $D$ at $Q_1$, and intersects with $L$ at point $C$. Now the power of $Q_1$ relative to the circle centered at $P_1$ can be expressed as

$|Q_1T_1|^2 = |Q_1F||Q_1F'|$

Draw semi-circles with diameter $FF'$ and diameter $Q_1C$. These intersect at point $R_1$ such that

• $\angle QR_1C = 90^\circ$ and
• $|Q_1R_1|^2 = |Q_1F||Q_1F'|$.

Therefore $|Q_1R_1| = |Q_1T|$

So a circular arc centered at $Q_1$ through point $R_1$ will intersect $D$ at the desired tangency point $T_1$. Just draw a perpendicular line through $T_1$ to find $P_1$.

To find $P_2$ extend the semi-circles to find another point of intersection $R_2$ (not shown) which will lie on the same arc as $R_1$ and $T$.

So just extend the arc centered at $Q_1$ until it intersects the diameter again at $T_2$. A perpendicular bisector yields point $P_2$.

Answer 2

The following image shows one way to do this. I'm sure it's not the most elegant but it's easy to understand:

The red parabola and line are given; the focus and directrix are in orange. (Clearly the red parabola isn't actually given, but it's there as a reference). Be aware that there are some extraneous lines and points on the image but the following instructions should walk you through it.

We begin by drawing any two non-overlapping circles tangent to the directrix (green). Construct the other external tangent (green dotted line). The point we seek is the centre of a circle tangent to these two lines passing through the focus.

Draw a line from the focus to the point where the tangents cross (blue). Produce a line from the intersection of this line with one of the circles (point W) to the point at which a tangent touches it (point U; line WU is dashed blue). Construct the parallel line through F (also dashed blue). Then construct the perpendicular bisector of FZ, where Z is the intersecton of this line with the tangent that isn't the directrix (solid purple). The intersection (G) of this point with the given line is on the parabola, and hence is one of the required points.

I appreciate this is an old question but maybe this will be useful or interesting to someone...

Answer 3

Two clarifications of helveticat's diagram:

• Point O is the intersection (if any) of the given line and the given directrix. It also ends up being the intersection of the two tangent lines (again, if any).

• The "any two non-overlapping circles tangent to the directrix" must also have their centers on the given line.

Answer 4

Here's a version of @helveticat's answer, which I think is even simpler.

Let $d$ be the directrix, $F$ is the focus and $a$ is the line that should intersect the parobla.

1. Extend $a$ and $d$ until they intersect in $O$ (if $a||d$ then draw a line through $F$ parallel to $a$ and $d$).
2. Draw helper circle with center $A$ on $a$ and tangent to $d$.
3. Intersect the helper circle with $FO$ at $C$ and $B$
4. Draw a line through $F$ parallel to $AB$, intersecting $a$ at $P$.
5. Draw a line through $F$ parallel to $AC$, intersecting $a$ at $Q$.

$P$ and $Q$ are the solutions.

For reference here is also the parabola and two red circles centered at $P$ and $Q$ which show that the points are in fact equidistant from $d$ and $F$:

We can also see that if $F$ is such that $FO$ does not intersect the helper circle there would be no solutions.

In case of $F \in a$ @helveticat's answer is more general and can be used.

Answer 5

Here is an analytic geometry solution, which can be converted to straightedge and compass construction:

We put F onto the origin of Cartesian coordinates, then get equations for P(x,y):

$\begin{cases}y=b-x\tan\alpha\\y+p=\sqrt{x^2+y^2}\end{cases}$

The solution is:

$x=-p\tan\alpha\pm\sqrt{p(p/\cos^{2}\alpha+2b)}$

Here let $q=p/\cos^{2}\alpha+2b$, then $\sqrt{pq}$ can be constructed by Geometric Mean Theorem.