Parabola with focus F and circle with center F (the same F) have two points in their intersection. Show that there are points A,B,C and D on the circle such that the lines containing AB, BC, CD and DA are tangent to the parabola. The hint was about choosing point A s.t. the line through AF and the line through the point A that is parallel to the axis of symmetry intersect the circle on symmetric (with respect to the axis of symmetry) points, and the take D and B to be the intersection of the tangents through A (to the parabola) with the circle and show that they are symmetric (again, with respect to the axis of symmetry). I understand why D and B being as described finishes the proof, however, I don't see why is it true. Thanks.

  • $\begingroup$ Parabola. It is written. $\endgroup$ – Yuval May 2 '17 at 16:29
  • $\begingroup$ Right. It is fixed now. $\endgroup$ – Yuval May 2 '17 at 16:32
  • $\begingroup$ There was one more hyperbola lurking in the text. I changed it to parabola. $\endgroup$ – Harald Hanche-Olsen May 2 '17 at 16:40
  • $\begingroup$ I don't understand the hint: the line through AF and the line through the point A that is parallel to the axis of symmetry intersect the circle on symmetric (with respect to the axis of symmetry) points for every position of A on the circle. $\endgroup$ – Aretino May 2 '17 at 21:14
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    $\begingroup$ According to my GeoGebra sketch, you can take $A$ to be any point on the circle but in the parabola's exterior: the two tangents from $A$ to the parabola meet the circle at $B$ and $D$, and then ---"magically"--- the other tangent from $B$ and the other tangent from $D$ meet at a common circle point, $C$. (There are some degenerate cases in which the points coincide.) The problem reduces to explaining the magic. $\endgroup$ – Blue May 2 '17 at 22:46

Let's begin with this:

If $P$ is a point on a parabola with focus $F$, and if $R$ is the intersection of the tangent at $P$ with the "vertex tangent", then $\overleftrightarrow{FR} \perp \overleftrightarrow{PR}$. Conversely, if $R$ is on the vertex tangent, then the line perpendicular to $\overleftrightarrow{FR}$ at $R$ is tangent to the parabola.

enter image description here

Proof. (Forward) Define $Q$ on the directrix as the foot of the perpendicular from $P$. By the definition of the parabola, $\overline{FP} \cong \overline{PQ}$. By the Reflection Property, the tangent at $P$ is the angle bisector of $\angle FPQ$, and therefore also the perpendicular bisector of $\overline{FQ}$. Since the vertex tangent is equidistant from the focus and directrix, it meets $\overline{FQ}$ at its midpoint. Thus, that midpoint is the intersection of the two tangents ($R$), and the tangent at $P$ meets $\overline{FR}$ at a right angle. (Reverse) Define $Q$ as the reflection of $F$ in $R$, noting that it necessarily lies on the directrix. Then erect a perpendicular to the directrix at $Q$, and define $P$ as its intersection with the parabola. As $\triangle FPQ$ is isosceles, $\overline{RP}$ bisects $\angle FPQ$, so that this line is tangent to the parabola. $\square$

Now ... Given a tangent at $P$ and the point $R$ where that tangent meets the vertex tangent, take $A$ to be any point on $\overline{PR}$, and define $B$ as the reflection of $A$ in $R$.

enter image description here

By the perpendicularity property proven above, we see that $\overline{AB}$ is a chord of some circle about $F$. Note that $\overline{A^\prime B^\prime}$, for $A^\prime$ and $B^\prime$ the reflections of $A$ and $B$ in the parabola's axis, is also a chord of that circle.

Joining $A$ to $B^\prime$ gives yet another chord whose midpoint, $S$, must lie on the vertex tangent. Thus, $\overline{FS}\perp \overline{AB}$, and our converse result implies that this chord is also tangent to the parabola.

enter image description here

Finally, $\overline{A^\prime B}$ (not shown) is a fourth tangent chord. $\square$

Note: This discussion starts with a tangent and derives a circle (and then more tangents). The reader is invited to adapt the argument to the question that assumes the circle and derives the tangents. Ultimately, the "magic" suggested in my comment to OP may boil-down to the perpendicularity property above.

  • $\begingroup$ Thanks. I believe this is exactly what is meant to be the continuation of the hint. Most elegant. $\endgroup$ – Yuval May 4 '17 at 10:15
  • $\begingroup$ Minor nit: when you write “… define $P$ as its intersection with the parabola …” you are implicitly assuming the existence of such an intersection. Makes perfect sense over the complex plane, but over the reals this might be seen as problematic. $\endgroup$ – MvG May 4 '17 at 20:57
  • $\begingroup$ @MvG: I am fairly confident that a line perpendicular to the directrix of a parabola meets that parabola in a (real) point. $\endgroup$ – Blue May 4 '17 at 22:30
  • $\begingroup$ For some reason I had a perpendicular at $R$ in mind, not in $Q$ as you wrote it. I stand corrected. $\endgroup$ – MvG May 4 '17 at 22:37
  • $\begingroup$ @MvG: I see that my description could be more clear. I'll update. Thanks! $\endgroup$ – Blue May 4 '17 at 23:05

I agree with the comment by Blue: the property you describe appears to be universal; the choice of $A$ does not matter.


So let's prove this. I like to use coordinates. Without loss of generality choose your coordinate system such that the focus is in the origin, the directrix is $y=-\frac12$ so the equation of the parabola is $y=x^2-\frac14$. The circle has arbitrary radius $r$. Using the tangent half-angle formulas, a point on the circle has coordinates $$\frac{r}{t^2+1}\begin{pmatrix}2t\\t^2-1\end{pmatrix}$$ with the exception of the point (excluding the point $(0,r)$ at the top, which corresponds to the limit $t\to\infty$). So now you start with any point


or in homogeneous coordinates


And likewise for $B,C,D$ using parameters $b,c,d$. The circle and parabola are described by

$$K=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-r^2\end{pmatrix}\qquad P=\begin{pmatrix}-4&0&0\\0&0&2\\0&2&1\end{pmatrix}$$

where a point lies on the respective conic iff multiplying the homogeneous coordinate vector of the point with this matrix from both sides (i.e. as a row vector from the left and a column vector from the right) returns zero.

Now things become technical, so I'd use a computer algebra system. Given homogeneous coordinate vectors for $A$ and $B$, the line joining these is the cross product of the vectors. That line is tangent to $P$ if multiplying the vector of the line with the adjunct of $P$ yields zero. That is the case iff

$$r^2 \cdot (a-b)^2 \cdot (4a^2b^2r + a^2b^2 + a^2 + b^2 - 4r + 1)=0$$

The first two factors describe degenerate situations, namely $r=0$ or $a=b$. The third factor characterizes the two possible choices of $b$ as

$$b^2=\frac{4r-a^2-1}{4a^2r+a^2+1}\qquad b=\pm\sqrt{\frac{4r-a^2-1}{4a^2r+a^2+1}}$$

So here we have two solutions, one of them is $b$ and the other is $d=-b$. A change in sign indicates a reflection in the axis of symmetry, so we know that $B$ and $D$, the points where the tangents to $P$ through $A$ intersect the circle again, have symmetric positions. And if they do, then for reasons of symmetry $C$ must be the reflection of $A$ in the axis of symmetry, and everything works out.

  • $\begingroup$ +1. You've saved me the trouble of posting a solution. :) Here's an interesting observation: consecutive points (say, $A$ and $B$) are equidistant from the line perpendicular to the axis of symmetry through the parabola's vertex. $\endgroup$ – Blue May 3 '17 at 1:55
  • $\begingroup$ Thank you. I was actually looking for a geometric solution, preferably one that uses the hint. If anyone has one in mind I would be glad to see it. $\endgroup$ – Yuval May 3 '17 at 4:40

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