4
$\begingroup$

I'm working through a classic Calculus book (Morris Kline), and one of the problems is:

Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex.

Basically, it's saying that if the parabola $y=x^2$ has a tangent line T at point $P$, and we draw a line $L$ perpendicular to $T$ that goes through the focus $F$, then $T$ meets $L$ at a point where $y=0$.

I managed to prove this by:

  1. Calculating the equation of $T$ (based on slope of $y'$ and point $P$)
  2. Calculating the equation of $L$ (based on slope of $\frac{-1}{y'}$ and point $F$)
  3. Putting $y=0$ in both equations and solving for $x$
  4. Observe that they both have the same $x$ at $y=0$ and therefore they meet on the $x$ axis.

HOWEVER, this seems like a very brute-force approach. It's almost like I cheated or I simulated it on the computer with $1000$ points and determined that it is so. I don't really understand WHY these $2$ lines must meet in this manner.

Is there any sort of geometric or intuitive proof?

$\endgroup$
6
  • $\begingroup$ I suppose that one possible theme for analytic geometry is "cheating is allowed". $\endgroup$
    – Lee Mosher
    Jun 7, 2014 at 15:54
  • $\begingroup$ As a computer scientist and I'm all for calculating things by brute force.. but it's nice to understand the theory behind thing sometimes :) $\endgroup$ Jun 7, 2014 at 15:56
  • $\begingroup$ @Captain, I don't get it: you differentiated, calculated equations, you did slope, substitution...where exactly does the "almost cheating" part kick in?? You did mathematics: that is not cheating! $\endgroup$
    – DonAntonio
    Jun 7, 2014 at 15:56
  • 1
    $\begingroup$ Given that this is an analysis textbook, I think you're most likely supposed to do it this way. On the other hand, I would be genuinely surprised if there was no clever synthetic way of doing this. It might help to describe parabola as the set of points equidistant to the focus and the directrix and see where you get from that. Maybe try working your way back: intersect the two tangents and show that the intersection is the foot? $\endgroup$
    – tomasz
    Jun 7, 2014 at 16:02
  • $\begingroup$ you might like to look at old mechanical linkages that drew parabolas. quadrivium.info/GGB/ParabRhombus.html for example and also kmoddl.library.cornell.edu/linkages . there is a nice example in museo.unimo.it/labmat/conparin.htm that shows a single linkage drawing three conics, depending on how the 'directrix' is chosen (imagine a circular directrix for example) $\endgroup$
    – don bright
    Jun 7, 2014 at 16:44

1 Answer 1

3
$\begingroup$

Suppose we have a parabola $P$ with focus $F$ ad directrix $l$ (so that $P$ is the set of points equidistant to $F$ and $l$), and let $A$ be a point on the parabola, and $A'$ be the projection of $A$ onto $l$.

Notice that the bisector of the interval $FA'$ intersects $P$ at $A$ (by the definition, as the distance from $A$ to $A'$ is the distance from $A$ to $l$), and it doesn't intersect $P$ at any other point: if it did, we would have, between the two intersection points, another point which is closer to $A'$ (and therefore to $l$) than to $F$. We conclude that the bisector is, in fact, tangent to $P$ at $A$: the only lines which intersect the parabola at exactly one point are the tangents and the vertical lines, and clearly the bisector isn't vertical. How you would prove it depends on how you define the tangent, exactly, but I think the best way would be to show that for any line through $P$ with slope distinct from vertical and tangential, you can find a secant with that slope.

Now, consider the triangle $FAA'$, and denote by $B$ the midpoint of $FA'$. Notice that because $FAA'$ is isosceles, $FB$ is perpendicular to $BA$, which is the tangent to $P$ at $A$, so it's enough to show that $B$ lies on the tangent to vertex of $P$.

To see that last part, notice that if you draw a line $l'$ parallel to $l$ through $B$, you get a line whose distance from $F$ is half of the distance from $F$ to $l'$ (by the intercept theorem, because $B$ is the bisetor of $FA'$ and $A'$ lies on $l$), so $l'$ is just the tangent to $P$ at vertex.

enter image description here

$\endgroup$
5
  • $\begingroup$ That's amazing, thanks! Do you mind if I ask where you learned to construct this sort of proof? $\endgroup$ Jun 7, 2014 at 17:52
  • 1
    $\begingroup$ @CaptainCodeman: It's kind of a matter of experience, I think. I'd been trying to do math olympiad some years ago, and while I wasn't very successful, I've picked up some intuition that apparently is not all gone. I'm not entirely content about this proof: the argument for the tangency is not quite as neat as I would have liked, but I'm glad you like it. :) $\endgroup$
    – tomasz
    Jun 7, 2014 at 18:24
  • $\begingroup$ OK that makes sense! By the way what software did you use to draw the diagram? $\endgroup$ Jun 7, 2014 at 18:28
  • 1
    $\begingroup$ Geogebra. You can also use this program to draw the thing you want to study and see some things that are not immediately obvious (for example, that the bisector is tangent) and would be hard to see in a crude drawing (well, you can see that in a good enough drawing and I did just that, but in Geogebra you can change the point $A$ to verify that the tangency is not a coincidence). $\endgroup$
    – tomasz
    Jun 7, 2014 at 18:32
  • $\begingroup$ That's some amazing software, thanks for the introduction! $\endgroup$ Jun 7, 2014 at 18:51

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.