I'm a maths teacher in Italian secondary school and I've been spending some time trying to construct "meaningful" problems about conic sections. I particularly like problems which focus on practical applications of the curves (in engineering, physics, finance, etc.), on their geometric properties, or both.

I know there's this really easy formula to find the latus rectum of a conic from its key parameters. It's quite nice and straightforward to derive it, but...

... What's its use? Why should anyone be interested in knowing the latus rectum of a curve? Is there any physical phenomenon, any architectural trick, or anything else where its position and/or length plays a key role?

I know just the relation between angular momentum, standard gravitational parameter and latus rectum in an astronomical orbit ($l=\frac{h^2}{GM}$, where $l$ is half the latus rectum, and $h$ is the ratio between the angular momentum associated to the orbiting body and its mass). It's cool, but a bit tricky, and probably too far-fetched for secondary school students... Is there anything simpler? Or just anything else, at any difficulty level?

Many thanks!

  • $\begingroup$ Is it an externally imposed requirements that your students have to learn about latera recta? If not, then your trouble finding applications is probably a good reason to use the time on something more useful and interesting instead. $\endgroup$ – Henning Makholm Jun 9 '15 at 19:13
  • $\begingroup$ I had to google "Latus rectum" to learn what it was. I do not think it likely has a prominent role in any applications, because, as you say, you can derive it from all the other key parameters of a conic section. I would guess that the terminology came up in classical European mathematics...probably for the astronomy orbit stuff that you already mentioned. $\endgroup$ – Christopher A. Wong Jun 9 '15 at 19:16
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    $\begingroup$ No, there are no specific requirements about this kind of stuff. As a matter of fact, I've never directly thaught that formula and I've just spotlighted it (and the name "lato retto") when some exercise made the students chance into it without noticing. I've treated it as a curiosity, and as a curiosity I'd always like to treat it... I'm just trying to understand how curious it is :D $\endgroup$ – wago Jun 9 '15 at 19:18
  • $\begingroup$ I think this question would be a better fit at Mathematics Educators Stack Exchange. $\endgroup$ – Joel Reyes Noche Jun 16 '15 at 1:31

The latus-rectum and eccentricity are together equally important in describing planetary motion of Newtonian conics.

It can be regarded as a principal lateral dimension. The semi-latus rectum equals radius of curvature at perigee, the fastest point near the sun. If extreme positions of planet from sun are a+c and a-c , then from the focus their arithmetic mean is at ellipse center, semi-major axis $b$ of ellipse is the the geometric mean and semi-latus rectum its harmonic mean.


I will comment on a different problem (since you asked also about something, anything else) which I think is nice for secondary school level.

Find (without differential calculus) the enveloping curve of a projectile. The answer is a parabola, and can be done (I don't remember exactly how) with geometry. This is called the safety bell, and clearly was important to know it for military applications of the last century.

I will leave it to you to have fun and find a solution.

  • $\begingroup$ Edited name of curve of projectile to parabola. $\endgroup$ – Narasimham Jun 9 '15 at 19:35
  • $\begingroup$ Thanks for the answer. It's a bit off topic, but I didn't know about it and I'll surely think on it. It seems quite fun. $\endgroup$ – wago Jun 9 '15 at 19:36
  • $\begingroup$ If a projectile of given velocity V is fired at all angles from a fixed point say a mountain top, then each trajectory is a parabola and the envelope touching all parabolas is another parabola. Is this the problem? But this can be obtained by p-discriminant method using differential calculus only I assumed. $\endgroup$ – Narasimham Jun 9 '15 at 19:41
  • $\begingroup$ Thanks, english is not my first language and I always confuse these two, parable and parabola! $\endgroup$ – Rogelio Molina Jun 9 '15 at 19:56
  • $\begingroup$ Indeed, that is the problem but it can be done without calculus. The idea, I recall, was to see the envelope generated by a family of free falling circles whose radius is time dependant. These circles come about by transforming the equation of motion of the projectile to one such circular trajectory, by a change of variables. $\endgroup$ – Rogelio Molina Jun 9 '15 at 20:00

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