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When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.

But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*

Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:

Is there an equally persuasive picture I could draw to convince high school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?

The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.


[*] The tedious algebraic details:

The foci lie at $(\pm c,0)$. Say $a>b$. Then hyperbola is the locus of points $(x,y)$ such that

$$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$

Assume for simplicity the expression between the absolute value bars is positive. (The algebra is even more tedious if we don't make such a simplifying assumption.) Then moving the right root to the other side and squaring gives

$$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$

Simplifying gives

$$xc-a^2=a\sqrt{(x-c)^2+y^2}$$

and upon squaring again we have

$$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$

which reduces to

$$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$

Defining $b^2:=c^2-a^2$ brings the equation into the form $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the definition of $b$.

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    $\begingroup$ As someone who believes there's a nice diagram for every nice mathematical relation, I think you've sparked my newest obsession. :) I have a vague suspicion that insights might be gleaned from the "Dandelin Spheres" configuration. After all, the spheres effectively provide picture-proofs ---from the definition of conic sections as sections of cones--- of the sum/difference-of-distances-from-foci and the focus-directrix properties of the curves. $\endgroup$ – Blue Jun 2 '13 at 6:02
  • $\begingroup$ I think you should spell it out. $\endgroup$ – Ryan Reich Jun 2 '13 at 6:03
  • $\begingroup$ @Blue: yes, I was hoping for a clever diagram but didn't have a sense of how to think about whether one might exist. It is the asymmetry between the two situations that really strikes me: as I commented to Ted below, I spent the afternoon trying to find the relevant picture -- assuming there had to be one on analogy with the ellipse -- before giving up and doing the algebra. $\endgroup$ – symplectomorphic Jun 2 '13 at 6:45
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Consider the hyperbola: $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\tag{1} $$ Given $f_1,f_2$, the two foci of a hyperbola, one property of a hyperbola is that there is a constant, $\Delta\text{ distance}$, so that any point on the hyperbola, p, satisfies $$ \Big||p-f_1|-|p-f_2|\Big|=\Delta\text{ distance}\tag{2} $$ Consider a point at the intersection of the hyperbola and the line between the foci. The $\Delta\text{ distance}$ given in $(2)$ is the distance between the two branches of the hyperbola.

$\hspace{3.3cm}$enter image description here

Using $(1)$, we get that the distance between the two branches of the hyperbola is $2a$. Therefore, $$ \Delta\text{ distance}=2a\tag{3} $$ Consider a point at an infinite distance on the upper right branch of the hyperbola. The $\Delta\text{ distance}$ given in $(2)$ is $$ \Delta\text{ distance}=|f_1-f_2|\cos(\theta)\tag{4} $$ Using $(1)$, we get that $$ \begin{align} \tan(\theta)&=\lim_{x,y\to\infty}\frac yx=\frac ba\\ \cos(\theta)&=\frac{a}{\sqrt{a^2+b^2}}\tag{5} \end{align} $$ Combining $(3)$, $(4)$, and $(5)$, we get $$ |f_1-f_2|=2\sqrt{a^2+b^2}\tag{6} $$ Thus, if $c$ is the distance from the center of the hyperbola to each of the foci, then $(6)$ gives $$ c^2=a^2+b^2\tag{7} $$

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    $\begingroup$ This is a much more conceptual argument doing the algebra of the limit I did earlier. I love it!!! Shame on me for not having thought of projection. $\endgroup$ – Ted Shifrin Jun 2 '13 at 15:00
  • $\begingroup$ Sorry, I have been reading this answer for a few hours and I just cant follow it. How is $\Delta = |f_1-f_2|cos(\theta)$ ? I see how $|f_1-f_2|cos(\theta)$ equals the top left side of the triangle in the figure, but I fail to see why that side of the triangle would be $\Delta$. Im a bit confused about the steps that follow but maybe if I get that step the rest will be clearer. Sorry maybe I'm missing something obvious. $\endgroup$ – Joaquin Brandan Aug 3 at 18:14
  • $\begingroup$ Im also having trouble seeing where does $\tan(\theta)$ come from in $(5)$, I an get algebraically from $(1)$ to $\lim_{x,y\to\infty}\frac yx=\frac ba\\$, but I dont quite see where does the $tan(\theta)$ come from. $\endgroup$ – Joaquin Brandan Aug 3 at 18:32
  • $\begingroup$ Nevermind the second comment, Just realized I can just see that the tangent of the triangle fromed between origin $(0,0)$ , $(x,0)$ and $(0,y)$ is indeed $tan(\theta)=\frac{y}{x}$. Still trying to figure out $(5)$ tho. :c $\endgroup$ – Joaquin Brandan Aug 3 at 19:01
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Here's a solution using for the special case of a hyperbola formed as a "vertical" section of a cone ... that is, for the case in which the cutting plane is parallel to the cone axis.

Take a "sideways" look at one of the hyperbola's Dandelin Spheres; let $C$ be its center, and let it meet the cone tangentially at $T$. Our view of the configuration is parallel to the cutting plane (and perpendicular to the cone axis), which here appears as the vertical line $FM$. Were we to look perpendicular to the plane, point $M$ would be the center of the hyperbola, point $F$ would be a focus (so that $|MF| = |OC| = c$ in traditional notation), and point $V$ would be a vertex (so that $|MV| = a$); importantly ---and this is something of a leap of faith--- the crossed lines would be the asymptotes (with "slope" $\pm b/a$, where Dandelin radius $b = |CF|$ gives the rise for a run of $a$).

Sideways view of a vertical conic section

Since right triangles $\triangle OTC$ and $\triangle VMO$ have matching legs ($CT \cong OM$, via mutual congruence with $CF$) and acute angles ($\angle COT \cong \angle OVM$), the triangles are themselves congruent, so that $|OT|=|MV| = a$ and $$|OT|^2 + |TC|^2 = |OC|^2 \qquad \implies \qquad a^2 + b^2 = c^2$$


For now, I'll leave it as an exercise for the reader to tweak the argument to cover hyperbolas created by oblique cutting planes (as well as to cover ellipses), where the situation is much more complicated. The reader should also justify the "leap of faith" (which, of course, has to be so, because we know how the story ends).

BTW, the figure was drawn using GeoGebra.


Regarding the oblique case ... If the cone's generator makes an acute angle $\theta$ with its axis, and the cutting plane makes an acute angle $\phi$ with that axis, then one can show $$a = r \cos\theta \qquad c = r \cos\phi$$

where $r$ is the radius of the circle through the Dandelin centers, centered at their midpoint. Interestingly, the consequent relation $$b^2 = c^2 - a^2 = r^2 \left( \cos^2\phi - \cos^2\theta \right) = r^2 \left( \sin^2\theta - \sin^2\phi \right)$$ reveals $b$ as the geometric mean of the Dandelin radii, $r \left( \sin\theta - \sin\phi \right)$ and $r \left( \sin\theta + \sin \phi \right)$.

While it's possible to find $a$, $b$, $c$ represented by segments in a "sideways" diagram analogous to the one above, the connection between $a$, $b$, and the hyperbola's asymptotes is far less obvious. After all, the asymptotes arise from intersecting the cone, at its apex, with a plane parallel to the cutting plane; those lines don't have the same "slope" as the cone's generators, so the geometry of the sideways diagram isn't sufficient for capturing their nature.

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You can solve this the same way as the ellipse, by taking the point with greatest y-value. However, that point lies at infinity, so instead of drawing an exact triangle, you take a limit of triangles with points on the curve (as you suggested) and scale them down so they don't go off to infinity themselves. This can be made precise in projective geometry, because then that point at infinity is a real point (there are 2, I believe). In fact, the hyperbola is just an ellipse turned inside out, so that it's greatest and least y-values lie at infinity (just like a line is a circle with a point at infinity).

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  • $\begingroup$ yes, thanks; you're right about the projective picture. by "geometric" reason I guess I was really asking for a purely "high-school" or "Euclidean" reason. I wouldn't convince many high school students by suddenly talking about points at infinity (tho I would certainly mention it to excite their imagination). what I'm asking is: is there a clever little plausibility argument to be given solely in terms of a limit/exhaustion of some kind of euclidean picture? $\endgroup$ – symplectomorphic Jun 2 '13 at 2:34
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    $\begingroup$ Well, yes, all nonsingular conics are projectively equivalent, but foci and distance therefrom are not projective notions. $\endgroup$ – Ted Shifrin Jun 2 '13 at 3:08
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Since the hyperbola is defined as the locus where the difference of the two distances is constant, I don't see any analogous argument. But certainly a limit argument does work: As $(x,y)\to \infty$ (with $x>0$), we have $\left|\frac yx\right|\to\frac ba$, and so \begin{align*} \sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=\\\left(\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right)&\frac{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\ &=\frac{4c}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\&=\frac{4c}{\sqrt{(1+c/x)^2+(y/x)^2}+\sqrt{(1-c/x)^2+(y/x)^2}}\\ &\to \frac{2ac}{\sqrt{a^2+b^2}}\,. \end{align*}

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  • $\begingroup$ right, thanks; I'm aware of the algebra. I was hoping for some kind of elegant picture to draw, but I guess there isn't one. I spent half the afternoon trying to draw the appropriate picture to immediately see the connection between $a$, $b$, and $c$ before giving up and just doing the algebra. $\endgroup$ – symplectomorphic Jun 2 '13 at 5:15
  • $\begingroup$ OK, sorry. I had actually never done this exercise before. When I try to make @Brian's suggestion work, I am confronted with the fact that the foci change. Working in $\mathbb P^2$, we are led to substitute $1/x$ for $x$ and $y/x$ for $y$, and we get $$\frac{x^2}{(1/a)^2} + \frac{y^2}{(b/a)^2} = 1\,,$$ and the algebra just doesn't work out. :( I'll look at Blue's solution later; it seems the most promising!! $\endgroup$ – Ted Shifrin Jun 2 '13 at 12:28

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