# A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes.

This may be a phenomenally stupid question, so apologies in advance. But when I teach conics, I show why $c^2=a^2-b^2$ for ellipses geometrically, just by drawing the obvious isosceles triangle from the two foci to the point lying on the ellipse with the greatest $y$ coordinate. But when it comes to hyperbolas, I can't think of a similarly convincing picture to draw; instead I have to show why $c^2=a^2+b^2$ algebraically, from the definition as a locus. Is there an equally persuasive picture I could draw to convince high school students (without having to get technical with points at infinity, as Brian correctly suggests in an answer)?

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 geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some kind of limiting argument would have to be made. But I thought I'd ask anyway. Whichever way the answer turns out, I am surprised: either by (1) the asymmetry between the two situations (the existence of an easy geometric picture for the ellipse, but none for the hyperbola) that I never seem to have noticed, or by (2) the fact that no one has ever mentioned to me how to make the symmetry explicit.

EDIT: Because RyanReich asked me to spell it out: $a$, $b$ and $c$ are the $a$ and $b$ that appear in the equation $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ for the hyperbola centered at the origin that opens horizontally and has foci at $(\pm c,0)$.

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also, I'm assuming everyone on MSE knows what I mean by $a$, $b$, and $c$, but if anyone doesn't, I'll spell it out. – symplectomorphic Jun 2 '13 at 1:41
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. – Blue Jun 2 '13 at 6:02
I think you should spell it out. – Ryan Reich Jun 2 '13 at 6:03
@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. – symplectomorphic Jun 2 '13 at 6:45

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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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? – symplectomorphic Jun 2 '13 at 2:34
Well, yes, all nonsingular conics are projectively equivalent, but foci and distance therefrom are not projective notions. – Ted Shifrin Jun 2 '13 at 3:08

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 $a$).

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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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, x, satisfies $$\Big||x-f_1|-|x-f_2|\Big|=\Delta\text{ distance}\tag{2}$$ 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.

$\hspace{3.3cm}$

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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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. – Ted Shifrin Jun 2 '13 at 15:00

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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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. – symplectomorphic Jun 2 '13 at 5:15
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!! – Ted Shifrin Jun 2 '13 at 12:28