Hello Carl,
I refer to pages 533534 of Precalculus where you (you and Jeff collectively) derive the equation x^2/a^2 – y^2/b^2 = 1 for the simplest case of a horizontal hyperbola centred at (0, 0). There is one sticking point for me in the derivation, namely, where you set c^2 = a^2 + b^2. You prove the analogous equation for horizontal ellipses (a^2 = b^2 + c^2) before relying on it to derive the basic equation of x^2/a^2 + y^2/b^2 = 1 for the latter. However, the justification for c^2 = a^2 + b^2 for hyperbolas is a mystery to me. I am aware that there is a natural reflective property relating to the foci of a hyperbola, so the relationship between a, b and c is something naturally occurring and there to be discovered – not something that is set by fiat. I've searched the Internet trying to find an answer, and all I've found is that there are other people like me who've done the same thing to no avail. It's as if nobody wants to talk about the answer, whatever it is. One site acknowledged the issue with a comment that it's all too hard, and just accept that c^2 = a^2 + b^2 for hyperbolas and move on. Every derivation of x^2/a^2 – y^2/b^2 = 1 that I've seen relies on the (to me) unproven proposition that c^2 = a^2 + b^2. I found a proof for c^2 = a^2 + b^2 on YouTube, but right at its end it relied on x^2/a^2 – y^2/b^2 = 1 as a prerequisite fact. I also devised my own rather simpler proof for c^2 = a^2 + b^2, but then realised that I was relying on the truth of y = +–b/a sqrt(x^2 – a^2), which of course is derived from x^2/a^2 – y^2/b^2 = 1. So, all I've seen so far is a chickenandegg relationship between the two derivations. You need to accept the truth of c^2 = a^2 + b^2 in order to derive x^2/a^2 – y^2/b^2 = 1, but you also need to accept the truth of x^2/a^2 – y^2/b^2 = 1 in order to prove c^2 = a^2 + b^2. There were some sticking points for me when I was working through the chapter on polynomials, but I was eventually able to resolve all of them to my satisfaction so that I felt "at peace" with the propositions in question, and could move on. However, I'm not seeing any light at the end of the tunnel with this particular issue with hyperbolas. Carl, can you shed some light on this, or are we dealing with an unspeakably awful Sasquatchesque monster here? 
Administrator

Hello!
Great question. Basically our derivation writes the equation of the hyperbola using the distance definition in terms of the parameters "a" and "c". (Our diagram uses the letter "b" in the label, but you can ignore that picture for now. It doesn't figure into the derivation in terms of "a" and "c".) We then use an endbehavior argument to show the ratio of y/x in the hyperbola is (+/) sqrt(c^2a^2) /a. This means the hyperbola is asymptotic to a rectangle whose diagonals have ratio (+/) sqrt(c^2a^2)/a. The key step is here: we *define* b = sqrt(c^2a^2), since the slope of the diagonals is (+/) sqrt(c^2a^2) /a = (+/) b/a, this means the endpoints of the conjugate axis are (0, (+/)b) . So... to recap: we use the definition of hyperbola to get the equation in terms of "a" and "c". We then note the hyperbola is asymptotic to the lines y =[ (+/) sqrt(c^2a^2) /a] x so we call the numerator of the slope, sqrt(c^2a^2) = b. Does that help? 
Thank you for your reply, Carl – and yes, it has been helpful.
The key point is that you have stressed that b is DEFINED in terms of a and c. In the case of the Pythagorean relationship between a/b/c in ellipses, the relationship is very visible just through the use of two congruent rightangled triangles, so one can see that the relationship is naturally motivated. It's just there waiting to be discovered and demonstrated. On the other hand, I (and obviously other people out there on the Internet) was looking for some analogous naturally motivated Pythagorean relationship between a/b/c in the case of hyperbolas – and was looking in vain for something that's really not there. Actually, I couldn't see any need for b in shaping a hyperbola because it seemed to me that the combination of a pair of vertices, together with a pair of foci that could be symmetrically slid in or out together, would be sufficient to distort the branches of a hyperbola to any possible height:width ratio. I wouldn't be surprised if part of the motivation for introducing b into the description of hyperbolas was to present them in a way that maximised an appearance of "family likeness" between hyperbolas and their elliptical cousins. 
Administrator

Hi again! Yeah, I guess we're just treating "b" as a convenient collection of parameters and recognizing it as a vertical stretch factor. All you need is "a" (vertices) and "c" (foci) for the equation. Another way I've thought of hyperbolas is to think of them as ellipses/circles that have been "turned inside out" via "imaginary scaling." That is, rewriting x^2y^2 = 1 as x^2+(iy)^2 = 1 where "i" is our friend the imaginary unit. So our "real" hyperbola "looks" like an "imaginary" circle. I have discussed this idea in the past with one of my colleagues who is more experienced in algebraic curves than I am and there seems to be something to this idea. As always, Jeff and I appreciate your careful reading of the texts as well as your comments! Carl From: MarkAU [via Stitz Zeager Open Source Mathematics] <[hidden email]> To: Carl Stitz <[hidden email]> Sent: Friday, April 13, 2018 7:33 AM Subject: Re: Hyperbolas: Chickenandegg problem with c^2 = a^2 + b^2
Thank you for your reply, Carl – and yes, it has been helpful.
The key point is that you have stressed that b is DEFINED in terms of a and c. In the case of the Pythagorean relationship between a/b/c in ellipses, the relationship is very visible just through the use of two congruent rightangled triangles, so one can see that the relationship is naturally motivated. It's just there waiting to be discovered and demonstrated. On the other hand, I (and obviously other people out there on the Internet) was looking for some analogous naturally motivated Pythagorean relationship between a/b/c in the case of hyperbolas – and was looking in vain for something that's really not there. Actually, I couldn't see any need for b in shaping a hyperbola because it seemed to me that the combination of a pair of vertices, together with a pair of foci that could be symmetrically slid in or out together, would be sufficient to distort the branches of a hyperbola to any possible height:width ratio. I wouldn't be surprised if part of the motivation for introducing b into the description of hyperbolas was to present them in a way that maximised an appearance of "family likeness" between hyperbolas and their elliptical cousins. If you reply to this email, your message will be added to the discussion below:
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