My question is somewhat long, but I will try to be concise and unambiguous in my explanation.
In my high school math class we were recently presented with the graph of a hyperbola, and given certain facts challenged to derive its equation.
My question only centers on the final part of the derivation, when the equation dealt with was:
$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$
Where $e$ represents the eccentricity, and $a$ represents a positive constant.
Simply divide by $a^2(1 - e^2)$, and use the substitution $b^2 = a^2(1 - e^2)$, where $b$ is another positive constant, and one finds the equation for a ellipse:
$x^2/a^2 + y^2/b^2 = 1$
We were shown that all one had to do to get the equation for a hyperbola instead was multiply the original equation by $-1$, so that it became:
$x^2(e^2 - 1) - y^2 = a^2(e^2 - 1)$
The substitution also changes slightly, to become $b^2 = a^2(e^2 - 1)$. Then the equation for a hyperbola can be found.
My question is simply why this multiplication was necessary. It seems to me that multiplying both sides of an equation by some factor does not change anything about it, and therefore the only real change was when $b^2 = a^2(1 - e^2)$ became $b^2 = a^2(e^2 - 1)$. Why was this necessary, and what does it imply about hyperbolas?
Was it necessary because for hyperbolas the eccentricity $e$ is greater than one, and so $b^2 = a^2(1 - e^2)$ would imply $b$ is a complex number? And if so, can the operation be thought of as restricting the domain and range of a hyperbola to the set of all real numbers?
That is my primary question, but I do have another, which is a little more vague:
If hyperbolas do have a complex component, does a plot of one in the complex plane form a cross-section of a three-dimensional or four-dimensional shape, the way an ellipse forms a cross-section of a football? Phrased another way: ellipses are bounded in the set of all real numbers – are hyperbolas bounded in the set of all complex numbers?