Hyperbolas were introduced in
Introduction to Conic Sections,
as one of
several different curves (‘conic sections’) that are formed by intersecting a plane with an
infinite double cone.
Identifying Conics by the
Discriminant
introduced the general equation for any conic
section,
and gave conditions under which the graph would be a hyperbola.
In this current section, we present and explore the standard definition of a hyperbola.
This definition facilitates the derivation of standard equations for hyperbolas.
Recall that the notation ‘$\,d(P,Q)\,$’ denotes the distance between points $\,P\,$ and $\,Q\,$.
DEFINITION
hyperbola
A hyperbola is the set of points in a plane such that
the difference of the distances from
two fixed points is constant.
More precisely:

$\,P\,$ is a general point on the hyperbola. $\,d(P,F_1)  d(P,F_2) = \text{constant}$ $ \color{red}{\text{red (longer)}}  \color{blue}{\text{blue (shorter)}} = \text{constant} $ 
You can play with hyperbolas using the dynamic JSXGraph at right:

foci: $\,F_1\,$ and $\,F_2\,$, marked with vertices: marked in red $\,P\,$: a typical point on the hyperbola $\,Q\,$: a free point (can be moved anywhere) 
In the definition of hyperbola, the hyperbola constant $\,k\,$ is
required to be positive,
and strictly less than the distance between the two foci:
$$
0 < k < d(F_1,F_2)
$$
Why?
As shown below, other values of $\,k\,$ don't give anything that
a reasonable person would want to call a hyperbola!
$k < 0\,$: No Points in the Solution SetAbsolute value is always nonnegative. That is, $\,x \ge 0\,$ for all real numbers $\,x\,$.Therefore, if $\,k < 0\,$, there are no points $\,P\,$ for which the hyperbola equation is true: $$ \overbrace{\strut \,d(P,F_1)  d(P,F_2)}^{\,\ge\, 0\,} = \overbrace{\strut k}^{\lt \, 0} \qquad \text{ is always false } $$ 

$k = 0\,$: Perpendicular Bisector of Segment Between the FociThe equation $\,x = 0\,$ is equivalent to $\,x = 0\,$, for all real numbers $\,x\,$.Thus, the following are equivalent: $$ \begin{gather} d(P,F_1)  d(P,F_2) = 0\cr\cr d(P,F_1)  d(P,F_2) = 0\cr\cr d(P,F_1) = d(P,F_2) \end{gather} $$ At right, $\,Q\,$ is the midpoint of the segment $\,\overline{F_1F_2}\,$ that connects the foci. When $\,k = 0\,$, the set of points satisfying the hyperbola equation is the perpendicular bisector of $\,\overline{F_1F_2}\,$. 

A ‘TWO RAY’ HYPERBOLA: $\,k = d(F_1,F_2)\,$Suppose the hyperbola constant, $\,k\,$, equals the distance between the foci.That is, $\,k = d(F_1,F_2)\,$. In this case, the solution set to the equation $$ \color{green}{d(P,F_1)}  \color{red}{d(P,F_2)} = k $$ is two rays, with $\,F_1\,$ and $\,F_2\,$ as endpoints, lying along the major axis. This degenerate hyperbola is shown at right, in black. $\,P\,$ is a point on the degenerate hyperbola; it can be dragged from one ray to the other. 

AN ‘EMPTY’ HYPERBOLA: $\,k > d(F_1,F_2)\,$Suppose $\,F_1\,$ and $\,F_2\,$ are distinct points in a plane, and let $\,k > d(F_1,F_2)\,$.Let $\,P\,$ be any point in the plane. Switch names (if needed) so that $\,P\,$ is closer to $\,F_1\,$ (or equidistant from both foci). This situation is illustrated at right. The shortest distance between any two points is a straight line. In particular (refer to the sketch at right), traveling from $\,P\,$ to $\,F_1\,$, and then from $\,F_1\,$ to $\,F_2\,$, must exceed (or equal) $\,\color{red}{d(P,F_2)}\,$: $$ d(P,F_1) + d(F_1,F_2) \ge d(P,F_2) $$ Rearranging: $$ d(P,F_2)  d(P,F_1) \le d(F_1,F_2)\qquad (*) $$ We have: $$ \begin{alignat}{2} d(P,F_2)  d(P,F_1) &\ =\ d(P,F_2)  d(P,F_1)&\qquad&\text{(for $\,x\ge 0\,$, $\,x = x\,$)}\cr &\ \le\ d(F_1,F_2)&&\text{by (*)}\cr &\ < \ k&&\text{(by hypothesis)} \end{alignat} $$ Consequently, there are no points $\,P\,$ for which the hyperbola equation is true: $$ \overbrace{\strut \,d(P,F_2)  d(P,F_1)}^{\text{strictly less than $\,k\,$}} = k \qquad \text{ is always false } $$ You might want to call this an empty hyperbola, an invisible hyperbola, or an imaginary hyperbola! There's nothing there! 

Most people don't want to call any of these situations—the empty set, a line, or two rays—a hyperbola!
This is why these values for $\,k\,$ are not allowed in the definition of hyperbola.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 