Definition 5.17.
A hyperbola is the set of points in a plane the difference of whose distances from two fixed points \(F_1\) and \(F_2\) (the foci) is constant.
Section 5.3 Hyperbola
For \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) the vertices are \((\pm a, 0)\text{,}\) the foci satisfy \(c^2 = a^2 + b^2\text{,}\) and the asymptotes are \(y = \pm\dfrac{b}{a}\,x\text{.}\)
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The formulas for \(a\) arise exactly as they did for the ellipse, with sums replaced by differences. Evaluating Definition 5.17 at the vertex \((a,0)\text{,}\) the two focal radii are \(c-a\) and \(c+a\text{,}\) so
\begin{equation*}
|r_1 - r_2| = (c+a) - (c-a) = 2a,
\end{equation*}
identifying the constant as \(2a\text{,}\) or \(a = \frac{1}{2}\,|r_1 - r_2|\text{.}\) For the second relation, draw the central box whose half-width is \(a\) and half-height is \(b\text{.}\) Its corner \((a,b)\) lies at distance \(c\) from the center, so the right triangle with legs \(a\) and \(b\) has hypotenuse \(c\text{,}\) and
\begin{equation*}
c^2 = a^2 + b^2 \qquad\Longleftrightarrow\qquad a^2 = c^2 - b^2.
\end{equation*}
Notice the contrast with the ellipse: there \(a\) was the hypotenuse, so \(c^2 = a^2 - b^2\text{;}\) here \(a\) is a leg, so \(c^2 = a^2 + b^2\text{.}\) That single difference is the source of the sign change between the two formulas.
The same box explains the asymptotes. Solving \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) for \(y\) gives
\begin{equation*}
y = \pm\frac{b}{a}\,\sqrt{x^2 - a^2}
= \pm\frac{b}{a}\,x\,\sqrt{1 - \frac{a^2}{x^2}}\qquad (x \gt 0).
\end{equation*}
The whole difference between the hyperbola and the pair of lines \(y = \pm\frac{b}{a}x\) is that square root. As \(|x|\) grows, \(a^2/x^2 \to 0\text{,}\) so the root tends to \(1\) and the curve tends to those lines. The gap closes but never shuts: since \(\sqrt{1 - a^2/x^2} \lt 1\) for every finite \(x\text{,}\) each branch stays strictly inside its asymptotes, approaching without ever touching. That is exactly the behavior of an asymptote—and it is why \(a\) and \(b\text{,}\) which never appear as distances to a focus, nonetheless control the shape at infinity.
Geometrically the lines are just the extended diagonals of the central box, as Figure 5.21 shows: the diagonal through the corner \((a,b)\) rises \(b\) over a run of \(a\text{,}\) so its slope is \(b/a\text{,}\) and the one through \((a,-b)\) has slope \(-b/a\text{.}\) So drawing the box gives the asymptotes for free, and the box plus the two vertices is enough to sketch the curve. For a vertical hyperbola \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\) the box is the same, but the transverse axis is vertical and the slopes become \(\pm\dfrac{a}{b}\text{.}\)
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Example 5.22 Horizontal hyperbola.
Find the foci and asymptotes of the hyperbola \(9x^2 - 16y^2 = 144\) and sketch its graph.
Solution.
Dividing by \(144\) gives standard form:
\begin{equation*}
\frac{x^2}{16} - \frac{y^2}{9} = 1
\;\Longrightarrow\; a = \sqrt{16} = 4,\quad b = \sqrt{9} = 3.
\end{equation*}
Since the \(x^2\)-term is positive, the hyperbola opens left and right. Then
\begin{equation*}
c^2 = a^2 + b^2 = 16 + 9 = 25 \;\Longrightarrow\; c = 5,
\end{equation*}
so the foci are \((\pm 5, 0)\text{,}\) the vertices are \((\pm 4, 0)\text{,}\) and the asymptotes are \(y = \pm\frac{b}{a}x = \pm\frac{3}{4}x\text{.}\)
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Example 5.25 A vertical hyperbola (opens up and down).
Find the vertices, foci, and asymptotes of \(16y^2 - 9x^2 = 144\) and sketch its graph.
Solution.
Dividing by \(144\) and being careful with signs,
\begin{equation*}
\frac{y^2}{9} - \frac{x^2}{16} = 1.
\end{equation*}
Now the \(y^2\)-term is the positive one, so the hyperbola opens up and down. In this orientation \(a^2\) is the denominator under \(y^2\text{,}\) giving \(a = 3\) (measured along the \(y\)-axis) and \(b = 4\text{.}\) Still using \(c^2 = a^2 + b^2\text{,}\)
\begin{equation*}
c^2 = 9 + 16 = 25 \;\Longrightarrow\; c = 5.
\end{equation*}
So the vertices are \((0,\pm 3)\) and the foci are \((0,\pm 5)\text{,}\) both on the \(y\)-axis. For a vertical hyperbola the asymptotes are \(y = \pm\dfrac{a}{b}\,x = \pm\dfrac{3}{4}x\)—note the slope uses \(a/b\) here, not \(b/a\text{.}\) The transverse axis has simply rotated from horizontal to vertical.
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