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. 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 \(y = \pm\dfrac{b}{a}\,x\text{.}\)
Figure5.15.A hyperbola opening left and right, with vertices \((\pm a,0)\text{,}\) foci \((\pm c,0)\text{,}\) and asymptotes \(y = \pm\frac{b}{a}x\text{.}\)
Figure5.16.Animation: as \(P\) moves along a branch, the absolute difference \(|r_1 - r_2|\) of its distances to the foci stays fixed at \(2a = 8\text{,}\) while the branch hugs its asymptotes.
The formulas for \(a\) arise exactly as they did for the ellipse, with sums replaced by differences. Evaluating the definition at the vertex \((a,0)\text{,}\) the two focal radii are \(c-a\) and \(c+a\text{,}\) so
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
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.
Figure5.17.Animation: deriving the formula for \(a\text{.}\) Sliding \(P\) to the vertex shows the constant equals \(2a\text{;}\) the central box then builds the right triangle giving \(c^2 = a^2 + b^2\text{.}\)
Figure5.18.Animation: in practice, put the equation in standard form and read \(a^2\) off the denominator under the positive term; then \(a\) is measured along the transverse axis, from the center to a vertex.
so the foci are \((\pm 5, 0)\text{,}\) the vertices are \((\pm 4, 0)\text{,}\) and the asymptotes are \(y = \pm\frac{b}{a}x = \pm\frac{3}{4}x\text{.}\)
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 \(c^2 = a^2 + b^2\text{,}\)
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.
Figure5.22.The vertical hyperbola \(\frac{y^2}{9}-\frac{x^2}{16}=1\text{:}\) it opens up and down, with vertices \((0,\pm3)\text{,}\) foci \((0,\pm5)\text{,}\) and asymptotes \(y=\pm\frac34 x\text{.}\)
