11.4 Hyperbolas

Hyperbolas

A hyperbola is a conic section formed by the difference of distances from two fixed points (foci). What sets it apart from ellipses and circles is the minus sign between its squared terms, which produces two separate branches that open away from each other and extend infinitely.

The key to working with hyperbolas is identifying the center, the direction the branches open, and the relationship $c^2 = a^2 + b^2$ (notice this is a sum, unlike the ellipse where it's a difference).

Graphing Hyperbolas and Key Features

Every hyperbola has a transverse axis (the axis that passes through both vertices) and a conjugate axis (perpendicular to it). The variable with the positive squared term tells you which axis is the transverse axis.

Hyperbolas centered at the origin:

• Transverse axis along the x-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
  • Vertices at $(\pm a, 0)$
  • Foci at $(\pm c, 0)$, where $c^2 = a^2 + b^2$
  • Asymptotes: $y = \pm \frac{b}{a}x$
  • Branches open left and right
• Transverse axis along the y-axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
  • Vertices at $(0, \pm a)$
  • Foci at $(0, \pm c)$, where $c^2 = a^2 + b^2$
  • Asymptotes: $y = \pm \frac{a}{b}x$
  • Branches open up and down

Hyperbolas centered at $(h, k)$:

• Transverse axis parallel to the x-axis: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
  • Vertices at $(h \pm a,\ k)$
  • Foci at $(h \pm c,\ k)$, where $c^2 = a^2 + b^2$
  • Asymptotes: $y - k = \pm \frac{b}{a}(x - h)$
• Transverse axis parallel to the y-axis: $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
  • Vertices at $(h,\ k \pm a)$
  • Foci at $(h,\ k \pm c)$, where $c^2 = a^2 + b^2$
  • Asymptotes: $y - k = \pm \frac{a}{b}(x - h)$

A quick way to sketch: draw a rectangle centered at $(h, k)$ with dimensions $2a$ by $2b$. The diagonals of that rectangle are the asymptotes. Then draw the two branches curving through the vertices and approaching those asymptotes.

Equations of Hyperbolas

Standard form (center at origin):

• Horizontal transverse axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
• Vertical transverse axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Shifted form (center at $(h, k)$):

• Horizontal transverse axis: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
• Vertical transverse axis: $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

Writing an equation from a graph or given information:

1. Find the center $(h, k)$. This is the midpoint between the two vertices.
2. Determine the direction the branches open. If they open left/right, the $x$-term is positive. If up/down, the $y$-term is positive.
3. Find $a$ by measuring the distance from the center to either vertex. The transverse axis has length $2a$.
4. Find $b$ using the asymptote slopes. For a horizontal transverse axis, the slopes are $\pm \frac{b}{a}$, so you can solve for $b$. For a vertical transverse axis, the slopes are $\pm \frac{a}{b}$.
5. Plug $h$, $k$, $a^2$, and $b^2$ into the correct form.

Example: A hyperbola has center $(2, -1)$, vertices at $(2, 2)$ and $(2, -4)$, and asymptote slopes of $\pm \frac{3}{5}$. The vertices are vertical (same $x$-value), so $a = 3$ (distance from center to vertex). The slope $\frac{a}{b} = \frac{3}{5}$ gives $b = 5$. The equation is $\frac{(y+1)^2}{9} - \frac{(x-2)^2}{25} = 1$.

Hyperbolas vs. Other Conic Sections

The fastest way to identify a conic is to look at the equation's structure:

Additional Hyperbola Properties

• Eccentricity: $e = \frac{c}{a}$, and for any hyperbola $e > 1$. The larger the eccentricity, the "flatter" the branches (they hug the asymptotes more closely).
• Latus rectum: A line segment perpendicular to the transverse axis through a focus, with endpoints on the hyperbola. Its length is $\frac{2b^2}{a}$.
• Degenerate hyperbola: When the equation reduces to two intersecting lines (for example, $\frac{x^2}{4} - \frac{y^2}{9} = 0$ factors into two lines: $y = \pm \frac{3}{2}x$). This happens when the right side equals 0 instead of 1.