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11.4 Hyperbolas

Updated March 2026Fiveable Content Team
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📘Intermediate Algebra Unit 11 Review

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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 c2=a2+b2c^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: x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
    • Vertices at (±a,0)(\pm a, 0)
    • Foci at (±c,0)(\pm c, 0), where c2=a2+b2c^2 = a^2 + b^2
    • Asymptotes: y=±baxy = \pm \frac{b}{a}x
    • Branches open left and right
  • Transverse axis along the y-axis: y2a2x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
    • Vertices at (0,±a)(0, \pm a)
    • Foci at (0,±c)(0, \pm c), where c2=a2+b2c^2 = a^2 + b^2
    • Asymptotes: y=±abxy = \pm \frac{a}{b}x
    • Branches open up and down

Hyperbolas centered at (h,k)(h, k):

  • Transverse axis parallel to the x-axis: (xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
    • Vertices at (h±a, k)(h \pm a,\ k)
    • Foci at (h±c, k)(h \pm c,\ k), where c2=a2+b2c^2 = a^2 + b^2
    • Asymptotes: yk=±ba(xh)y - k = \pm \frac{b}{a}(x - h)
  • Transverse axis parallel to the y-axis: (yk)2a2(xh)2b2=1\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1
    • Vertices at (h, k±a)(h,\ k \pm a)
    • Foci at (h, k±c)(h,\ k \pm c), where c2=a2+b2c^2 = a^2 + b^2
    • Asymptotes: yk=±ab(xh)y - k = \pm \frac{a}{b}(x - h)

A quick way to sketch: draw a rectangle centered at (h,k)(h, k) with dimensions 2a2a by 2b2b. The diagonals of that rectangle are the asymptotes. Then draw the two branches curving through the vertices and approaching those asymptotes.

Graphing hyperbolas and key features, Graphing Hyperbolas | College Algebra

Equations of Hyperbolas

Standard form (center at origin):

  • Horizontal transverse axis: x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
  • Vertical transverse axis: y2a2x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

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

  • Horizontal transverse axis: (xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
  • Vertical transverse axis: (yk)2a2(xh)2b2=1\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)(h, k). This is the midpoint between the two vertices.
  2. Determine the direction the branches open. If they open left/right, the xx-term is positive. If up/down, the yy-term is positive.
  3. Find aa by measuring the distance from the center to either vertex. The transverse axis has length 2a2a.
  4. Find bb using the asymptote slopes. For a horizontal transverse axis, the slopes are ±ba\pm \frac{b}{a}, so you can solve for bb. For a vertical transverse axis, the slopes are ±ab\pm \frac{a}{b}.
  5. Plug hh, kk, a2a^2, and b2b^2 into the correct form.

Example: A hyperbola has center (2,1)(2, -1), vertices at (2,2)(2, 2) and (2,4)(2, -4), and asymptote slopes of ±35\pm \frac{3}{5}. The vertices are vertical (same xx-value), so a=3a = 3 (distance from center to vertex). The slope ab=35\frac{a}{b} = \frac{3}{5} gives b=5b = 5. The equation is (y+1)29(x2)225=1\frac{(y+1)^2}{9} - \frac{(x-2)^2}{25} = 1.

Graphing hyperbolas and key features, Deriving the Equation of a Hyperbola Centered at the Origin | College Algebra

Hyperbolas vs. Other Conic Sections

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

ConicEquation PatternKey Feature
Hyperbola(xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 (or yy first)Minus sign between squared terms; two open branches
Ellipse(xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1Plus sign between squared terms; closed oval shape
Circle(xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2Plus sign, and both denominators are equal (special ellipse)
Parabola(xh)2=4p(yk)(x-h)^2 = 4p(y-k) or (yk)2=4p(xh)(y-k)^2 = 4p(x-h)Only one variable is squared

A common mistake is confusing hyperbolas and ellipses. Remember: minus means hyperbola, plus means ellipse. Also, for hyperbolas c2=a2+b2c^2 = a^2 + b^2, while for ellipses c2=a2b2c^2 = a^2 - b^2. The foci of a hyperbola are farther from the center than the vertices, which is the opposite of an ellipse.

Additional Hyperbola Properties

  • Eccentricity: e=cae = \frac{c}{a}, and for any hyperbola e>1e > 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 2b2a\frac{2b^2}{a}.
  • Degenerate hyperbola: When the equation reduces to two intersecting lines (for example, x24y29=0\frac{x^2}{4} - \frac{y^2}{9} = 0 factors into two lines: y=±32xy = \pm \frac{3}{2}x). This happens when the right side equals 0 instead of 1.