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Vertices of a Hyperbola — Definition, Formula & Examples

Vertices of a Hyperbola

The points at which a hyperbola makes its sharpest turns. The vertices are on the major axis (the line through the foci).

Two hyperbola curves opening left and right, each with a labeled "vertex" at their closest points on the major axis.

 

 

See also

Vertex, directrices of a hyperbola, vertices of an ellipse, vertex of a parabola

Key Formula

Horizontal hyperbola: x2a2−y2b2=1⇒Vertices at (±a, 0)\text{Horizontal hyperbola: } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad \text{Vertices at } (\pm a,\, 0) Vertical hyperbola: y2a2−x2b2=1⇒Vertices at (0, ±a)\text{Vertical hyperbola: } \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \quad \Rightarrow \quad \text{Vertices at } (0,\, \pm a) General center (h,k): Vertices at (h±a, k) or (h, k±a)\text{General center } (h, k)\text{: Vertices at } (h \pm a,\, k) \text{ or } (h,\, k \pm a)
Where:
  • aa = Distance from the center to each vertex along the transverse axis
  • bb = Distance from the center to each co-vertex direction (related to the conjugate axis)
  • (h,k)(h, k) = Center of the hyperbola

Worked Example

Problem: Find the vertices of the hyperbola given by the equation x²/25 − y²/9 = 1.
Step 1: Identify the form of the equation. The positive term is x², so this is a horizontal hyperbola of the form x²/a² − y²/b² = 1.
x225−y29=1\frac{x^2}{25} - \frac{y^2}{9} = 1
Step 2: Read off a² from the denominator under the positive term. Here a² = 25, so a = 5.
a2=25⇒a=5a^2 = 25 \quad \Rightarrow \quad a = 5
Step 3: Since the center is at the origin (0, 0) and the transverse axis is horizontal, the vertices are located at (±a, 0).
Vertices=(±5, 0)\text{Vertices} = (\pm 5,\, 0)
Answer: The vertices are (5, 0) and (−5, 0).

Another Example

This example differs by using a vertical hyperbola with a shifted center (h, k) ≠ (0, 0), showing how to handle translations.

Problem: Find the vertices of the hyperbola (y − 3)²/16 − (x + 2)²/4 = 1.
Step 1: Identify the form. The positive term involves y, so this is a vertical hyperbola with center (h, k).
(y−3)216−(x+2)24=1\frac{(y - 3)^2}{16} - \frac{(x + 2)^2}{4} = 1
Step 2: Read the center from the equation. Here h = −2 and k = 3.
(h,k)=(−2,3)(h, k) = (-2, 3)
Step 3: Find a from the denominator under the positive (y) term. Since a² = 16, we get a = 4.
a2=16⇒a=4a^2 = 16 \quad \Rightarrow \quad a = 4
Step 4: For a vertical hyperbola, the vertices are at (h, k ± a). Substitute the values.
Vertices=(−2, 3±4)=(−2, 7) and (−2, −1)\text{Vertices} = (-2,\, 3 \pm 4) = (-2,\, 7) \text{ and } (-2,\, -1)
Answer: The vertices are (−2, 7) and (−2, −1).

Frequently Asked Questions

How many vertices does a hyperbola have?
A hyperbola has exactly two vertices. They are the two points where the curve intersects its transverse axis. Each branch of the hyperbola contains one vertex, and the two vertices are symmetric about the center.
What is the difference between vertices and foci of a hyperbola?
The vertices are the points on the hyperbola closest to the center, located at distance aa from the center. The foci are points inside each branch at distance cc from the center, where c>ac > a and c2=a2+b2c^2 = a^2 + b^2. The foci are used to define the hyperbola (the difference of distances from any point on the curve to the two foci is constant), while the vertices mark the endpoints of the transverse axis.
How do you find the vertices from a hyperbola equation not in standard form?
First, rewrite the equation in standard form by completing the square for both variables. Once in the form (x−h)2/a2−(y−k)2/b2=1(x-h)^2/a^2 - (y-k)^2/b^2 = 1 or (y−k)2/a2−(x−h)2/b2=1(y-k)^2/a^2 - (x-h)^2/b^2 = 1, identify the center (h,k)(h, k) and the value of aa. Then apply the vertex formula based on whether the transverse axis is horizontal or vertical.

Vertices of a Hyperbola vs. Vertices of an Ellipse

Vertices of a HyperbolaVertices of an Ellipse
DefinitionThe two points closest to the center, on the transverse axisThe two endpoints of the major axis (farthest from center) and two endpoints of the minor axis
Number of vertices2 vertices2 major vertices (sometimes 4 total if co-vertices are counted)
Relationship to focic² = a² + b², so foci are farther from center than vertices (c > a)c² = a² − b², so foci are closer to center than major vertices (c < a)
Formula (centered at origin, horizontal)(±a, 0) from x²/a² − y²/b² = 1(±a, 0) from x²/a² + y²/b² = 1, where a > b

Why It Matters

Vertices are one of the first features you identify when graphing a hyperbola — they anchor the curve and help you sketch the two branches. In precalculus and analytic geometry courses, finding vertices is a standard part of analyzing conic sections. The distance aa from center to vertex also appears in the relationships c2=a2+b2c^2 = a^2 + b^2 and in deriving the asymptote equations y=±(b/a)xy = \pm (b/a)x, making it essential for fully describing the hyperbola.

Common Mistakes

Mistake: Confusing which denominator gives a². Students often assume a² is always the larger denominator.
Correction: For hyperbolas, a² is always the denominator under the positive term, regardless of whether it is larger or smaller than b². This differs from ellipses, where a² is the larger denominator.
Mistake: Using the ellipse relationship c² = a² − b² instead of the hyperbola relationship c² = a² + b².
Correction: For hyperbolas, the foci are farther from the center than the vertices, so c² = a² + b². Remember the plus sign for hyperbolas.

Related Terms