Book A Free Math Class

# Vertex Of Hyperbola - Definition, Formula, and Examples

TL;DR

The vertices of a hyperbola are the two points where each branch is closest to the centre, lying on the transverse axis a distance aaa from the centre. For \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), they sit at \((\pm a,0)\), and the distance between them is \( 2a \). This article defines the vertex, derives its coordinates from any standard equation, and works through examples.

## The Two Turning Points Where A Hyperbola Bends Back On Itself

A hyperbola runs off to infinity in four directions, yet each of its two curves has one exact point where it stops approaching the centre and turns away - and that single point controls the whole shape. Miss it and you cannot graph the curve; find it and the transverse axis, the width, and the direction all fall into place.

The **vertex of a hyperbola** is the point on a branch that is closest to the centre; a hyperbola has **two** vertices, one per branch, both lying on the transverse axis. They are equidistant from the centre, separated by a distance of \( 2a \), and they are the points where the curve crosses its transverse axis. The vertices, together with the foci, are collinear on that axis. A **hyperbola** is the set of points whose _difference_ of distances to two fixed foci is constant; for the broader family it belongs to, see the [conic section](/content/math/geometry/conic-sections/index.html).

By the end you will read the vertices straight off any standard equation, handle both horizontal and vertical hyperbolas, and shift them correctly when the centre is not at the origin.

## Reading The Vertices From The Standard Equation

Everything starts with which variable carries the **positive** term. That single sign tells you the axis direction, and the number under it gives you \( a \).

**Case 1 - horizontal (opens left/right):**
\[
x^2/a^2 - y^2/b^2 = 1 \Rightarrow \text{vertices } (\pm a, 0)\n\]

**Case 2 - vertical (opens up/down):**
\[
y^2/a^2 - x^2/b^2 = 1 \Rightarrow \text{vertices } (0, \pm a)\n\]

Here \( a^2 \) is always the denominator of the **positive** term, and \( a \) is its positive square root. Two things students mix up, so pin them now: the positive term decides the direction (not the larger denominator, as with an ellipse), and \( a \) is under that positive term regardless of whether \( a>b \) or \( a<b \).

For a hyperbola centred at \( (h,k) \), replace \( x \) with \( (x-h) \) and \( y \) with \( (y-k) \), then **shift the origin vertices by (h,k)**. A horizontal hyperbola centred at \( (h,k) \) has vertices \( (h\pm a,k) \); a vertical one has \( (h,k\pm a) \). The related landmarks - foci, directrices, and asymptotes - all sit relative to the same centre.

## Examples Of The Vertex Of A Hyperbola

Six worked cases, from a clean origin-centred read to a shifted vertical hyperbola you must complete.

### Example 1

**Find the vertices of \( \frac{x^2}{25} - \frac{y^2}{16} = 1 \).**

The \( x^2 \) term is positive, so the hyperbola is horizontal and the vertices lie on the \( x \)-axis. \( a^2=25 \), so \( a=5 \). The vertices are \((\pm 5,0)\), that is \((5,0)\) and \((-5,0)\). The distance between them is \( 2a=10 \).

### Example 2

**Find the vertices of \( \frac{y^2}{9} - \frac{x^2}{49} = 1 \).**

The larger denominator is 49, but the y^2 term is positive, so the hyperbola is vertical and \( a^2=9 \), giving \( a=3 \). The vertices are \((0,\pm 3)\), that is \((0,3)\) and \((0,-3)\).

### Example 3

**A hyperbola has vertices \((\pm 6,0)\). Write \( a \) and the distance between the vertices.**

The vertices lie on the \( x \)-axis at \( \pm 6 \), so \( a=6 \). The distance between the two vertices is \( 2a=12 \).

### Example 4

**Find the vertices of \( \frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1 \).**

The centre is \( (h,k)=(3,-2) \). The \( x \)-term is positive, so the hyperbola is horizontal with \( a^2=16 \), giving \( a=4 \). Shift the origin vertices \((\pm a,0)\) by the centre: vertices are \((3\pm 4,-2)\). So the vertices are \((7,-2)\) and \((-1,-2)\).

### Example 5

**Find the vertices of \( 9x^2 - 4y^2 = 36 \).**

The equation is not yet in standard form; divide every term by 36 to make the right side 1. \( \frac{9x^2}{36} - \frac{4y^2}{36} = 1 \) gives \( a^2=4 \) and \( a=2 \). The vertices are \((\pm 2,0)\).

### Example 6

**A vertical hyperbola is centred at \((-1,4)\) with \( a=5 \). Give its vertices.**

For a vertical hyperbola, the vertices are \((h,k\pm a)\). Substitute \( h=-1, k=4, a=5 \): \((-1,9)\) and \((-1,-1)\).

## Where The Vertex Earns Its Keep: "The Point Of Closest Approach"

The vertex is not just a graphing landmark. It is the mathematical name for a **point of closest approach**.

- **The defining property.** Of every point on a hyperbola's branch, the vertex is the one nearest the centre and nearest its focus.
- **Real-world stakes.** When objects follow hyperbolic paths, the vertex marks their closest pass - the _perigee_ of a flyby. Mission planners compute this vertex precisely as it fixes the speed and closeness during an encounter.
- **Where it goes next.** Vertices anchor the whole conic family: a parabola has one vertex, an ellipse has two on each axis, and a hyperbola has two on the transverse axis.

## Common Mistakes With Hyperbola Vertices

### Mistake 1: Using the larger denominator instead of the positive term

**Where it slips in:** Deciding which axis the vertices lie on.

**Don't do this:** Pick \( a^2 \) as the bigger of the two denominators, as you would for an ellipse.

**The correct way:** For a hyperbola, \( a^2 \) is the denominator of the **positive** term; the sign decides the direction.

### Mistake 2: Forgetting to shift by the centre

**Where it slips in:** Hyperbolas with \( (x-h) \) and \( (y-k) \).

**Don't do this:** Read vertices as \((\pm a,0)\) and stop, ignoring \( (h,k) \).

**The correct way:** The vertices' form is only for a centre at the origin.

### Mistake 3: Not converting to standard form first

**Where it slips in:** Equations not in standard form, like \( 9x^2 - 4y^2 = 36 \).

**The correct way:** Divide through so the right side equals 1 before reading any denominators.

## Conclusion

- The **vertices of a hyperbola** are the two points where each branch is closest to the centre, along the transverse axis.
- For \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) the vertices are \((\pm a,0)\); for \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) they are \((0,\pm a)\).
- The **positive term** decides the axis direction regardless of denominator size.
- The distance between the vertices is \( 2a \), the length of the transverse axis.
- For a centre at \( (h,k) \), shift the vertices accordingly.