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 conic section.
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):
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad\Rightarrow\quad \text{vertices } (\pm a,\ 0)$$
Case 2 - vertical (opens up/down):
$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \quad\Rightarrow\quad \text{vertices } (0,\ \pm a)$$
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; see foci of hyperbola for how far the foci lie beyond the vertices.
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 $\dfrac{x^2}{25} - \dfrac{y^2}{16} = 1$.
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 $\dfrac{y^2}{9} - \dfrac{x^2}{49} = 1$.
Wrong path first: the larger denominator is $49$, so a student may write $a^2 = 49$ and place vertices on the $x$-axis, copying the ellipse rule "big denominator wins." That is wrong for a hyperbola. The axis is set by the positive term, not the bigger number. Here 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 $(0, -3)$. The lesson: for a hyperbola, follow the sign, not the size.
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 $2a = 12$. This is a horizontal hyperbola centred at the origin, since the vertices are symmetric about $(0,0)$.
Example 4
Find the vertices of $\dfrac{(x-3)^2}{16} - \dfrac{(y+2)^2}{9} = 1$.
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 $1$. $\dfrac{9x^2}{36} - \dfrac{4y^2}{36} = 1$ $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$ The $x^2$ term is positive, so $a^2 = 4$ 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 $h = -1$, $k = 4$, $a = 5$. $(-1,\ 4 + 5) = (-1, 9)$ and $(-1,\ 4 - 5) = (-1, -1)$. The two vertices are $(-1, 9)$ and $(-1, -1)$, both directly above and below the centre.
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, and that idea reaches far past a textbook curve.
The defining property. Of every point on a hyperbola's branch, the vertex is the one nearest the centre and nearest its focus. That "closest point" is exactly what an engineer or physicist wants to locate when two objects follow hyperbolic paths.
Real-world stakes. An object flying past a star or planet fast enough to escape follows a hyperbolic trajectory, and the vertex marks its closest pass - the perigee of a flyby. Mission planners for spacecraft gravity-assist manoeuvres compute this vertex precisely, because it fixes how much speed the flyby adds and how close the craft comes to the body. NASA describes the technique in its overview of the gravity assist manoeuvre.
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. Learning to read the vertex from the equation is the same skill across all three, and the vertex-to-focus spacing is what sets each conic's eccentricity.
That is why the vertex is worth finding first: it is the anchor point every other feature of the curve is measured from.
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, full stop, whether or not it is larger. The sign decides the direction; the size is irrelevant. The memoriser who carried over the ellipse rule places the vertices on the wrong axis, as Example 2 shows.
Mistake 2: Forgetting to shift by the centre
Where it slips in: Hyperbolas written 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 $(\pm a, 0)$ form is only for a centre at the origin. When the centre is $(h, k)$, the vertices are $(h \pm a, k)$ (horizontal) or $(h, k \pm a)$ (vertical). The rusher who skips the shift lands the vertices at the origin's positions instead of the true ones.
Mistake 3: Not converting to standard form first
Where it slips in: Equations like $9x^2 - 4y^2 = 36$ where the right side is not $1$.
The correct way: Divide through so the right side equals $1$ before reading any denominators, as in Example 5. Reading $a^2 = 9$ straight off $9x^2$ gives a vertex that is wrong by a factor.
Conclusion
The vertices of a hyperbola are the two points where each branch is closest to the centre, lying on the transverse axis.
For $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ the vertices are $(\pm a, 0)$; for $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ they are $(0, \pm a)$.
The positive term - not the larger denominator - decides the axis direction, and $a^2$ is its denominator.
The distance between the two vertices is $2a$, the length of the transverse axis.
For a centre at $(h, k)$, shift the vertices to $(h \pm a, k)$ or $(h, k \pm a)$.
To take hyperbolas and conics further with a teacher, explore Bhanzu's geometry tutor, our high school math tutor sessions, or math tutoring online. To see a trainer graph a hyperbola from its equation live, you can book a free demo class.
A practical next step
Work through the exercises below. For each, put the equation in standard form first, then read off the vertices.
Find the vertices of $\dfrac{y^2}{36} - \dfrac{x^2}{13} = 1$. (Answer to Question 1: vertical hyperbola, $a = 6$, vertices $(0, 6)$ and $(0, -6)$.)
Find the vertices of $\dfrac{(x+2)^2}{9} - \dfrac{(y-1)^2}{4} = 1$. (Answer to Question 2: horizontal, centre $(-2, 1)$, $a = 3$, vertices $(1, 1)$ and $(-5, 1)$.)
Read More
Hyperbola equation and graphing - the full curve, its equation, and how to graph it.
Eccentricity - the shape parameter that classifies each conic.
Eccentricity of ellipse - the shape parameter measured from vertex-to-focus spacing.
Foci of ellipse - the two-focus definition, for the closed conic.
Directrix of parabola - the single-vertex conic's guiding line.
Latus rectum of parabola, ellipse, hyperbola - the chord through the focus, across all conics.
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