What Is The Vertex Of An Ellipse?
An ellipse is the smooth oval curve of points whose distances from two fixed points (the foci) add to a constant. The vertices of an ellipse are the two endpoints of the major axis - the longest diameter of the ellipse - and they are the two points on the curve farthest from the center. Each endpoint of the shorter minor axis is called a co-vertex. An ellipse therefore has exactly two vertices and two co-vertices. For the underlying curve, see ellipse.
The key distinction, and the source of nearly every mistake on this topic, is that vertices lie on the major (longer) axis while co-vertices lie on the minor (shorter) axis. Since the major axis is longer, the vertices are always farther from the center than the co-vertices - that is, $a > b$ always.
The Formula: Finding The Vertices
For an ellipse centered at the origin, the standard equation is one of two forms, and which denominator is larger decides where the vertices sit. Write $a^2$ for the larger denominator and $b^2$ for the smaller.
Horizontal ellipse (major axis along the $x$-axis, larger number under $x^2$):
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad \text{vertices } (\pm a, 0), \ \text{co-vertices } (0, \pm b)$$
Vertical ellipse (major axis along the $y$-axis, larger number under $y^2$):
$$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \quad \Rightarrow \quad \text{vertices } (0, \pm a), \ \text{co-vertices } (\pm b, 0)$$
For an ellipse centered at $(h, k)$, shift everything by the center:
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad \Rightarrow \quad \text{vertices } (h \pm a, k)$$
The recipe is always the same three steps: put the equation in standard form, find the larger denominator and call its square root $a$, then place the vertices $a$ units from the center along the axis of the larger denominator. The distance $a$ is the semi-major axis; $b$ is the semi-minor axis. The foci sit between the center and vertices, a detail covered in foci of ellipse.
Examples Of The Vertex Of An Ellipse
Each example builds from a direct read-off to a fuller reasoning task. The problem statement is bold; the working is not.
Example 1
Find the vertices of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
The larger denominator is 25, under $x^2$, so the ellipse is horizontal and $a^2 = 25$, giving $a = 5$.
$$\text{vertices} = (\pm a, 0) = (5, 0) \text{ and } (-5, 0)$$
The co-vertices, from $b^2 = 9$, are $(0, 3)$ and $(0, -3)$.
Example 2
Find the vertices of $\frac{x^2}{16} + \frac{y^2}{49} = 1$.
Here the larger denominator is 49, under $y^2$, so the major axis is vertical and $a^2 = 49$, giving $a = 7$.
$$\text{vertices} = (0, \pm a) = (0, 7) \text{ and } (0, -7)$$
The co-vertices, from $b^2 = 16$, are $(\pm 4, 0)$.
Example 3: The trap of always reading a from the x-term
A student meets $\frac{x^2}{9} + \frac{y^2}{36} = 1$ and writes "vertices at $(\pm 3, 0)$," taking $a^2$ from under $x^2$ out of habit.
Following that habit gives $(\pm 3, 0)$. But check which axis is longer: the number under $y^2$ is 36, which is larger than the 9 under $x^2$. The larger denominator marks the major axis, so the major axis here is vertical, not horizontal.
Correctly, $a^2 = 36$ (the larger), so $a = 6$ along the $y$-axis:
$$\text{vertices} = (0, \pm 6) = (0, 6) \text{ and } (0, -6)$$
The points $(\pm 3, 0)$ are the co-vertices, not the vertices. The rescue is to compare the denominators first and let the larger one - not the $x$-term - decide the major axis.
Example 4
An ellipse is centered at $(2, -1)$ with equation $\frac{(x-2)^2}{49} + \frac{(y+1)^2}{4} = 1$. Find its vertices.
The larger denominator is 49 under the $x$-term, so the major axis is horizontal, $a = 7$, and the center is $(2, -1)$. Shift $a$ units left and right of the center:
$$\text{vertices} = (2 \pm 7, -1) = (9, -1) \text{ and } (-5, -1)$$
Example 5
The vertices of an ellipse are $(0, 8)$ and $(0, -8)$, and the co-vertices are $(5, 0)$ and $(-5, 0)$. Write its standard equation.
The vertices are on the $y$-axis, so the ellipse is vertical with $a = 8$ ($a^2 = 64$) and $b = 5$ ($b^2 = 25$):
$$\frac{x^2}{25} + \frac{y^2}{64} = 1$$
Example 6
A whispering-gallery dome has an elliptical cross-section 40 m wide and 24 m tall. Find the vertices, taking the center as the origin.
The width spans the major axis, so the full major axis is 40 m and the semi-major axis is $a = 20$. The vertices sit at the ends of the horizontal major axis:
$$\text{vertices} = (\pm 20, 0) = (20, 0) \text{ and } (-20, 0)$$
The semi-minor axis is $b = 12$, giving co-vertices $(0, \pm 12)$. The dome's acoustic focus points depend on these dimensions.
Where The Vertex Of An Ellipse Earns Its Keep
The vertices fix the size and orientation of every real ellipse, so they matter wherever ellipses appear.
Planetary orbits. Planets travel in ellipses with the Sun at one focus. The vertices mark perihelion (closest approach) and aphelion (farthest point) - the two extreme distances from the Sun, which sit exactly at the ends of the major axis.
Whispering galleries. In an elliptical dome such as the one described in this whispering gallery overview, sound from one focus reflects to the other. The vertices set the major axis, which fixes where the two focus points fall, as in Example 6.
Engineering and optics. Elliptical gears, arches, and reflectors are all specified by their semi-major axis $a$ (to the vertex) and semi-minor axis $b$ (to the co-vertex).
The deeper reason the vertices matter is that, together with the co-vertices, they pin down the entire curve: $a$ and $b$ determine the ellipse's shape, its eccentricity, and the position of its foci. That chain - vertices to axes to foci - is what Johannes Kepler used to describe planetary motion. For how stretched-out an ellipse is, see eccentricity, and for the broader family of curves, conic sections. The standard definitions are laid out in the OpenStax section on the ellipse.
The Mistakes Students Make Most Often
Mistake 1: Assuming the x-term always gives the vertices
Where it slips in: Reading $a^2$ from under $x^2$ by habit, ignoring which denominator is larger, as in Example 3.
Don't do this: State horizontal vertices for an ellipse whose larger denominator is under $y^2$.
The correct way: Compare the two denominators first. The larger one is $a^2$ and marks the major axis; the vertices lie on that axis. Only when the larger denominator is under $x^2$ are the vertices horizontal.
Mistake 2: Swapping vertices and co-vertices
Where it slips in: Calling the minor-axis endpoints "vertices," or forgetting that $a > b$.
Don't do this: Report the co-vertices as the ellipse's vertices.
The correct way: Vertices are on the major (longer) axis at distance $a$; co-vertices are on the minor (shorter) axis at distance $b$, and $a$ is always the larger of the two. If your "vertex" is closer to the center than your "co-vertex," you have swapped them.
Mistake 3: Forgetting to shift by the center
Where it slips in: Applying the origin formulas to an ellipse centered at $(h, k)$.
Don't do this: Write vertices as $(\pm a, 0)$ when the center is not the origin.
The correct way: For a center at $(h, k)$, the vertices are $(h \pm a, k)$ (horizontal) or $(h, k \pm a)$ (vertical). Always add the center coordinates back in
Conclusion
The vertices of an ellipse are the two endpoints of the major axis, at distance $a$ from the center.
The co-vertices are the endpoints of the minor axis, at distance $b$, and $a > b$ always.
Find the vertices by taking $a$ from the larger denominator and placing them along that axis.
The most common mistake is reading $a$ from the $x$-term instead of the larger denominator.
The vertices, foci, and eccentricity together fix the ellipse's full shape and orientation.
A Practical Next Step
Work through the exercises below to solidify your understanding, and check each answer as you go.
Find the vertices of $\frac{x^2}{100} + \frac{y^2}{36} = 1$. (Answer to Question 1: $(\pm 10, 0)$.)
Find the vertices of $\frac{x^2}{9} + \frac{y^2}{81} = 1$. (Answer to Question 2: $(0, \pm 9)$.)
An ellipse centered at $(1, 2)$ has $a = 4$ along the $x$-axis. Give its vertices. (Answer to Question 3: $(5, 2)$ and $(-3, 2)$.)
To take the vertex of an ellipse further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor for conic sections, or browse math classes online. Want to work these through with a live trainer? Try a free class.
Read More
Vertex of a Hyperbola - the same idea for the hyperbola's two branches.
Area of an Ellipse - how the semi-axes $a$ and $b$ set the enclosed area.
Foci of a Hyperbola - the focus points of the ellipse's conic cousin.
Parabola - another conic and how its vertex is defined.
Directrix of a Parabola - the guiding line that pairs with a conic's focus.
Was this article helpful?
Your feedback helps us write better content