What Is the Directrix of an Ellipse?
The directrix of an ellipse is a fixed line, perpendicular to the major axis, that pairs with a focus to define the ellipse. Specifically, an ellipse is the set of all points $P$ for which the distance to a fixed point (the focus) divided by the distance to the directrix is a constant $e$ less than 1:
$$\frac{PF}{PM} = e, \qquad 0 < e < 1$$
Here $PF$ is the distance from a point on the curve to the focus, $PM$ is its perpendicular distance to the directrix, and $e$ is the eccentricity. Because an ellipse has two foci, it has two directrices - one associated with each focus, each lying outside the curve beyond the nearer vertex. This focus-directrix rule is the same one that defines a parabola and a hyperbola; only the value of $e$ changes.
The Two Lines an Ellipse Never Touches but Cannot Exist Without
Every ellipse hides two straight lines it never touches, yet they fix its exact shape. They sit outside the curve entirely, one beyond each end of the long axis, and no point of the ellipse ever reaches them. Still, every single point on the ellipse obeys a strict rule about its distance to one of these lines and its distance to a focus. Those hidden lines are the directrices, and they are what turn a vague oval into a precisely defined conic.
What Is the Formula for the Directrix of an Ellipse?
For the standard ellipse with a horizontal major axis,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b),$$
the two directrices are the vertical lines
$$x = \frac{a}{e} \qquad \text{and} \qquad x = -\frac{a}{e},$$
where $e$ is the eccentricity. The variables mean: $a$ is the semi-major axis (half the long axis), $b$ is the semi-minor axis, and $e = \dfrac{c}{a}$ with $c = \sqrt{a^2 - b^2}$ the focal distance. An equivalent form is $x = \pm\dfrac{a^2}{c}$, since $\dfrac{a}{e} = \dfrac{a}{c/a} = \dfrac{a^2}{c}$.
If instead the major axis is vertical, $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$, the directrices are horizontal lines $y = \pm\dfrac{a}{e}$. In every case the directrix is perpendicular to the major axis and lies outside the ellipse.
What Are the Properties of the Directrix of an Ellipse?
The directrix is a fixed line, and a handful of properties pin down exactly where it sits and how it behaves:
An ellipse has two directrices. One pairs with each focus, placed symmetrically on either side of the centre.
Each lies outside the curve. Because $e < 1$, the distance $\dfrac{a}{e}$ exceeds $a$, so every directrix falls beyond the nearer vertex and never touches the ellipse.
Both are perpendicular to the major axis. A horizontal-major-axis ellipse has vertical directrices; a vertical one has horizontal directrices.
They run parallel to the minor axis and to the latus rectum. All three are square to the major axis, so they share one direction.
They are equidistant from the centre. The two directrices sit at $\pm\dfrac{a}{e}$, mirror images across the minor axis.
They carry the focus-directrix ratio. For every point $P$ on the ellipse, $\dfrac{PF}{PM} = e$, the property that defines the curve.
Together these say the directrix is not free-floating: its orientation is fixed by the major axis and its distance is fixed by the eccentricity.
Where Does the Formula x = a/e Come From?
The directrix location is not arbitrary - it drops out of the focus-directrix definition applied at a vertex. Take the right vertex of a horizontal ellipse, the point $(a, 0)$, which lies on the curve, so it must satisfy $\dfrac{PF}{PM} = e$.
Its distance to the right focus $F(ae, 0)$ is
$$PF = a - ae = a(1 - e).$$
Let the right directrix be the line $x = d$. The vertex's perpendicular distance to it is
$$PM = d - a.$$
The focus-directrix rule requires $\dfrac{PF}{PM} = e$:
$$\frac{a(1 - e)}{d - a} = e$$
Solve for $d$:
$$a(1 - e) = e(d - a)$$
$$a - ae = ed - ea$$
$$a = ed$$
$$d = \frac{a}{e}$$
There it is: the right directrix is $x = \dfrac{a}{e}$, and by symmetry the left is $x = -\dfrac{a}{e}$. Since $e < 1$ for an ellipse, $\dfrac{a}{e} > a$, which confirms the directrix sits beyond the vertex, outside the curve.
How Is The Directrix of an Ellipse Different From a Parabola's?
The focus-directrix idea is shared across conics, but the count and formula differ. A parabola has one focus and one directrix of a parabola; an ellipse has two of each.
Conic | Eccentricity | Number of directrices | Directrix (standard form) |
|---|---|---|---|
Parabola | $e = 1$ | 1 | $x = -a$ (for $y^2 = 4ax$) |
Ellipse | $0 < e < 1$ | 2 | $x = \pm\dfrac{a}{e}$ |
Hyperbola | $e > 1$ | 2 | $x = \pm\dfrac{a}{e}$ |
The ellipse and hyperbola share the $\pm\dfrac{a}{e}$ form because both have two foci, but the ellipse keeps its directrices outside the curve (since $\tfrac{a}{e} > a$), while the value of $e$ separates the three conic sections entirely.
Examples of Directrix of Ellipse
Six examples, from reading off the directrices to recovering an ellipse from them.
Example 1
Find the directrices of $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.
Here $a^2 = 25$, $b^2 = 16$, so $a = 5$. Find $c$ and then $e$:
$c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = 3, \qquad e = \frac{c}{a} = \frac{3}{5}$
Directrices $x = \pm\dfrac{a}{e} = \pm\dfrac{5}{3/5} = \pm\dfrac{25}{3}$.
Final answer: $x = \dfrac{25}{3}$ and $x = -\dfrac{25}{3}$.
Example 2
A student finds the directrix of $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ by writing "$x = \pm ae = \pm 5 \times \tfrac{3}{5} = \pm 3$." Where does this go wrong?
The tempting move mixes up two formulas: $ae$ is the location of the focus, and it gets used where the directrix belongs.
The result $x = \pm 3$ is actually the foci $(\pm 3, 0)$, which lie inside the ellipse — but a directrix must lie outside it. The directrix formula divides by $e$; it does not multiply.
The correct method uses $x = \dfrac{a}{e}$:
$x = \pm\frac{a}{e} = \pm\frac{5}{3/5} = \pm\frac{25}{3} \approx \pm 8.33$
Final answer: $x = \pm\dfrac{25}{3}$. Foci use $ae$ (inside); directrices use $\dfrac{a}{e}$ (outside).
Example 3
An ellipse has $a = 6$ and eccentricity $e = \tfrac{1}{2}$. Find its directrices.
Apply the formula directly.
$x = \pm\frac{a}{e} = \pm\frac{6}{1/2} = \pm 12$
Final answer: $x = 12$ and $x = -12$. Both lie well beyond the vertices at $(\pm 6, 0)$.
Example 4
Find the directrices of $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$.
Here the larger denominator is under $y^2$, so the major axis is vertical: $a^2 = 25$, $b^2 = 16$, $a = 5$.
$c = \sqrt{25 - 16} = 3, \qquad e = \frac{3}{5}$
Since the major axis is vertical, the directrices are horizontal:
$y = \pm\frac{a}{e} = \pm\frac{25}{3}$
Final answer: $y = \pm\dfrac{25}{3}$. Orientation decides whether the directrix is $x = \dots$ or $y = \dots$.
Example 5
An ellipse has directrices $x = \pm 8$ and eccentricity $e = \tfrac{3}{4}$. Find $a$.
The directrix formula is $\dfrac{a}{e} = 8$.
$a = 8 \times e = 8 \times \frac{3}{4} = 6$
Final answer: $a = 6$. Working backward, the semi-major axis is the directrix value times the eccentricity.
Example 6
For the ellipse $\dfrac{x^2}{100} + \dfrac{y^2}{64} = 1$, verify the vertex satisfies the focus-directrix ratio.
First the constants: $a = 10$, $b = 8$, $c = \sqrt{100 - 64} = 6$, $e = \tfrac{6}{10} = \tfrac{3}{5}$. Right focus $(6, 0)$, right directrix $x = \dfrac{a}{e} = \dfrac{50}{3}$.
At the vertex $(10, 0)$: distance to focus $= 10 - 6 = 4$; distance to directrix $= \dfrac{50}{3} - 10 = \dfrac{20}{3}$.
$$\frac{PF}{PM} = \frac{4}{20/3} = \frac{4 \times 3}{20} = \frac{12}{20} = \frac{3}{5} = e \checkmark$$
Final answer: the ratio equals $e = \tfrac{3}{5}$, confirming the definition. The reasoning step is checking the ratio, not just the numbers.
Why the Directrix Matters: It Encodes Eccentricity
The directrix is not a decorative line; it is the geometric home of eccentricity, the single number that says how "stretched" an ellipse is. Its position packages that information into a distance.
It defines the ellipse. The focus-directrix ratio $\dfrac{PF}{PM} = e$ is an alternative definition of the ellipse, equivalent to the two-foci "sum of distances" definition but stated with one focus and one line.
It reads eccentricity as a location. Because $x = \dfrac{a}{e}$, a directrix far from the centre means small $e$ (nearly circular); a directrix close to the vertex means $e$ near 1 (very elongated).
It unifies the conics. The same ratio, with $e < 1$, $e = 1$, $e > 1$, produces ellipse, parabola, and hyperbola - the directrix is what makes that one continuous story possible.
What Are the Most Common Mistakes With the Directrix of an Ellipse?
Three errors account for most wrong answers, and two of them come from confusing the directrix with the focus.
Mistake 1: Using ae instead of a/e
Where it slips in: Recalling that "$ae$" appears in ellipse formulas and applying it to the directrix.
Don't do this: Writing the directrix as $x = \pm ae$ (that is the focus).
The correct way: The focus is at $x = \pm ae$ (inside the ellipse); the directrix is at $x = \pm\dfrac{a}{e}$ (outside it). Students first meeting conics mix the two because both use $a$ and $e$. A quick sanity check settles it: the directrix must be farther from the centre than the vertex, so it divides by $e$, giving a value greater than $a$.
Mistake 2: Forgetting the ellipse has two directrices
Where it slips in: Reporting a single directrix, as if the ellipse behaved like a parabola.
Don't do this: Giving only $x = \dfrac{a}{e}$ and stopping.
The correct way: An ellipse has two foci and therefore two directrices, $x = +\dfrac{a}{e}$ and $x = -\dfrac{a}{e}$. The memoriser who ports the parabola's single-directrix habit lands here. Always give both, with the $\pm$.
Mistake 3: Not checking which axis is the major axis
Where it slips in: Assuming the major axis is always horizontal, so the directrix is always $x = \dots$.
Don't do this: Writing $x = \pm\dfrac{a}{e}$ for $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$, whose major axis is vertical.
The correct way: The larger denominator sits under the major axis. If it is under $y^2$, the major axis is vertical and the directrices are horizontal, $y = \pm\dfrac{a}{e}$. The rusher who ignores orientation swaps the axis. Identify $a^2$ (the larger denominator) first.
Conclusion
The directrix of an ellipse is the line $x = \pm\dfrac{a}{e}$, perpendicular to the major axis and lying outside the curve.
An ellipse has two directrices, one per focus, tied together by the focus-directrix ratio $\dfrac{PF}{PM} = e < 1$.
The focus is at $x = \pm ae$ (inside); the directrix at $x = \pm\dfrac{a}{e}$ (outside) — never swap them.
The larger denominator marks the major axis and decides whether the directrix is an $x =$ or $y =$ line.
To take conic sections further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor, or math classes online.
Practice These to Solidify Your Understanding
Work through these, then check your answers:
Find the directrices of $\dfrac{x^2}{169} + \dfrac{y^2}{144} = 1$. (Answer to Question 1: $a = 13$, $c = 5$, $e = \tfrac{5}{13}$, so $x = \pm\tfrac{169}{5}$.)
An ellipse has $a = 8$, $e = \tfrac{1}{4}$. Find its directrices. (Answer to Question 2: $x = \pm 32$.)
An ellipse has directrices $y = \pm 20$ and $e = \tfrac{2}{5}$. Find $a$ and state the axis orientation. (Answer to Question 3: $a = 8$, vertical major axis.)
If Question 3 tripped you, revisit Example 4 on orientation. Want a trainer to walk conics through with your child? Book a free demo class.
Read More
Latus rectum of ellipse — the focal chord whose length is $\tfrac{2b^2}{a}$.
Vertex of ellipse — the endpoints of the major axis the directrix sits beyond.
Area of an ellipse — the $\pi a b$ measure of the space inside.
Eccentricity — the number $e$ that positions every directrix.
Latus rectum of parabola, ellipse, and hyperbola — the three conics compared side by side.
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