Find a Parabola’s Focus, Vertex, and Directrix: Formulas & Examples

To find a parabola’s vertex, focus, and directrix, rewrite its equation in a standard form with $4p$. The vertex is $(h,k)$. For a vertical parabola, the focus is $(h,k+p)$ and the directrix is $y=k-p$; for a horizontal parabola, the focus is $(h+p,k)$ and the directrix is $x=h-p$.

Quick answer: Write the equation as (x − h)² = 4p(y − k). The vertex is (h, k), the focus is (h, k + p), and the directrix is the line y = k − p. If the equation is y = ax² + bx + c, then p = 1/(4a). Example: x² = 8y means 4p = 8, so p = 2. Focus (0, 2), directrix y = −2.

Find a Parabola’s Focus, Vertex, and Directrix: Formulas & Examples

What the vertex, focus, and directrix mean

A parabola is the set of points equally far from a fixed point, the focus, and a fixed line, the directrix. Distance to a line means the perpendicular distance. The focus does not lie on the directrix.

The axis of symmetry passes through the vertex and focus and is perpendicular to the directrix. The vertex lies halfway between the focus and the directrix along that axis. A parabola has one focus and one vertex; unlike a circle or ellipse, it has no center of symmetry.

Here $p=1$. The focus is 1 unit above the vertex, and the directrix is 1 unit below it.

Focus and directrix formulas: match the standard form

These formulas apply to parabolas whose axes are vertical or horizontal. In both forms, $p\ne0$, and the distance from the vertex to the focus is $|p|$. Keep the sign of $p$ when locating the focus and directrix.

Vertical parabola: opens up or down

\[(x-h)^2=4p(y-k)\]

Vertex: $(h,k)$
Focus: $(h,k+p)$
Directrix: $y=k-p$
Axis: $x=h$

If $p>0$, it opens up. If $p<0$, it opens down. The squared expression contains $x$, but the focus moves in the $y$-direction.

Horizontal parabola: opens right or left

\[(y-k)^2=4p(x-h)\]

Vertex: $(h,k)$
Focus: $(h+p,k)$
Directrix: $x=h-p$
Axis: $y=k$

If $p>0$, it opens right. If $p<0$, it opens left. Notice the expression $y-k$: for $(y+1)^2$, use $k=-1$.

Find $p$, not just $4p$. If the coefficient is 8, then $4p=8$, so $p=2$. If the equation is $y=a(x-h)^2+k$, first divide by $a$ to obtain the standard form. This gives $p=\frac{1}{4a}$, not $\frac a4$.

Worked examples: vertical and horizontal parabolas

Example 1: find all three features of $y=3x^2+6x$

1. Complete the square to locate the vertex.

[begin{aligned}y&=3(x^2+2x)\&=3big((x+1)^2-1big)\&=3(x+1)^2-3.end{aligned}]

The vertex is $(-1,-3)$. Check it by substituting $x=-1$ into the original equation: $y=3-6=-3$.

2. Rewrite with the squared expression alone.

\[(x+1)^2=\frac13(y+3)\]

Match $(x-h)^2=4p(y-k)$: $h=-1$, $k=-3$, and $4p=\frac13$, so $p=\frac1{12}$.

3. Move along the vertical axis. Add $p$ to the vertex’s $y$-coordinate for the focus; subtract it for the directrix.

Vertex: $(-1,-3)$
Focus: $\left(-1,-\frac{35}{12}\right)$
Directrix: $y=-\frac{37}{12}$
Axis: $x=-1$; the parabola opens up.

Example 2: $(y+1)^2=-8(x-3)$

Here the $y$-expression is squared, so the axis is horizontal. Read $h=3$, $k=-1$, and $4p=-8$, giving $p=-2$.

Vertex: $(3,-1)$
Focus: $(h+p,k)=(1,-1)$
Directrix: $x=h-p=5$
Axis: $y=-1$; the parabola opens left.

The focus is 2 units left of the vertex, while the directrix is 2 units right of it. This checks both the sign and the opening direction.

What if the equation is $y=ax^2+bx+c$?

For $a\ne0$, calculate $h=-\frac b{2a}$, then substitute into the original equation to get $k=ah^2+bh+c$. Use $p=\frac1{4a}$. The vertex is $(h,k)$, the focus is $(h,k+p)$, and the directrix is $y=k-p$. Completing the square gives the same result and makes the signs easier to check.

Practice: find the vertex, focus, and directrix

Try each equation before opening its answer. Identify the axis and the signed value of $p$ as part of your work.

$y=3(x-5)^2+2$
$y=4x^2+x-1$
$y=x^2+2x+3$
$x=y^2-4y$

Show answers and standard forms

1. $(x-5)^2=\frac13(y-2)$, so $p=\frac1{12}$. Vertex $(5,2)$; focus $\left(5,\frac{25}{12}\right)$; directrix $y=\frac{23}{12}$.

2. $y=4\left(x+\frac18\right)^2-\frac{17}{16}$, so $p=\frac1{16}$. Vertex $\left(-\frac18,-\frac{17}{16}\right)$; focus $\left(-\frac18,-1\right)$; directrix $y=-\frac98$.

3. $(x+1)^2=y-2$, so $p=\frac14$. Vertex $(-1,2)$; focus $\left(-1,\frac94\right)$; directrix $y=\frac74$.

4. $(y-2)^2=x+4$, so $p=\frac14$. Vertex $(-4,2)$; focus $\left(-\frac{15}4,2\right)$; directrix $x=-\frac{17}4$.

Check the equation before using a parabola formula. An equation such as $(y-2)^2=3(x-5)^2$ describes two intersecting lines, not a parabola. A standard vertical or horizontal parabola has one squared variable and one linear variable.

Continue practicing parabolas

Once you can read the features from an equation, work backward with writing the equation of a parabola. Use the same vertex and signed focal distance to choose the correct standard form.

For broader course review, view Pre-Calculus for Beginners and inspect its available preview before choosing a book.

Pre-Calculus for Beginners 2027 The Ultimate Step by Step Guide to Acing Precalculus

Formula reference: OpenStax Precalculus, The Parabola. The worked examples and practice explanations above were checked using the standard forms.

Need more practice? Pre-Calculus for Beginners 2027 covers Pre-Calculus step by step, with practice problems and answers. Look inside · $17.99

Focus and Directrix Formulas at a Glance

The focus is a point inside the curve, and the directrix is a line outside it. Every point on the parabola is the same distance from both. The number p is the distance from the vertex to the focus, with a sign that shows the direction.

Equation	Opens	Vertex	Focus	Directrix
(x − h)² = 4p(y − k), p > 0	Up	(h, k)	(h, k + p)	y = k − p
(x − h)² = 4p(y − k), p < 0	Down	(h, k)	(h, k + p)	y = k − p
(y − k)² = 4p(x − h), p > 0	Right	(h, k)	(h + p, k)	x = h − p
(y − k)² = 4p(x − h), p < 0	Left	(h, k)	(h + p, k)	x = h − p
y = a(x − h)² + k	Up if a > 0, down if a < 0	(h, k)	(h, k + 1/(4a))	y = k − 1/(4a)

The squared variable tells you the direction: x² means the parabola opens up or down, and y² means it opens left or right.

Find the Equation from the Focus and Directrix

You can also work backward. Suppose the focus is (2, 5) and the directrix is y = 1.

The vertex is halfway between the focus and the directrix. Halfway between y = 5 and y = 1 is y = 3, so the vertex is (2, 3).
p is the distance from the vertex to the focus: 5 − 3 = 2. The focus is above the vertex, so p is positive and the parabola opens up.
Substitute into (x − h)² = 4p(y − k): (x − 2)² = 8(y − 3).

Check with a point. If y = 5, then (x − 2)² = 16, so x = 6. The point (6, 5) is 4 units from the focus (2, 5) and 5 − 1 = 4 units from the directrix. The distances match.

For a vertical directrix like x = −1 with focus (3, −1), the same steps give vertex (1, −1), p = 2, and (y + 1)² = 8(x − 1).

Frequently Asked Questions

How do you find the focus and directrix of a parabola?

Rewrite the equation in a standard form with 4p, then read off h, k and p. For a vertical parabola (x − h)² = 4p(y − k), the focus is (h, k + p) and the directrix is y = k − p. For (x − 2)² = 12(y + 1), 4p = 12 so p = 3, the vertex is (2, −1), the focus is (2, 2) and the directrix is y = −4.

What is the focus of the parabola y² = 4x?

The focus of y² = 4x is (1, 0). This parabola opens to the right, so compare it with y² = 4px: 4p = 4 gives p = 1, the vertex is (0, 0) and the directrix is x = −1. The same method works for y² = 8x, where p = 2 and the focus is (2, 0), and for y² = 12x, where the focus is (3, 0).

What is the focus of the parabola y = 4x²?

The focus of y = 4x² is (0, 1/16). Rewrite it as x² = (1/4)y, so 4p = 1/4 and p = 1/16. The vertex is (0, 0), and because a = 4 is positive, the parabola opens up with the directrix at y = −1/16. A large a makes a narrow parabola with the focus very close to the vertex.

What is the focus of the parabola x² = 16y?

The focus of x² = 16y is (0, 4). Match it to x² = 4py: 4p = 16, so p = 4. The vertex is the origin, the parabola opens upward, and the directrix is the line y = −4. Every point on the curve is the same distance from (0, 4) as from that line; for example, (8, 4) is 8 units from both.

Is the focus of a parabola the vertex?

No. The vertex lies on the parabola, but the focus is a point inside the curve, on the axis of symmetry, |p| units away from the vertex. The vertex sits exactly halfway between the focus and the directrix. For x² = 4y, the vertex is (0, 0), the focus is (0, 1) and the directrix is y = −1.

How do you find the focus of a parabola in vertex form?

For y = a(x − h)² + k, the vertex is (h, k) and p = 1/(4a), so the focus is (h, k + 1/(4a)). Take y = 2(x − 1)² + 3: p = 1/8, giving the focus (1, 25/8) and the directrix y = 23/8. If a is negative, p is negative too, and the focus lies below the vertex.

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Find a Parabola’s Focus, Vertex, and Directrix: Formulas & Examples

What the vertex, focus, and directrix mean