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 \(x=h-p\).
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.
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.
\[\begin{aligned}y&=3(x^2+2x)\\&=3\big((x+1)^2-1\big)\\&=3(x+1)^2-3.\end{aligned}\]
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.
Formula reference: OpenStax Precalculus, The Parabola. The worked examples and practice explanations above were checked using the standard forms.
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