Parabola Equation Calculator – Standard, Vertex & Focus Forms
A parabola is the set of points whose distance to a fixed point (the focus)
equals its distance to a fixed line (the directrix).
This tool focuses on axis-aligned parabolas (no rotation / no \(xy\) term).
1) Focus-directrix (standard) forms
These are the most geometry-friendly forms because the vertex, focus, directrix, and opening direction
are all visible immediately.
-
Vertical axis (opens up or down):
\[
(x-h)^2 = 4p(y-k)
\]
Vertex \(V=(h,k)\), focus \(F=(h,k+p)\), directrix \(y=k-p\), axis of symmetry \(x=h\).
-
Horizontal axis (opens left or right):
\[
(y-k)^2 = 4p(x-h)
\]
Vertex \(V=(h,k)\), focus \(F=(h+p,k)\), directrix \(x=h-p\), axis of symmetry \(y=k\).
The parameter \(p\) is a signed distance from the vertex to the focus.
• Vertical: \(p>0\) opens upward, \(p<0\) opens downward.
• Horizontal: \(p>0\) opens right, \(p<0\) opens left.
A useful geometric fact is the latus rectum length:
\[
\text{latus rectum length} = 4|p|
\]
2) Vertex (function) forms
If you divide the standard form by \(4p\), you get the common “vertex form” used in algebra.
In vertex form, the sign of \(a\) tells the opening direction:
• Vertical: \(a>0\) opens up, \(a<0\) opens down.
• Horizontal: \(a>0\) opens right, \(a<0\) opens left.
3) Quadratic polynomial → vertex, focus, directrix
Many problems start from a quadratic polynomial. For an axis-aligned parabola, you can convert it to vertex form
by completing the square.
3a) Vertical parabola: \(y=ax^2+bx+c\)
Step 1. Factor out \(a\) from the quadratic terms.
\[
\begin{aligned}
y &= ax^2 + bx + c \\
&= a\left(x^2 + \frac{b}{a}x\right) + c
\end{aligned}
\]
Step 2. Complete the square inside the parentheses.
\[
\begin{aligned}
x^2 + \frac{b}{a}x
&= \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2
\end{aligned}
\]
Step 3. Substitute back to get vertex form.
\[
\begin{aligned}
y
&= a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c \\
&= a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)
\end{aligned}
\]
From this, the vertex form \(y=a(x-h)^2+k\) is obtained with:
\[
\begin{aligned}
h &= -\frac{b}{2a} \\
k &= c - \frac{b^2}{4a}
\end{aligned}
\]
To connect with focus-directrix form, compare \(y-k=a(x-h)^2\) with \((x-h)^2=4p(y-k)\):
\[
\begin{aligned}
y-k &= a(x-h)^2 \\
(x-h)^2 &= \frac{1}{a}(y-k) \\
4p &= \frac{1}{a} \\
p &= \frac{1}{4a}
\end{aligned}
\]
So for \(y=ax^2+bx+c\) (with \(a\neq 0\)):
• Vertex: \(V=(h,k)\)
• Focus: \(F=(h,k+p)\)
• Directrix: \(y=k-p\)
• Axis: \(x=h\)
3b) Horizontal parabola: \(x=ay^2+by+c\)
The same idea applies, but now you complete the square in \(y\).
\[
\begin{aligned}
x &= ay^2 + by + c \\
&= a\left(y^2 + \frac{b}{a}y\right) + c \\
&= a\left(y + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)
\end{aligned}
\]
Thus the vertex parameters are:
\[
\begin{aligned}
k &= -\frac{b}{2a} \\
h &= c - \frac{b^2}{4a}
\end{aligned}
\]
Comparing with \(x = a(y-k)^2 + h\) and \((y-k)^2 = 4p(x-h)\) again gives:
\[
p = \frac{1}{4a}
\]
So for \(x=ay^2+by+c\) (with \(a\neq 0\)):
• Vertex: \(V=(h,k)\)
• Focus: \(F=(h+p,k)\)
• Directrix: \(x=h-p\)
• Axis: \(y=k\)
4) Expanding to a “general” (polynomial) form
Sometimes you want a form with expanded powers. Starting from vertical vertex form:
\[
\begin{aligned}
y &= a(x-h)^2 + k \\
&= a(x^2 - 2hx + h^2) + k \\
&= ax^2 + (-2ah)x + (ah^2 + k)
\end{aligned}
\]
This shows how the coefficients relate when you expand.
A similar expansion holds for the horizontal form \(x=a(y-k)^2+h\).
5) Quick interpretation checklist
-
Axis-aligned only: if an \(xy\) term appears, the parabola is rotated and needs a different method.
-
Opening direction: determined by the sign of \(p\) (or the sign of \(a\) in vertex form).
-
“Wider vs narrower”: smaller \(|a|\) means a wider parabola; larger \(|a|\) means a narrower parabola.
6) Worked example
Example: \(y=\dfrac{x^2}{4}\). Rewrite it as \(y = \dfrac{1}{4}x^2\), so \(a=\dfrac{1}{4}\).
Step 1. Identify \(p\) using \(p=\dfrac{1}{4a}\).
\[
\begin{aligned}
p &= \frac{1}{4a} \\
&= \frac{1}{4\cdot \left(\frac{1}{4}\right)} \\
&= 1
\end{aligned}
\]
Step 2. Here \(h=0\), \(k=0\), so the standard form is:
\[
\begin{aligned}
(x-0)^2 &= 4\cdot 1\,(y-0) \\
x^2 &= 4y
\end{aligned}
\]
Step 3. Focus and directrix:
\[
\begin{aligned}
F &= (0,0+p) = (0,1) \\
\text{directrix: } y &= 0-p = -1
\end{aligned}
\]
Summary: Vertex \((0,0)\), focus \((0,1)\), directrix \(y=-1\), opens upward.
