by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

If a parabola is placed in a coordinate plane in a simple way, then a simple equation is obtained.


Place a parabola with its vertex at the origin, as shown at right.

  • If we put the focus on the $\,y$-axis,
    then the directrix will be parallel to the $\,x$-axis.
  • Or, if we put the directrix parallel to the $\,x$-axis,
    then the focus will be on the $\,y$-axis.
In either case, let $\,p \ne 0\,$ denote the $\,y$-value of the focus.
Thus, the focus has coordinates $\,(0,p)\,$.

Although the sketch at right shows the situation where $\,p\gt 0\,$,
the following derivation also holds for $\,p \lt 0\,$.

parabola with vertex at origin, directrix parallel to x-axis

Notice that:

$p\gt 0$ if and only if the focus is above the vertex if and only if the parabola is concave up (holds water)
$p\lt 0$ if and only if the focus is below the vertex if and only if the parabola is concave down (sheds water)

For example, if the focus is above the vertex, then $\,p\,$ must be greater than zero, and the parabola must be concave up.
As a second example, if the parabola is concave down, then $\,p\,$ must be less than zero, and the focus is below the vertex.

Since the vertex is a point on the parabola, the definition of parabola dictates that it must be the same distance from the focus and the directrix.

Thus, the directrix must cross the $\,y\,$-axis at $\,-p\,$; indeed, every $\,y$-value on the directrix equals $\,-p\,$.

Let $\,(x,y)\,$ denote a typical point on the parabola.

The distance from $\,(x,y)\,$ to the focus $\,(0,p)\,$ is found using the distance formula: $$ \tag{1} \cssId{s36}{\sqrt{(x-0)^2 + (y-p)^2 } = \sqrt{x^2 + (y-p)^2}} $$

To find the distance from $\,(x,y)\,$ to the directrix, first drop a perpendicular from $\,(x,y)\,$ to the directrix.
This perpendicular intersects the directrix at $\,(x,-p)\,$.
The distance from $\,(x,y)\,$ to the directrix is therefore the distance from $\,(x,y)\,$ to $\,(x,-p)\,$: $$ \tag{2} \cssId{s40}{\sqrt{(x-x)^2 + ((y-(-p))^2} = \sqrt{(y+p)^2}} $$

From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: $$ \cssId{s42}{\sqrt{x^2 + (y-p)^2} = \sqrt{(y+p)^2}} $$

This equation simplifies considerably, as follows:

Squaring both sides: $ x^2 + (y-p)^2 = (y+p)^2 $
Multiplying out: $ x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2 $
Subtracting $\,y^2 + p^2\,$ from both sides: $ x^2 - 2py = 2py $
Adding $\,2py\,$ to both sides: $ x^2 = 4py $
Dividing by $\,4p\,$ and rearranging: $ \displaystyle y = \frac{1}{4p} x^2 $

Such a beautiful, simple description for our parabola!
The most critical thing to notice is the coefficient of $\,x^2\,$, since it holds the key to locating the focus of the parabola.

As an example, consider the equation $\,y = 5x^2\,$.
Comparing $\,y = 5x^2\,$ with $\displaystyle \,y = \frac1{4p}x^2\,$, we see that $\displaystyle 5 = \frac{1}{4p}\,$.
Solving for $\,p\,$ gives:

$\displaystyle 5 = \frac{1}{4p}$ original equation
$20p = 1$ multiply both sides by $\,4p$
$\displaystyle p = \frac{1}{20}$ divide both sides by $\,20$

Thus, $\,y = 5x^2\,$ graphs as a parabola with vertex at the origin, and focus $\displaystyle\,(0,\frac{1}{20})\,$.
So easy!

In summary, we have:


Every equation of the form $\,y= ax^2\,$ (for $\,a\ne0\,$) is a parabola
with vertex at the origin, directrix parallel to the $\,x\,$-axis, and focus on the $\,y\,$-axis.

If $\,a\gt 0\,$, then the parabola is concave up (holds water).
If $\,a\lt 0\,$, then the parabola is concave down (sheds water).

If $\,p\,$ denotes the $\,y$-value of the focus, then $\displaystyle\,a = \frac{1}{4p}\,$.
Solving for $\,p\,$ gives $\,\displaystyle p = \frac{1}{4a}\,$, and thus the coordinates of the focus are $\displaystyle \,(0,\frac{1}{4a})\,$.

MEMORY DEVICE: ‘one over four pee pairs’

Notice that if $\displaystyle\,a = \frac{1}{4p}\,$, then $\displaystyle\,p = \frac{1}{4a}\,$.   Or, if $\displaystyle\,p = \frac{1}{4a}\,$, then $\displaystyle\,a = \frac{1}{4p}\,$.   What a beautiful symmetric relationship!

Thus, I fondly refer to $\,a\,$ and $\,p\,$ with the catchy phrase:   ‘one over four pee pairs’. (Try to say this quickly ten times in a row!)
If you know either $\,a\,$ or $\,p\,$, then it's easy to find the other—just multiply by $\,4\,$ and then flip (take the reciprocal).

For example, if you know that $\,a = 5\,$, then $\,p\,$ is found as follows:

Or, if you know that $\displaystyle\,p = \frac{1}{20}\,$, then $\,a\,$ is found as follows:


Now, here's some very good news.
By using graphical transformations, knowledge of this one simple equation $\,y = ax^2\,$
actually gives full understanding of all parabolas with directrix parallel to the $\,x$-axis!

The results are summarized next:

Shift the parabola (together with its focus and directrix) horizontally by $\,h\,$, and vertically by $\,k\,$.
This yields the following information:

original equation: $y=ax^2$ shifted equation: $y=a(x-h)^2 + k$
original vertex: $(0,0)$ new vertex: $(h,k)$
original focus: $\displaystyle (0,p) = (0,\frac{1}{4a})$ new focus: $\displaystyle (h,p+k)= (h,\frac{1}{4a}+k)$
original directrix: $y=-p$ new directrix: $y=-p+k$
EXAMPLE graphing a shifted parabola

In this example, I illustrate the approach that I usually take when asked to give
complete information about the parabola $\,y = a(x-h)^2 + k\,$.

Completely describe the graph of the equation $\,y = -3(x+5)^2 + 1\,$.


For fun, zip up to WolframAlpha and type in any of the following:

vertex of y = -3(x+5)^2 + 1

focus of y = -3(x+5)^2 + 1

directrix of y = -3(x+5)^2 + 1

How easy is that?!

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Quadratic Functions and
the Completing the Square Technique

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3 4 5 6 7 8 9 10 11 12

(MAX is 12; there are 12 different problem types.)