EQUATIONS OF SIMPLE PARABOLAS

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

Place a parabola with its vertex at the origin.

  • If we put the focus on the $\,y$-axis,
    the 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 [beautiful math coming... please be patient] $\,p \ne 0\,$ denote the $\,y$-value of the focus.
Thus, the focus has coordinates $\,(0,p)\,$.

Although the sketch at right shows the situation where [beautiful math coming... please be patient] $\,p\gt 0\,$,
the following derivation also holds for $\,p \lt 0\,$.

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

Notice that:

[beautiful math coming... please be patient] $p\gt 0$ if and only if the focus is above the vertex if and only if the parabola is concave up (holds water)
AND
[beautiful math coming... please be patient] $p\lt 0$ 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: [beautiful math coming... please be patient] $$ \tag{1} \sqrt{(x-0)^2 + (y-p)^2 } = \sqrt{x^2 + (y-p)^2} $$

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 $\,(x,-p)\,$: [beautiful math coming... please be patient] $$ \tag{2} \sqrt{(x-x)^2 + ((y-(-p))^2} = \sqrt{(y+p)^2} $$

From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: [beautiful math coming... please be patient] $$ \sqrt{x^2 + (y-p)^2} = \sqrt{(y+p)^2} $$

This equation simplifies considerably, as follows:

Squaring both sides: [beautiful math coming... please be patient] $ x^2 + (y-p)^2 = (y+p)^2 $
Multiplying out: [beautiful math coming... please be patient] $ 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: [beautiful math coming... please be patient] $ x^2 = 4py $
Dividing by $\,4p\,$ and rearranging: [beautiful math coming... please be patient] $ \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:

[beautiful math coming... please be patient] $ \begin{alignat}{2} 5\ &= \frac{1}{4p} &\qquad&\text{original equation} \cr 20p\ &= 1 &&\text{multiply both sides by } 4p\cr p\ &= \frac{1}{20} &&\text{divide both sides by } 20 \end{alignat} $

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:

EQUATIONS OF SIMPLE PARABOLAS

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 [beautiful math coming... please be patient] $\,a\gt 0\,$, then the parabola is concave up (holds water).
If $\,a\lt 0\,$, the 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 [beautiful math coming... please be patient] $\displaystyle \,(0,\frac{1}{4a})\,$.

MEMORY DEVICE: ‘one over four pee pairs’

Notice that if [beautiful math coming... please be patient] $\displaystyle\,a = \frac{1}{4p}\,$, 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:

SHIFTING THE PARABOLA

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: [beautiful math coming... please be patient] $y=ax^2$ shifted equation: $y=a(x-h)^2 + k$
original vertex: $(0,0)$ new vertex: $(h,k)$
original focus: [beautiful math coming... please be patient] $\displaystyle (0,p) = (0,\frac{1}{4a})$ 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 $\,y = a(x-h)^2 + k\,$.

Question:
Completely describe the graph of the equation [beautiful math coming... please be patient] $\,y = -3(x+5)^2 + 1\,$.
Solution:


[beautiful math coming... please be patient] $\,y=-3(x+5)^2+1\,$

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