Equations of Simple Parabolas

You may want to review the prior web exercise, Parabolas.

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

Derivation: parabola with vertex at origin, focus on $y$-axis
Equations of simple parabolas
Memory Device: ‘one over four pee pairs’
Shifting the parabola
Graphing a shifted parabola

Derivation

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

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

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 above shows the situation where $\,p\gt 0\,,$ the following derivation also holds for $\,p \lt 0\,.$

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)
AND
$p\lt 0$	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 $\,y = \frac1{4p}x^2\,,$ we see that $\, 5 = \frac{1}{4p}\,.$ 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 $\,(0,\frac{1}{20})\,.$ 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 $\,a\gt 0\,,$ 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 $\,a = \frac{1}{4p}\,.$ Solving for $\,p\,$ gives $\, p = \frac{1}{4a}\,,$ and thus the coordinates of the focus are $\,(0,\frac{1}{4a})\,.$

Memory Device: ‘one over four pee pairs’

Notice that if $\,a = \frac{1}{4p}\,,$ then $\,p = \frac{1}{4a}\,.$ Or, if $\,p = \frac{1}{4a}\,,$ then $\,a = \frac{1}{4p}\,.$ 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:

$4\times 5 = 20\,$ (multiply by $\,4\,$)
$p=\frac{1}{20}$ (flip)

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

$4\times \frac{1}{20} = \frac{4}{20} = \frac{1}{5}\,$ (multiply by $\,4\,$)
$a = (\text{the reciprocal of } \frac{1}{5}) = 5\,$ (flip)

Read-Through, Part 2

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: $y=ax^2$

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 $\,y = a(x-h)^2 + k\,.$

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

Solution:

Concave up or down? Since $\,a=-3\lt 0\,,$ the parabola is concave down (sheds water). Since the focus is always inside a parabola, we know the focus is under the vertex.
Find the vertex: The vertex of $\,y = a(x-h)^2 + k\,$ is the point $\,(h,k)\,$:
$$ \cssId{sb34}{y = a( \overset{\text{What value of } \ x\ }{\overset{\text{ makes this zero?}}{\overbrace{ \underset{\text{answer: } \ h}{\underbrace{x-h}}}}})^2\ \ \ + \overset{y\text{-value of the vertex}}{\overbrace{\ \ k\ \ }}} $$
For us, what makes $\,x+5\,$ equal to zero? Answer: $\,-5\,$

Thus, $\,-5\,$ is the $x$-value of the vertex. When $\,x = -5\,,$ the corresponding $y$-value is $\,1\,.$ Thus, the vertex is $\,(-5,1)\,.$
Find the distance from the vertex to the focus: Using the ‘one over four pee pair’ memory device:
$$\cssId{sb42}{p = \frac{1}{4a} = \frac{1}{4(-3)} = -\frac{1}{12}}$$
Thus, the distance from the focus to the vertex is:
$$\cssId{sb44}{|p| = \left|-\frac1{12}\right| = \frac{1}{12}}$$
Find the focus: We already determined that the focus lies below the vertex. So, the focus is:
$$\cssId{sb48}{(-5,1-\frac{1}{12}) =(-5,\frac{11}{12})}$$
Find the equation of the directrix: Every horizontal line has an equation of the form:
$$\cssId{sb51}{y = \text{(some #)}}$$
In our example, the focus lies $\,\frac{1}{12}\,$ below the vertex, so the directrix lies $\,\frac{1}{12}\,$ above the vertex. Thus, the equation of the directrix is $\,y=1+\frac{1}{12}\,,$ that is, $\,y=\frac{13}{12}\,.$
Plot an additional point: Plotting an additional point gives a sense of the ‘width’ of the parabola. For example, when $\,x = -4\,$ we have:

$$ \begin{align} y &= -3(-4+5)^2 + 1\cr &= -3(1) + 1\cr &= -2 \end{align} $$

In this parabola, the vertex is close to the focus, so the parabola is narrow.

graph of the parabola y = -3(x+5)^2 + 1 $\,y=-3(x+5)^2+1\,$

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