Equations of Simple Parabolas (Part 1)

(This page is Part 1. Click here for Part 2.)

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.

• 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:

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 $\,(x,-p)\,$:

From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal:

This equation simplifies considerably, as follows:

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:

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:

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)