EQUATIONS OF SIMPLE PARABOLAS

LESSON READ-THROUGH (Part 1 of 2)
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.

Derivation

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)
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 $\displaystyle \,y = \frac1{4p}x^2\,$, 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:

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 $\displaystyle\,a = \frac{1}{4p}\,$.
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:


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:


$\,y=-3(x+5)^2+1\,$

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.
PROBLEM TYPES:
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AVAILABLE MASTERED IN PROGRESS

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