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.
Thus, the focus has coordinates $\,(0,p)\,$.
Although the sketch at right shows the situation where $\,p\gt 0\,$, |
|
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 $\displaystyle \,y = \frac1{4p}x^2\,$,
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:
- $\,4\times 5 = 20\,$ (multiply by $\,4\,$)
- $\displaystyle p=\frac{1}{20}$ (flip)
- $\displaystyle\,4\times \frac{1}{20} = \frac{4}{20} = \frac{1}{5}\,$ (multiply by $\,4\,$)
- $\displaystyle a = (\text{the reciprocal of } \frac{1}{5}) = 5\,$ (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 $\,y = -3(x+5)^2 + 1\,$.
Completely describe the graph of the equation $\,y = -3(x+5)^2 + 1\,$.
Solution:
$\,y=-3(x+5)^2+1\,$
-
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 $\,(h,k)\,$:
$$ \cssId{sb34}{y = a( \overset{\text{What value of } \ x\ }{\overset{\text{ makes this zero?}}{\overbrace{ \underset{\text{Answer: } \ h}{\underbrace{x-h}}}}})^2\ \ \ + \overset{\text{this is the } y\text{-value of the vertex}}{\overbrace{\ \ k\ \ }}} $$ 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,
$\displaystyle\,p = \frac{1}{4a} = \frac{1}{4(-3)} = -\frac{1}{12}\,$.Thus, the distance from the focus to the vertex is $\displaystyle\,|p| = \left|-\frac1{12}\right| = \frac{1}{12}\,$. -
Find the focus:
We already determined that the focus lies below the vertex.
So, the focus is:$\displaystyle\,(-5,1-\frac{1}{12}) =(-5,\frac{11}{12})\,$ -
Find the equation of the directrix:
Every horizontal line has an equation of the form: $\,y = \text{(some #)}\,$
In our example, the focus lies $\displaystyle\,\frac{1}{12}\,$ below the vertex, so the directrix lies $\displaystyle\,\frac{1}{12}\,$ above the vertex.
Thus, the equation of the directrix is $\displaystyle\,y=1+\frac{1}{12}\,$, that is, $\displaystyle\,y=\frac{13}{12}\,$. -
Plot an additional point:
Plotting an additional point gives a sense of the ‘width’ of the parabola.
For example, when $\,x = -4\,$ we have:$\,y = -3(-4+5)^2 + 1 = -3(1) + 1 = -2\,$In this parabola, the vertex is close to the focus, so the parabola is narrow.
$\,y=-3(x+5)^2+1\,$
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
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