If a parabola is placed in a coordinate plane in a simple way, then a simple equation is obtained.
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: $$ \tag{1} \cssId{s36}{\sqrt{(x0)^2 + (yp)^2 } = \sqrt{x^2 + (yp)^2}} $$
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 $\,(x,p)\,$:
$$
\tag{2}
\cssId{s40}{\sqrt{(xx)^2 + ((y(p))^2}
=
\sqrt{(y+p)^2}}
$$
From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: $$ \cssId{s42}{\sqrt{x^2 + (yp)^2} = \sqrt{(y+p)^2}} $$
This equation simplifies considerably, as follows:
Squaring both sides:  $ x^2 + (yp)^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 $\,20$ 
In summary, we have:
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 $\displaystyle \,(0,\frac{1}{4a})\,$.
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:
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:  $y=a(xh)^2 + k$ 
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:  $y=p+k$ 
In this example, I illustrate the approach that I usually take when asked to give
complete information about the parabola $\,y = a(xh)^2 + k\,$.
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?!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
