The prior section,
Parabolas: Definition, Reflectors/Collectors, Derivation of Equations,
culminated with going from a given equation of a parabola to its graph.
This section covers the reverse direction:
going from a graph (or information describing the graph) of a parabola
to the equation of the parabola.
This section assumes knowledge of parabolas in standard form, defined as:
When you're given information describing a parabola, you should always
start by asking yourself:
Does the information uniquely describe a parabola?
In other words, is there only one parabola that satisfies the given information?
Here are a couple scenarios that do not uniquely define a parabola:
Find ‘the’ parabola with vertex at the origin
and directrix parallel to the $x$axis.
This is a faulty question; it does not describe a unique parabola, as shown at right. Indeed, there are infinitely many parabolas with vertex at the origin and directrix parallel to the $x$axis. 

Find ‘the’ parabola in standard form passing
through the point $\,(1,1)\,.$
This is a faulty question; it does not describe a unique parabola, as shown at right. There are two parabolas matching the given information: $\,y = x^2\,$ and $\,x = y^2\,.$ 

The
general conic equation has six unknowns: $A\,,$ $B\,,$ $C\,,$ $D\,,$ $E\,,$ and $F\,$:
$$
\cssId{s32}{Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0} \qquad
\cssId{s33}{(*)}
$$
One might initially think that six points are needed to solve for these six unknowns.
However, only five points are actually needed, as follows:
For example, consider any five points on the $x$axis. Any equation of the form $\,Bxy + Cy^2 + Ey = 0\,$ contains the $x$axis ($\,y = 0\,$) as part of its solution set. (See an example at right.) Thus, there is not a unique ‘conic’ through these points. 
Suppose you are given five points $\,(x_1,y_1)\,$ through ($\,x_5,y_5)\,,$ no three of which are collinear.
It is highly algebraically intensive to solve (*) for the coefficients $\,A\,$ through $\,F\,.$
(Try it!
I suspect you'll throw up your arms in defeat after several pages of calculations!)
If you did complete the calculation, however,
you'd find that one of the
coefficients can be any real number
—then, the other five are determined by it.
What does this mean?
What's going on here?
To understand, let's look at a simpler and
familiar analogous situation:
At right, the blue parabola ($\,y = x^2  4\,$) and the green parabola ($\,x = y^2  4\,$) intersect in the four points shown. An additional point (making a fifth point) is needed to uniquely specify the parabola. This illustrates that four points are not enough to uniquely define a general parabola. Under certain circumstances, however, we can get away with fewer points. Keep reading! 

If a parabola is known to be in standard form ($\,x^2 = 4py\,$ or $\,y^2 = 4px\,$),
then only one point is needed to determine the single unknown $\,p\,.$
Example:
Find the equation of the parabola with vertex at the origin, directrix parallel
to the $y$axis,
passing through the point $\,(1,3)\,.$
Find the focus and directrix of the parabola.
Solution: A quick sketch shows that the desired parabola has all $x$values positive. Therefore, the parabola must be of the form $\,y^2 = 4px\,.$ Why? Since $\,y^2\,$ is always nonnegative, $\,4px\,$ must also always be nonnegative; this forces all the $x$values to have the same sign. Since the point $(1,3)\,$ must satisfy the equation, we have: $$\begin{gather} \cssId{s137}{3^2 = 4p(1)}\cr \cssId{s138}{p = \frac 94} \end{gather} $$ Thus, the desired equation is $\,y^2 = 4(\frac 94)x\,$; that is, $\,y^2 = 9x\,.$ Since $\,p\,$ gives the $x$value of the focus, the focus is $\,(\frac 94,0)\,.$ Since the vertex (the origin) is halfway between the focus and directrix, the equation of the directrix is $\,x = \frac 94\,.$ 

Safest Approach:

Using Knowledge of the Standard Forms
This approach is quicker, but requires knowledge of the standard forms and the significance of $\,p\,$ in these standard forms. Since the focus is on the $y$axis, the form is $\,x^2 = 4py\,,$ where $\,p\,$ is the $y$value of the focus. So, $\,p = 2\,.$ Thus, the equation of the parabola is: $$\begin{gather} \cssId{s181}{x^2 = 4py}\cr \cssId{s182}{x^2 = 4(2)y}\cr \cssId{s183}{x^2 = 8y} \end{gather} $$ Of course, both approaches give the same solution! 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 