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:
 vertex at the origin

one of these two pieces of information:
 focus on $x$axis or $y$axis
 directrix parallel to $y$axis or $x$axis
(These are equivalent, when the vertex is at the origin.)
As shown in the
the prior section,
such parabolas have equations of the form $\,x^2 = 4py\,$ or $\,y^2 = 4px\,$,
where $\,p\,$ gives the distance from the origin to the focus.
Stop and Think:
Does the Given Information Uniquely Identify the Parabola?
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 $\,x = y^2\,$.


What Information Does Uniquely Define a Parabola?
Five Points are Needed to Determine a General Conic
The general conic equation has six unknowns: $A\,$, $B\,$, $C\,$, $D\,$, $E\,$, and $F\,$:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \qquad (*)$$
One might initially think that six points are needed to solve for these six unknowns.
However, only five points are actually needed, as follows:
 If $\,A = B = C = D = E = F = 0\,$, then the solution set is the entire plane,
since all points $\,(x,y)\,$ satisfy the equation $\,0x^2 + 0xy + 0y^2 + 0x + 0y + 0 = 0\,$.
To get something other than the entire plane as the solution set,
at least one of $\,A\,$, $\,B\,$, $\,C\,$, $\,D\,$ or $\,E\,$ must be nonzero.
(If $\,F\,$ is the only nonzero constant, then the equation is always false.)
 Dividing both sides of (*) by a nonzero coefficient gives an
equivalent equation (an equation with the same graph) that has only five unknown coefficients.
For example, if $\,A\ne 0\,$, then dividing both sides of (*) by $\,A\,$
gives
$$x^2 + \frac BAxy + \frac CAy^2 + \frac DAx +
\frac EAy + \frac FA = 0$$
or (more simply)
$$x^2 + \hat Bxy + \hat Cy^2 + \hat Dx + \hat Ey + \hat F = 0$$
Five pieces of noncontradictory, nonoverlapping information
are needed to solve uniquely for these five unknowns.

An example of contradictory information would be: $\,\hat B = 1\,$ and $\,\hat B = 2\,$.
These can't both be true at the same time.

An example of overlapping information would be: $\,\hat B = 1\,$ and $\,\hat B  1 = 0\,$.
No new information is given with the second equation.

[optional, if you're worried]
You might be thinking: ‘Hey! The
equation
$$x^2 + \frac BAxy + \frac CAy^2 + \frac DAx +
\frac EAy + \frac FA = 0$$
still has six unknowns! We didn't
really get rid of one of the unknowns!’
However, this is the situation:

Let $\,S_1\,$ denote the set of equations generated by
$$x^2 + \frac BAxy + \frac CAy^2 + \frac DAx +
\frac EAy + \frac FA = 0$$
as $\,A\,$ through $\,F\,$ (six unknowns) vary over all real numbers (with $\,A\ne 0\,$).

Let $\,S_2\,$ denote the set of equations generated by
$$x^2 + \hat Bxy + \hat Cy^2 + \hat Dx +
\hat Ey + \hat F = 0$$
as $\,\hat B\,$ through $\,\hat F\,$ (five unknowns) vary over all real numbers.
 Then, $\,S_1 = S_2\,$. How?
 Take any equation in $\,S_1\,$.
Let $\,\hat B = \frac BA\,$, $\,\hat C = \frac CA\,$, etc.
So, this equation is also in $\,S_2\,$.
Thus, $\,S_1\,$ is a subset of $\,S_2\,$.

Take any equation in $\,S_2\,$.
Let $\,B = \hat BA\,$, so that $\,\frac{B}{A} = \frac{\hat BA}{A} = \hat B\,$ (and so on).
So, this equation is also in $\,S_1\,$.
Thus, $\,S_2\,$ is a subset of $\,S_1\,$.
 When both $\,S_1\subset S_2\,$ and $\,S_2\subset S_1\,$, we have $\,S_1 = S_2\,$.
 Note: To get a ‘real’ conic (not a degenerate form or limiting case), no three
of the five points can be collinear.
Otherwise, you can get things like
a line, intersecting lines, or parallel lines; and, you may lose uniqueness.
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.


Given five points $\,(x_1,y_1)\,$ through ($\,x_5,y_5)\,$,
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
parameters can be any real number—then, the other five are uniquely determined.
What does this mean? What's going on here?
To understand, let's look at a simpler and
familiar analogous situation:
 As long as $\,A\,$ or $\,B\,$ is nonzero, then the equation $\,Ax + By + C = 0\,$
graphs as a line (possibly vertical or horizontal).
 Although there are three unknowns in this
equation ($\,A\,$, $\,B\,$ and $\,C\,$), only two points
are needed to determine a line.
 What happens if we try to find $\,A\,$, $\,B\,$ and $\,C\,$ so
that the line contains two known distinct points, $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$?
The approach below solves for both $\,A\,$ and $\,C\,$ in terms of $\,B\,$:
$$
\begin{align}
(1)\quad Ax_1 + By_1 + C = 0 \qquad\qquad&\text{$\,(x_1,y_1)\,$ on the line}\cr
(2)\quad Ax_2 + By_2 + C = 0\qquad\qquad &\text{$\,(x_2,y_2)\,$ on the line}\cr\cr
A(x_1x_2) + B(y_1y_2) = 0\qquad\qquad&\text{use elimination: subtract (2) from (1)}\cr\cr
(3)\quad A = B\cdot\frac{y_1y_2}{x_1x_2} = B\cdot\frac{y_2y_1}{x_2x_1}\qquad\qquad&\text{solve for $\,A\,$ in terms of $\,B\,$, assuming $\,x_1 \ne x_2\,$}
\end{align}
$$
Note: if $\,x_1 = x_2\,$ then the line is vertical.
Continuing:
$$\begin{alignat}{2}
C &= Ax_1  By_1 &\qquad&\text{solve for $\,C\,$ in (1)}\cr\cr
&= B\cdot\frac{y_2y_1}{x_2x_1}x_1  By_1&&\text{use (3) for $\,A\,$}\cr\cr
&= B\left(\frac{y_2y_1}{x_2x_1}x_1  y_1\right)&&\text{simplify; this gives $\,C\,$ in terms of $\,B\,$} \end{alignat}
$$
Substitution of the expressions for $\,A\,$ and $\,C\,$ in terms of $\,B\,$ into $\,Ax + By + C = 0\,$ gives:
$$
(4)\quad B\cdot\frac{y_2y_1}{x_2x_1}x + By + B\left(\frac{y_2y_1}{x_2x_1}x_1  y_1\right) = 0
$$
So, $\,B\,$ can be any real number.
If $\,B = 0\,$, then $\,A\,$ and $\,C\,$ are also zero.
In this case, the equation $\,0x + 0y + 0 = 0\,$ is true for all real numbers $\,x\,$ and $\,y\,$.
For $\,B\ne 0\,$:
$$
\begin{align}
\,\frac{y_2y_1}{x_2x_1}x + y + \frac{y_2y_1}{x_2x_1}x_1  y_1 = 0\qquad&\text{divide both sides of (4) by $\,B\,$}\cr\cr
(y_2y_1)x + y(x_2x_1) + (y_2y_1)x_1  y_1(x_2x_1) = 0\qquad&\text{clear fractions}\cr\cr
(x_2x_1)(y  y_1)  (y_2y_1)(x  x_1) = 0\qquad&\text{factor}\cr\cr
y  y_1 = \frac{y_2y_1}{x_2x_1}(xx_1)\qquad&\text{solve for $\,y  y_1\,$}
\end{align}
$$
With $\displaystyle\,m := \frac{y_2y_1}{x_2x_1}\,$ (the slope of the line),
this is of course the familiar pointslope equation of a line.
For our purposes here, the important understanding is that only two points were needed, not three;
and one of the unknowns (for us, $\,B\,$) became a degree of freedom in the equation.
Four Points Aren't Enough to Uniquely Define a General Parabola
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! 

Only One Point Needed to Determine a Parabola in Standard Form
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:
$$\begin{gather}
3^2 = 4p(1)\cr
p = \frac 94
\end{gather}
$$
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 $\,x = \frac 94\,$.


Finding the Equation of a Parabola: Two Approaches
Problem:
Find the equation of the parabola in standard form with focus $\,(0,2)\,$.
Solutions:
Since the equation is in standard form, the vertex is at the origin.
Safest Approach:
Use the Definition of Parabola
This approach does not require memorization/knowledge of the standard forms.
However, it takes longer, because you're going through the
derivation of the standard form for a special case.
By definition, a parabola is the set of all points equidistant
from the focus and directrix.
Since the vertex is halfway between the focus and directrix,
the directrix is the line $\,y = 2\,$.
Let $\,(x,y)\,$ be a typical point on the parabola.
Using the distance formula, the distance from $\,(x,y)\,$ to the focus is:
$$\sqrt{(x0)^2 + (y2)^2} = \sqrt{x^2 + (y2)^2}$$
The distance from $\,(x,y)\,$ to the directrix is $\,y  (2) = y + 2\,$.
(The points have the same $x$value—just compute distance along a line.)
Equating the distances gives:
$$\begin{gather}
\sqrt{x^2 + (y2)^2} = y+2\cr\cr
x^2 + (y2)^2 = (y+2)^2\cr\cr
x^2 + y^2  4y + 4 = y^2 + 4y + 4\cr\cr
x^2 = 8y
\end{gather}
$$

Using Knowledge of the Standard Forms
$x^2 = 4py$
(focus on $y$axis)
Since $\,x^2\,$ is nonnegative,
so is $\,4py\,$.
This forces $\,y\,$ to always
have the same sign.

$y^2 = 4px$
(focus on $x$axis)
Since $\,y^2\,$ is nonnegative,
so is $\,4px\,$.
This forces $\,x\,$ to always
have the same sign.

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:
$$\begin{gather}
x^2 = 4py\cr
x^2 = 4(2)y\cr
x^2 = 8y
\end{gather}
$$
Of course, both approaches give the same solution!

Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
?