Finding the Equation of a Parabola

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:

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:

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:

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: $$\sqrt{(x-0)^2 + (y-2)^2} = \sqrt{x^2 + (y-2)^2}$$ 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: $$\begin{gather} \sqrt{x^2 + (y-2)^2} = y+2\cr\cr x^2 + (y-2)^2 = (y+2)^2\cr\cr x^2 + y^2 - 4y + 4 = y^2 + 4y + 4\cr\cr x^2 = 8y \end{gather} $$

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:
Definition of an Ellipse


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