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This lesson will explore the geometric and algebraic representations of a parabola.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Try your knowledge on these topics.

a In the diagram, the distance between points and is the same as the distance between points and

Use the Distance Formula to find the coordinate of If necessary, round your answer to decimal places.

b The graph of is translated units to the right and units up. Which of the following functions represent the equation of the resulting graph?
Explore

Investigating Points that Are Equidistant from a Pair of Points

In the following applet, points and are plotted. By moving point all the points that are equidistant from and can be seen.
As it can be seen above, the locus of these points form a line. Similarly, the locus of all the points that are equidistant from two parallel lines also form a line. In fact, they form a line that is also parallel to the given lines.
Now, think about the locus of all points equidistant from a line and a point not on the line. What type of curve do you think the locus of these points forms?
Discussion

Parabola

A parabola is a curve that is geometrically defined as the locus of all points equidistant from a line and a point not on the line. The line is called the directrix, and the point is the focus of the parabola. In the following applet, point is equidistant from the directrix and the focus

parabola
Example

Properties of a Parabola

In the applet below, point is equidistant from point and line
Zosia claims that the points equidistant from point and line form a parabola. Her friend Heichi, however, claims that the points form some kind of closed curve. Who is correct?

Hint

Is it possible to find a point directly above so that the distance from to and the distance from to are the same?

Solution

It is not possible to find a point directly above so that the distance from to is the same as the distance from to Therefore, the curve formed by the points that are equidistant from and cannot be a closed curve.
Considering point and line there are infinitely many points are equidistant from and These points extend infinitely to the left, right, and upward. Therefore, Heichi's claim is false. Furthermore, as defined earlier, all points on a plane equidistant from a point and a line form a parabola. This means that Zosia is correct.
Example

Using the Distance Formula To Calculate Points on a Parabola

The equation of a parabola can be found using the Distance Formula.

Consider the previous parabola, this time drawn on a coordinate plane. The focus of the parabola is and its directrix is the line with the equation Consider also a point with an coordinate of lying on the parabola.

What is the coordinate of
Repeat the process for any point on the parabola. Write in terms of

Hint

Use the Distance Formula to express the distance from the focus to the point

Solution

Note that the directrix of the parabola is a horizontal line. Therefore, the distance between and the directrix is given by the length of the vertical segment that connects and the line. This length is the difference between the coordinate of and which is the coordinate of every point on the line.
Recall that a parabola is the set of all the points in a plane that are equidistant from a point called focus and a line called directrix. Since point is on the parabola, it is equidistant from the directrix and the focus.
Therefore, the distance between and is also These values can be substituted into the Distance Formula.
Solve for
The coordinate of point is Generalizing this process for any point will produce the equation of the parabola.
Solve for
The equation of the parabola is
Discussion

Equation of a Parabola From Focus and Directrix

Keeping the previous example in mind, consider a parabola with focus and directrix How can its equation be obtained?

By definition, any point on the parabola must be equidistant from the focus and the directrix. This means that and are congruent segments. Therefore, they have the same length. The Distance Formula can be used to write an expression for each length.

Points Substitution Simplififcation
and
and
As it has already been said, and must be equal. Setting the expressions equal to each other makes it possible to solve for
Solve for
The obtained quadratic equation corresponds to a parabola with focus and directrix Notice in the graph that the vertex of this parabola is the origin Furthermore, paying close attention, it can be seen that the equation represents a quadratic function with coefficients and
If given a directrix other than the axis and a focus not on the axis, the corresponding parabola's equation can be written either using the definition of a parabola or using transformations of functions.
Example

Finding the Equation of a Parabola

In the graph, a parabola with the focus at and directrix is shown.

a Use the definition of a parabola to find its equation.
b Consider a parabola with focus at and directrix Determine the transformations that could be applied to the graph of this parabola so that the given graph is obtained.
c Use the equation of the parabola in Part B and the transformations applied to write the equation of the given parabola. Is it the same equation as the equation from Part A?

Answer

a
b The given graph is the graph of translated unit to the right and units up.
c Equation:
Is It the Same Equation as in Part A? Yes, when the right-hand side of this equation is simplified, it results in the same equation as in Part A.

Hint

a The distance between any point on the parabola and the directrix is
b What is the transformation that maps the point onto the point
c How does the equation of a quadratic function change when it is translated horizontally or vertically?

Solution

a Let be any point on the given parabola. Then, the distance form this point to the directrix is

Since is equidistant from and the line it is known that and are equal. Therefore, is also By substituting this information together with the points and into the Distance Formula, the equation of the parabola can be obtained.

Solve for
b Consider a parabola with focus at and directrix
A parabola opens upwards with a focus at (1,2) and a directrix at y=4.

Notice that after a translation unit to the right and units up, the image of the focus of this parabola is the focus of the given parabola. The parabola can be obtained by translating the the above curve unit to the right and units up.

c In a previous example, the equation of a parabola with focus and directrix was obtained. Using this information, the equation of a parabola with focus and directrix can be obtained.
Focus Directrix Equation




Therefore, the equation of the parabola with focus at and directrix is

parabola

Recall the general form for translations of functions.

Transformations of
Horizontal Translations
Vertical Translations
The given parabola is the image of the parabola with equation after a translation unit to the right and units up. With that as a guide, the equation of the given parabola can be written.
Expanding the square of the binomial and simplifying will give the equation found in Part A.
Simplify right-hand side