Level 2 - The Focus and Directrix of Parabolas - Connecting Algebra and Geometry Through Coordinates (Geometry)

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

A

and

B

is the same as the distance between points

A

and

C .

Use the Distance Formula to find the $y -$ coordinate of $B .$ If necessary, round your answer to $2$ decimal places.

b The graph of

f (x)

is translated

3

units to the right and

3

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

A

and

B

are plotted. By moving point

C,

all the points that are equidistant from

A

and

B

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

Example

Properties of a Parabola

In the applet below, point

P

is equidistant from point

F

and line

d .

Zosia claims that the points equidistant from point

F

and line

d

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 $P$ directly above $F$ so that the distance from $P$ to $F$ and the distance from $P$ to $d$ are the same?

Solution

It is not possible to find a point

P

directly above

F

so that the distance from

F

P

is the same as the distance from

d

P .

Therefore, the curve formed by the points that are equidistant from

F

and

d

cannot be a closed curve.

Considering point

F

and line

d,

there are infinitely many points are equidistant from

F

and

d .

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 $F (0, 2),$ and its directrix is the line with the equation $y = - 2 .$ Consider also a point $P$ with an $x -$ coordinate of $3,$ lying on the parabola.

What is the

y -

coordinate of

P ?

Repeat the process for any point

P (x, y)

on the parabola. Write

y

in terms of

x .

Hint

Use the Distance Formula to express the distance from the focus $F (0, 2)$ to the point $P (3, y) .$

Solution

Note that the directrix of the parabola is a horizontal line. Therefore, the distance between

P (3, y)

and the directrix is given by the length of the vertical segment that connects

P

and the line. This length is the difference between the

y -

coordinate of

P

and

- 2,

which is the

y -

coordinate of every point on the line.

d = y - (- 2) \Leftrightarrow d = y + 2

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

P

is on the parabola, it is equidistant from the directrix and the focus.

Therefore, the distance between

F (0, 2)

and

P (3, y)

is also

y + 2 .

These values can be substituted into the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstituteValues

Substitute values

y + 2 = (3 - 0)^{2} + (y - 2)^{2}

Solve for

y

SubTerm

Subtract term

y + 2 = 3^{2} + (y - 2)^{2}

CalcPow

Calculate power

y + 2 = 9 + (y - 2)^{2}

RaiseEqn

$LHS^{2} = RHS^{2}$

(y + 2)^{2} = 9 + (y - 2)^{2}

ExpandPosNegPerfectSquare

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

y^{2} + 4 y + 4 = 9 + y^{2} - 4 y + 4

SubEqn

$LHS - y^{2} = RHS - y^{2}$

4 y + 4 = 9 - 4 y + 4

SubEqn

$LHS - 4 = RHS - 4$

4 y = 9 - 4 y

AddEqn

$LHS + 4 y = RHS + 4 y$

8 y = 9

DivEqn

$LHS / 8 = RHS / 8$

y = \frac{9}{8}

The

y -

coordinate of point

P

\frac{9}{8} .

Generalizing this process for any point

P (x, y)

will produce the equation of the parabola.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstituteValues

Substitute values

y + 2 = (x - 0)^{2} + (y - 2)^{2}

Solve for

y

SubTerm

Subtract term

y + 2 = x^{2} + (y - 2)^{2}

RaiseEqn

$LHS^{2} = RHS^{2}$

(y + 2)^{2} = x^{2} + (y - 2)^{2}

ExpandPosNegPerfectSquare

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

y^{2} + 4 y + 4 = x^{2} + y^{2} - 4 y + 4

SubEqn

$LHS - y^{2} = RHS - y^{2}$

4 y + 4 = x^{2} - 4 y + 4

SubEqn

$LHS - 4 = RHS - 4$

4 y = x^{2} - 4 y

AddEqn

$LHS + 4 y = RHS + 4 y$

8 y = x^{2}

DivEqn

$LHS / 8 = RHS / 8$

y = \frac{x ^{2}}{8}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

y = \frac{1}{8} x^{2}

The equation of the parabola is

y = \frac{1}{8} x^{2} .

Discussion

Equation of a Parabola From Focus and Directrix

Keeping the previous example in mind, consider a parabola with focus $(0, p)$ and directrix $y = - p .$ How can its equation be obtained?

By definition, any point $P (x, y)$ on the parabola must be equidistant from the focus and the directrix. This means that $\overline{F P}$ and $\overline{R P}$ are congruent segments. Therefore, they have the same length. The Distance Formula can be used to write an expression for each length.

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Points	Substitution	Simplififcation
$F (0, p)$ and $P (x, y)$	$F P$ $=$ $(x - 0)^{2} + (y - p)^{2}$	$F P$ $=$ $x^{2} + (y - p)^{2}$
$R (x, - p)$ and $P (x, y)$	$R P$ $=$ $(x - x)^{2} + (y - (- p))^{2}$	$R P$ $=$ $(y + p)^{2}$

As it has already been said,

F P

and

R P

must be equal. Setting the expressions equal to each other makes it possible to solve for

y .

x^{2} + (y - p)^{2} = (y + p)^{2}

Solve for

y

RaiseEqn

$LHS^{2} = RHS^{2}$

x^{2} + (y - p)^{2} = (y + p)^{2}

ExpandPosNegPerfectSquare

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

x^{2} + y^{2} - 2 y p + p^{2} = y^{2} + 2 y p + p^{2}

SubEqn

$LHS - y^{2} = RHS - y^{2}$

x^{2} - 2 y p + p^{2} = 2 y p + p^{2}

SubEqn

$LHS - p^{2} = RHS - p^{2}$

x^{2} - 2 y p = 2 y p

AddEqn

$LHS + 2 y p = RHS + 2 y p$

x^{2} = 4 y p

DivEqn

$LHS / 4 p = RHS / 4 p$

\frac{x ^{2}}{4 p} = y

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

\frac{1}{4 p} x^{2} = y

RearrangeEqn

Rearrange equation

y = \frac{1}{4 p} x^{2}

The obtained quadratic equation corresponds to a parabola with focus

F (0, p)

and directrix

y = - p .

Notice in the graph that the vertex of this parabola is the origin

(0, 0) .

Furthermore, paying close attention, it can be seen that the equation represents a quadratic function with coefficients

a = \frac{1}{4 p},

b = 0,

and

c = 0 .

y = a x^{2} + b x + c ⇓ y = \frac{1}{4 p} x^{2} + 0 x + 0 y = \frac{1}{4 p} x^{2} + 0 x + 0

If given a directrix other than the

x -

axis and a focus not on the

y -

axis, the corresponding parabola's equation can be written either using the definition of a parabola or using transformations of functions.

Example

Investigating Horizontal Parabolas

Up to this point, parabolas whose directrices are parallel to the $x -$ axis have been discussed. Next, parabolas whose directrices are parallel to the $y -$ axis will be examined.

Izabella is making an original video game character. She wants a force field in the shape of a parabola. When the character is at $F (- 4, - 2)$ and the opposing team's army is on vertical line $x = 2,$ the force field will appear as shown in the graph.

a Help Izabella to determine whether the parabola can be the graph of a function.

b Knowing the equation of the parabola will help Izabella in designing the force field shape. Use the definition of a parabola to find its equation.

Answer

a The parabola is not the graph of a function.

x = - \frac{1}{1 2} (y + 2)^{2} - 1

Hint

a Use the Vertical Line Test to check whether the curve can be the graph of a function.

b Use the definition of a parabola and the Distance Formula.

Solution

a The Vertical Line Test can be used to determine if the parabola can represent the graph of a function.

The vertical line intersects the graph more than once several times. Therefore, this parabola cannot be the graph of a function. Accordingly, the equation of this parabola is not a function. Furthermore, its equation does not have the same form as the equations of parabolas whose directrices are parallel to the

x -

axis.

b For any point

P (x, y)

on the parabola, the distance from the directrix

x = 4

must be the same as the distance from the focus

F (- 4, - 2) .

To be specific,

\overline{F P}

and

\overline{R P}

must be congruent. To write an expression for each length, the Distance Formula can be used.

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Points	Substitution	Simplififcation
$F (- 4, - 2)$ and $P (x, y)$	$F P$ $=$ $(x - (- 4))^{2} + (y - (- 2))^{2}$	$F P$ $=$ $(x + 4)^{2} + (y + 2)^{2}$
$R (2, y)$ and $P (x, y)$	$R P$ $=$ $(x - 2)^{2} + (y - y)^{2}$	$R P$ $=$ $(x - 2)^{2}$

Now, setting the expressions equal to each other makes it possible to write an equation.

(x + 4)^{2} + (y + 2)^{2} = (x - 2)^{2}

RaiseEqn

$LHS^{2} = RHS^{2}$

(x + 4)^{2} + (y + 2)^{2} = (x - 2)^{2}

ExpandPosNegPerfectSquare

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

x^{2} + 8 x + 16 + (y + 2)^{2} = x^{2} - 4 x + 4

SubEqn

$LHS - x^{2} = RHS - x^{2}$

8 x + 16 + (y + 2)^{2} = - 4 x + 4

It can be seen that the equation does not include any

x^{2} -

term. Therefore, it will be easier to solve for

x .

8 x + 16 + (y + 2)^{2} = - 4 x + 4

Solve for

x

AddEqn

$LHS + 4 x = RHS + 4 x$

12 x + 16 + (y + 2)^{2} = 4

SubEqn

$LHS - 16 = RHS - 16$

12 x + (y + 2)^{2} = - 12

SubEqn

$LHS - (y + 2)^{2} = RHS - (y + 2)^{2}$

12 x = - (y + 2)^{2} - 12

DivEqn

$LHS / 12 = RHS / 12$

x = \frac{- ( y + 2 ) ^{2} - 1 2}{1 2}

WriteDiffFrac

Write as a difference of fractions

x = \frac{- ( y + 2 ) ^{2}}{1 2} - \frac{1 2}{1 2}

MoveNegNumToFrac

Put minus sign in front of fraction

x = - \frac{( y + 2 ) ^{2}}{1 2} - \frac{1 2}{1 2}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

x = - \frac{1}{1 2} (y + 2)^{2} - \frac{1 2}{1 2}

QuotOne

$\frac{a}{a} = 1$

x = - \frac{1}{1 2} (y + 2)^{2} - 1

Alas! Izabella has found the equation. It can be simplified if desired. However, since it is acceptable to write the equation as it is, further simplifications will not be performed. This quadratic equation corresponds to the given parabola, whose directrix is parallel to the

y -

axis.

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Catch-Up and Review

Hint

Solution

Hint

Solution

Answer

Hint

Solution