5. The Focus and Directrix of Parabolas
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Chapter 4
5.
The Focus and Directrix of Parabolas
| Student Learning Objectives: |
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Why
Parabolas are unique curves that have a geometric definition based on a set of points equidistant from a fixed point, known as the focus, and a fixed line, termed the directrix. This relationship between the focus and directrix gives rise to the equation of the parabola. The lesson delves into the algebraic and geometric representations of parabolas, highlighting their reflective properties. These properties have practical applications, such as in satellite dish designs, where signals from space reflect off the dish and converge at the focus. The shape of a parabola and its equation can vary based on the position of the focus and directrix, leading to different forms of the equation. The exploration also touches upon transformations that can change the focus and directrix, providing insights into the versatile nature of parabolas in mathematics.
Lesson Settings & Tools
| | 9 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The Focus and Directrix of Parabolas
Slide of 9
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.
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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?
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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.
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Discussion
Parabola
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Example
Properties of a Parabola
In the applet below, point P is equidistant from point F and line d.
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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 to P is the same as the distance from d to 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.
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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?
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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) ⇔ 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 = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteValues
Substitute values
y+2 = sqrt(( 3- 0)^2+( y- 2)^2)
▼
Solve for y
SubTerm
Subtract term
y+2 = sqrt(3^2+(y-2)^2)
CalcPow
Calculate power
y+2 = sqrt(9+(y-2)^2)
RaiseEqn
LHS^2=RHS^2
(y+2)^2 = 9+(y-2)^2
ExpandPosNegPerfectSquare
(a± b)^2=a^2± 2ab+b^2
y^2+4y+4 = 9+y^2-4y+4
SubEqn
LHS-y^2=RHS-y^2
4y+4 = 9-4y+4
SubEqn
LHS-4=RHS-4
4y=9-4y
AddEqn
LHS+4y=RHS+4y
8y=9
DivEqn
.LHS /8.=.RHS /8.
y=9/8
The y-coordinate of point P is 98. Generalizing this process for any point P( x, y) will produce the equation of the parabola.
d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteValues
Substitute values
y+2 = sqrt(( x- 0)^2+( y- 2)^2)
▼
Solve for y
SubTerm
Subtract term
y+2 = sqrt(x^2+(y-2)^2)
RaiseEqn
LHS^2=RHS^2
(y+2)^2 = x^2+(y-2)^2
ExpandPosNegPerfectSquare
(a± b)^2=a^2± 2ab+b^2
y^2+4y+4 = x^2+y^2-4y+4
SubEqn
LHS-y^2=RHS-y^2
4y+4 = x^2-4y+4
SubEqn
LHS-4=RHS-4
4y=x^2-4y
AddEqn
LHS+4y=RHS+4y
8y=x^2
DivEqn
.LHS /8.=.RHS /8.
y=x^2/8
MoveNumRight
a/b=1/b* a
y=1/8x^2
The equation of the parabola is y= 18x^2.
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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 FP and RP are congruent segments. Therefore, they have the same length. The Distance Formula can be used to write an expression for each length.
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | ||
|---|---|---|
| Points | Substitution | Simplififcation |
| F( 0, p) and P( x, y) | FP = sqrt(( x- 0)^2+( y- p)^2) | FP = sqrt(x^2+(y-p)^2) |
| R( x, - p) and P( x, y) | RP = sqrt(( x- x)^2+( y-( - p) )^2) | RP = sqrt((y+p)^2) |
As it has already been said, FP and RP must be equal. Setting the expressions equal to each other makes it possible to solve for y.
sqrt(x^2+(y-p)^2)=sqrt((y+p)^2)
▼
Solve for y
RaiseEqn
LHS^2=RHS^2
x^2+(y-p)^2=(y+p)^2
ExpandPosNegPerfectSquare
(a± b)^2=a^2± 2ab+b^2
x^2+y^2-2yp+p^2=y^2+2yp+p^2
SubEqn
LHS-y^2=RHS-y^2
x^2-2yp+p^2=2yp+p^2
SubEqn
LHS-p^2=RHS-p^2
x^2-2yp=2yp
AddEqn
LHS+2yp=RHS+2yp
x^2= 4yp
DivEqn
.LHS /4p.=.RHS /4p.
x^2/4p=y
MoveNumRight
a/b=1/b* a
1/4px^2=y
RearrangeEqn
Rearrange equation
y=1/4px^2
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Example
Finding the Equation of a Parabola
In the graph, a parabola with the focus at (1,2) and directrix y=4 is shown.
a Use the definition of a parabola to find its equation.
b Consider a parabola with focus at (0,- 1) and directrix y=1. 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 y=- 1/4x^2+1/2x+11/4
b The given graph is the graph of y=- 14x^2 translated 1 unit to the right and 3 units up.
c Equation: y=- 1/4(x-1)^2+3
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.
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 P(x,y) on the parabola and the directrix is 4-y.
b What is the transformation that maps the point (0,- 1) onto the point (1,2)?
c How does the equation of a quadratic function change when it is translated horizontally or vertically?
Solution
a Let P(x,y) be any point on the given parabola. Then, the distance form this point to the directrix y=4 is 4-y. 
Since P(x,y) is equidistant from F(1,2) and the line y=4, it is known that FP and RP are equal. Therefore, FP is also 4-y. By substituting this information together with the points F( 1, 2) and P( x, y) into the Distance Formula, the equation of the parabola can be obtained.
d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteValues
Substitute values
4-y = sqrt(( x- 1)^2+( y- 2)^2)
▼
Solve for y
RaiseEqn
LHS^2=RHS^2
(4-y)^2=(x-1)^2+(y-2)^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
16-8y+y^2=x^2-2x+1+y^2-4y+4
AddTerms
Add terms
16-8y+y^2=x^2-2x+y^2-4y+5
SubEqn
LHS-y^2=RHS-y^2
16-8y=x^2-2x-4y+5
SubEqn
LHS-16=RHS-16
- 8y=x^2-2x-4y-11
AddEqn
LHS+4y=RHS+4y
- 4y=x^2-2x-11
DivEqn
.LHS /(- 4).=.RHS /(- 4).
y=x^2-2x-11/- 4
WriteDiffFrac
Write as a difference of fractions
y=x^2/- 4-2x/- 4-11/- 4
MovePartNumRight
a* b/c=a/c* b
y=1/- 4x^2-2/- 4x-11/- 4
MoveNegDenomToFrac
Put minus sign in front of fraction
y=- 1/4x^2-(- 2/4)x-(- 11/4)
SubNeg
a-(- b)=a+b
y=- 1/4x^2+2/4x+11/4
ReduceFrac
a/b=.a /2./.b /2.
y=- 1/4x^2+1/2x+11/4
b Consider a parabola with focus at (0,- 1) and directrix y=1.
Notice that after a translation 1 unit to the right and 3 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 1 unit to the right and 3 units up.
c In a previous example, the equation of a parabola with focus (0, p) and directrix y= - p was obtained. Using this information, the equation of a parabola with focus (0, - 1) and directrix y= 1 can be obtained.
It is noteworthy that the transformations can change the focus and the directrix.
| Focus | Directrix | Equation |
|---|---|---|
| (0, p) | y= - p | y=1/4 px^2 |
| (0, - 1) | y= - (- 1) ⇕ y=1 |
y=1/4( - 1)x^2 ⇕ y=- 1/4x^2 |
Therefore, the equation of the parabola with focus at (0,- 1) and directrix y=1 is y=- 14x^2.
Recall the general form for translations of functions.
| Transformations of f(x) | |
|---|---|
| Horizontal Translations | Translation right h units, h>0 y=f(x- h) |
| Translation left h units, h>0 y=f(x+ h) | |
| Vertical Translations | Translation up k units, k>0 y=f(x)+ k |
| Translation down k units, k>0 y=f(x)- k | |
The given parabola is the image of the parabola with equation y=- 14x^2 after a translation 1 unit to the right and 3 units up. With that as a guide, the equation of the given parabola can be written. y =- 1/4(x- 1)^2+ 3 Expanding the square of the binomial and simplifying will give the equation found in Part A.
y =- 1/4(x-1)^2+3
▼
Simplify right-hand side
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
y =- 1/4(x^2-2x+1)+3
Distr
Distribute - 1/4
y =- 1/4x^2+2x/4-1/4+3
MovePartNumRight
a* b/c=a/c* b
y =- 1/4x^2+2/4x-1/4+3
ReduceFrac
a/b=.a /2./.b /2.
y =- 1/4x^2+1/2x-1/4+3
NumberToFrac
a = 4* a/4
y =- 1/4x^2+1/2x-1/4+12/4
AddFrac
Add fractions
y =- 1/4x^2+1/2x+11/4 ✓
| Equation | Focus | Directrix |
|---|---|---|
| y=- 1/4x^2 | (0,- 1) | y=1 |
| y=- 1/4(x-1)^2 | (1,- 1) | y=1 |
| y =- 1/4(x-1)^2+3 | (1,2) | y=4 |
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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.
b x = - 1/12(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, FP and RP must be congruent. To write an expression for each length, the Distance Formula can be used.
It can be seen that the equation does not include any x^2-term. Therefore, it will be easier to solve for x.
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.
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | ||
|---|---|---|
| Points | Substitution | Simplififcation |
| F( - 4, - 2) and P( x, y) | FP = sqrt(( x-( - 4))^2+( y-( - 2))^2) | FP = sqrt((x+4)^2+(y+2)^2) |
| R( 2, y) and P( x, y) | RP = sqrt(( x- 2)^2+( y- y )^2) | RP = sqrt((x-2)^2) |
Now, setting the expressions equal to each other makes it possible to write an equation.
sqrt((x+4)^2+(y+2)^2) = sqrt((x-2)^2)
RaiseEqn
LHS^2=RHS^2
(x+4)^2+(y+2)^2 = (x-2)^2
ExpandPosNegPerfectSquare
(a± b)^2=a^2± 2ab+b^2
x^2+8x+16+(y+2)^2 = x^2-4x+4
SubEqn
LHS-x^2=RHS-x^2
8x+16+(y+2)^2 = - 4x+4
8x+16+(y+2)^2 = - 4x+4
▼
Solve for x
AddEqn
LHS+4x=RHS+4x
12x+16+(y+2)^2 =4
SubEqn
LHS-16=RHS-16
12x+(y+2)^2 =- 12
SubEqn
LHS-(y+2)^2=RHS-(y+2)^2
12x=- (y+2)^2-12
DivEqn
.LHS /12.=.RHS /12.
x=- (y+2)^2-12/12
WriteDiffFrac
Write as a difference of fractions
x=- (y+2)^2/12-12/12
MoveNegNumToFrac
Put minus sign in front of fraction
x=- (y+2)^2/12-12/12
MoveNumRight
a/b=1/b* a
x=- 1/12(y+2)^2-12/12
QuotOne
a/a=1
x=- 1/12(y+2)^2-1
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Closure
Parabolas in the Real World
What is the purpose of designing a satellite dish in the form of a paraboloid, or a three-dimensional parabola?
The shape of a parabola brings along an important reflective property. This property is used to collect or project light, sound, or radio waves. For this reason, satellite dishes are designed in the form of paraboloids — surfaces generated by the rotation of a parabola around its axis of symmetry.
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The Focus and Directrix of Parabolas
Exercises
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A parabola can be viewed as every point in a plane that is equidistant from a fixed point called the focus and a fixed line called the directrix. From the diagram, we can identify both the focus and the directrix of the parabola.
Using the Distance Formula, we can write one expression for FP and one for PD. Since these distances are congruent, we can equate these expressions and solve for y.
Distance of FP
By substituting the endpoints of FP we can create an expression for this distance.
Distance of PD
To determine PD, we will substitute D(x,-1) and P(x,y) into the Distance Formula.
Solving for y
As already mentioned, these distances are congruent. Therefore, we can equate them and then solve for y to obtain the equation of the parabola.
Like in Part A, we will first identify the focus and directrix of the given parabola.
Next, we need expressions for FP and PD.
Distance of FP
Distance of PD
Solving for y
As was done in Part A, we will equate the distances and solve for y.
Let's draw a parabola with the given characteristics.
As was done in previous parts, we will now create expressions for FP and PD.
Distance of FP
Distance of PD
Solving for y
Let's equate the distances and solve for y.
Remember that any segment between the focus and an arbitrary point P on the parabola is congruent with the segment from point P to the directrix.
This means the vertex of the parabola must bisect the vertical line segment that we can draw between the directrix and the focus. Therefore, if the directrix is at y=- 9, the focus must be at F(0,9).
Now we can proceed in the same way as in the previous parts.
Distance of FP
Distance of PD
Solving for y
Like in Parts A through C, we will equate the distances and solve for y to obtain the equation of the parabola.
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Hint & Answer
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Solution
Which of the given characteristics describe a parabola that opens downward? Explain your reasoning. A: &Focus: (0,-6) &Directrix: y=6 [0.6em] B: &Focus: (0,-2) &Directrix: y=2 [0.6em] C: &Focus: (0,6) &Directrix: y=-6 [0.6em] D: &Focus: (0,-1) &Directrix: y=1
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Every parabola that opens downward has a focus that is below the directrix. In the diagram it is possible to graph parabolas with the given characteristics.
We can see that the characteristics A, B, and D result in parabolas that opens downward.
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Hint & Answer
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Solution
Consider the following parabola with focus at F(0,-9) and vertex at the origin V(0,0).
Which of the following options are possible coordinates of point P that we see in the graph. Explain your reasoning. A: &(6,-1) && B: (- 3,- 14) [0.4em] C: &(- 4,- 49) && D: (- 1, 136) [0.4em] E: &(-8, 169) && F: (- 2,- 118)
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Notice that point P is in the third quadrant of the coordinate plane. This means it has a negative x- and y-coordinate. Therefore we can immediately discard the points A, D, and E. &A: (6,-1) && B: (- 3,- 14 ) [0.4em] &C: (- 4,- 49) && D: (- 1, 136) [0.4em] &E: (-8, 169) && F: (- 2,- 118) Since we have been given the focus and vertex of the parabola, we can find its equation using the following formula. y=1/4px^2 Keep in mind that this formula describes a parabola with its vertex at the origin and with a vertical axis of symmetry. In the formula, p represents the y-coordinate of the focus, which we know is -9. Therefore, by substituting - 9 for p we can obtain the equation of the parabola.
Now we have to substitute each of the remaining points into the equation and see if we end up with a true statement.
| Point | Substitute | Evaluate |
|---|---|---|
| B( 3, -1/4) | -1/4? =- 1/36( 3)^2 | -1/4= -1/4 ✓ |
| C( -4, -4/9) | -4/9? =- 1/36( - 4)^2 | -4/9= -4/9 ✓ |
| F( -2, -1/8) | -1/8? =- 1/36( - 2)^2 | -1/8= -1/9 * |
Now we can strike the remaining option that does not fall on the parabola. &A: (6,-1 ) && B: (- 3,- 14) [0.4em] & C: (- 4,- 49) && D: (- 1, 136) [0.4em] &E: (-8,- 169) && F: (- 2,- 118) The points B and C fall on the parabola.
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Hint & Answer
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Solution
The distance from point A to the directrix is 2 units. Write an equation of the parabola.
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In the diagram we see that the parabola has its vertex in the origin and has a vertical axis of symmetry. This means we can describe the parabola with the following equation. y=1/4px^2 In this equation, p represents the y-coordinate of the focus. We can also obtain p by finding the opposite number to the directrix. This is because the focus and directrix is equidistant from the vertex. Focus:& (0,p) Directrix:& y=- p Since we know the distance from A to the directrix, we can figure out the equation of the directrix by subtracting 2 from 1. y=1-2 ⇔ y=-1 When we know the equation of the directrix, we can find p by solving the equation y=- p.
Now we can determine the equation of the parabola.
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Solution
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In the given equation, the variable that has been raised to the power of two is x. y= 1/12x^2 This means the parabola will have a vertical axis of symmetry. We also see that the vertex of the parabola is at the origin. This means the equation of the parabola can be described with the following equation. y= 1/4 px^2 The variable p represents the y-coordinate of the focus. Since the vertex is at the origin, we can write the coordinates of the focus as (0, p). Therefore, to find p, we can equate the coefficient of x in the formula, with the coefficient of x in the given equation. l Given equation → 1/12&= 1/4 p ← Formula Let's solve for p.
The focus is at (0, 3).
From Part A, we know that p=3. However, the opposite number to p also describes the directrix of the parabola. This is because the focus and directrix are equidistant from the vertex. Focus:& (0,3) Directrix:& y=- 3 This means the directrix is y=-3.
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The Focus and Directrix of Parabolas
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