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| 9 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Try your knowledge on these topics.
Use the Distance Formula to find the y-coordinate of B. If necessary, round your answer to 2 decimal places.
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?
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.
Use the Distance Formula to express the distance from the focus F(0,2) to the point P(3,y).
Substitute values
Subtract term
LHS2=RHS2
(a±b)2=a2±2ab+b2
LHS−y2=RHS−y2
LHS−4=RHS−4
LHS+4y=RHS+4y
LHS/8=RHS/8
ba=b1⋅a
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=(x2−x1)2+(y2−y1)2 | ||
---|---|---|
Points | Substitution | Simplififcation |
F(0,p) and P(x,y) | FP = (x−0)2+(y−p)2 | FP = x2+(y−p)2 |
R(x,-p) and P(x,y) | RP = (x−x)2+(y−(-p))2 | RP = (y+p)2 |
LHS2=RHS2
(a±b)2=a2±2ab+b2
LHS−y2=RHS−y2
LHS−p2=RHS−p2
LHS+2yp=RHS+2yp
LHS/4p=RHS/4p
ba=b1⋅a
Rearrange equation
In the graph, a parabola with the focus at (1,2) and directrix y=4 is shown.
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.
Substitute values
LHS2=RHS2
(a−b)2=a2−2ab+b2
Add terms
LHS−y2=RHS−y2
LHS−16=RHS−16
LHS+4y=RHS+4y
LHS/(-4)=RHS/(-4)
Write as a difference of fractions
ca⋅b=ca⋅b
Put minus sign in front of fraction
a−(-b)=a+b
ba=b/2a/2
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.
Focus | Directrix | Equation |
---|---|---|
(0,p) | y=-p | y=4p1x2 |
(0,-1) | y=-(-1) ⇕ y=1 |
y=4(-1)1x2 ⇕ y=-41x2 |
Therefore, the equation of the parabola with focus at (0,-1) and directrix y=1 is y=-41x2.
Recall the general form for translations of functions.
Transformations of f(x) | |
---|---|
Horizontal Translations | Translation right h units, h>0y=f(x−h)
|
Translation left h units, h>0y=f(x+h)
| |
Vertical Translations | Translation up k units, k>0y=f(x)+k
|
Translation down k units, k>0y=f(x)−k
|
(a−b)2=a2−2ab+b2
Distribute -41
ca⋅b=ca⋅b
ba=b/2a/2
a=44⋅a
Add fractions
Equation | Focus | Directrix |
---|---|---|
y=-41x2 | (0,-1) | y=1 |
y=-41(x−1)2 | (1,-1) | y=1 |
y=-41(x−1)2+3 | (1,2) | y=4 |
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.
d=(x2−x1)2+(y2−y1)2 | ||
---|---|---|
Points | Substitution | Simplififcation |
F(-4,-2) and P(x,y) | FP = (x−(-4))2+(y−(-2))2 | FP = (x+4)2+(y+2)2 |
R(2,y) and P(x,y) | RP = (x−2)2+(y−y)2 | RP = (x−2)2 |
LHS2=RHS2
(a±b)2=a2±2ab+b2
LHS−x2=RHS−x2
LHS+4x=RHS+4x
LHS−16=RHS−16
LHS−(y+2)2=RHS−(y+2)2
LHS/12=RHS/12
Write as a difference of fractions
Put minus sign in front of fraction
ba=b1⋅a
aa=1
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.
Use the Distance Formula to write the equation of the parabola.
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.
By substituting the endpoints of FP we can create an expression for this distance.
To determine PD, we will substitute D(x,-1) and P(x,y) into the Distance Formula.
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.
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.
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.
Like in Parts A through C, we will equate the distances and solve for y to obtain the equation of the parabola.
Consider the following parabola with focus at F(0,-9) and vertex at the origin V(0,0).
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.
The distance from point A to the directrix is 2 units. Write an equation of the parabola.
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.
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.