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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.
What is the y-coordinate of P?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 | ||
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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) | |
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Horizontal Translations | Translation right h units, h>0y=f(x−h)
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Translation left h units, h>0y=f(x+h)
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Vertical Translations | Translation up k units, k>0y=f(x)+k
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Translation down k units, k>0y=f(x)−k
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(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−x |
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