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 ycoordinate 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 xcoordinate of 3, lying on the parabola.
What is the ycoordinate of P?Use the Distance Formula to express the distance from the focus F(0,2) to the point P(3,y).
Substitute values
Substitute values
Subtract term
$LHS_{2}=RHS_{2}$
(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=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $  

Points  Substitution  Simplififcation 
F(0,p) and P(x,y)  FP = $(x−0)_{2}+(y−p)_{2} $  FP = $x_{2}+(y−p)_{2} $ 
$R(x,p)$ and P(x,y)  RP = $(x−x)_{2}+(y−(p))_{2} $  RP = $(y+p)_{2} $ 
$LHS_{2}=RHS_{2}$
(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
$LHS_{2}=RHS_{2}$
(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)  $y=p$  $y=4p1 x_{2}$ 
$(0,1)$  $y=(1)$ $⇕$ y=1 
$y=4(1)1 x_{2}$ $⇕$ $y=41 x_{2}$ 
Therefore, the equation of the parabola with focus at (0,1) and directrix y=1 is $y=41 x_{2}.$
Recall the general form for translations of functions.
Transformations of f(x)  

Horizontal Translations  $Translation righthunits,h>0y=f(x−h) $

$Translation lefthunits,h>0y=f(x+h) $
 
Vertical Translations  $Translation upkunits,k>0y=f(x)+k $

$Translation downkunits,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 

$y=41 x_{2}$  (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 xaxis have been discussed. Next, parabolas whose directrices are parallel to the yaxis 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=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{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} $ 
$LHS_{2}=RHS_{2}$
(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 threedimensional 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.