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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.
A parabola is a curve that is geometrically defined as the locus of all points equidistant from a line and a point not on the line. The line is called the directrix, and the point is the focus of the parabola. In the following applet, point $P$ is equidistant from the directrix $d$ and the focus $F.$
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
Calculate power
$LHS_{2}=RHS_{2}$
$(a±b)_{2}=a_{2}±2ab+b_{2}$
$LHS−y_{2}=RHS−y_{2}$
$LHS−4=RHS−4$
$LHS+4y=RHS+4y$
$LHS/8=RHS/8$
Substitute values
Subtract term
$LHS_{2}=RHS_{2}$
$(a±b)_{2}=a_{2}±2ab+b_{2}$
$LHS−y_{2}=RHS−y_{2}$
$LHS−4=RHS−4$
$LHS+4y=RHS+4y$
$LHS/8=RHS/8$
$ba =b1 ⋅a$
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} $ 
$LHS_{2}=RHS_{2}$
$(a±b)_{2}=a_{2}±2ab+b_{2}$
$LHS−y_{2}=RHS−y_{2}$
$LHS−p_{2}=RHS−p_{2}$
$LHS+2yp=RHS+2yp$
$LHS/4p=RHS/4p$
$ba =b1 ⋅a$
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
$LHS_{2}=RHS_{2}$
$(a−b)_{2}=a_{2}−2ab+b_{2}$
Add terms
$LHS−y_{2}=RHS−y_{2}$
$LHS−16=RHS−16$
$LHS+4y=RHS+4y$
$LHS/(4)=RHS/(4)$
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 $

$(a−b)_{2}=a_{2}−2ab+b_{2}$
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 $x$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.
$d=(x_{2}−x$ 
