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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The equation of a parabola can be determined if its focus and directrix is known.

Given any point on the parabola, the distance from that point to the focus is equal to the closest point on the directrix. Then, by using the distance formula, the lengths can be compared to get the equation for the parabola.Use the distance formula

Substitute points

$PD=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

$PD=(x−x)_{2}+(y−1)_{2} $

DiffZero$x−x=0$

$PD=0_{2}+(y−1)_{2} $

CalcPowCalculate power

$PD=(y−1)_{2} $

$PF=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

$PF=(x−(-2))_{2}+(y−(-5))_{2} $

NegNeg$-(-a)=a$

$PF=(x+2)_{2}+(y+5)_{2} $

Compare line segments

$(y−1.5)_{2} =(x+2)_{2}+(y+4.5)_{2} $

RaiseEqn$LHS_{2}=RHS_{2}$

$(y−1.5)_{2}=(x+2)_{2}+(y+4.5)_{2}$

ExpandNegPerfectSquare$(a−b)_{2}=a_{2}−2ab+b_{2}$

$y_{2}−3y+1.5_{2}=(x+2)_{2}+(y+4.5)_{2}$

ExpandPosPerfectSquare$(a+b)_{2}=a_{2}+2ab+b_{2}$

$y_{2}−3y+1.5_{2}=(x+2)_{2}+y_{2}+9y−4.5_{2}$

$y_{2}−3y+1.5_{2}=(x+2)_{2}+y_{2}+9y−4.5_{2}$

Solve for $y$

SubEqn$LHS−1.5_{2}=RHS−1.5_{2}$

$y_{2}−3y=(x+2)_{2}+y_{2}+9y−4.5_{2}−1.5_{2}$

SubEqn$LHS−y_{2}−9y=RHS−y_{2}−9y$

$y_{2}−3y−y_{2}−9y=(x+2)_{2}+9y+4.5_{2}$

CommutativePropAddCommutative Property of Addition

$y_{2}−y_{2}−3y−9y=(x+2)_{2}+9y+4.5_{2}$

SubTermsSubtract terms

$-12y=(x+2)_{2}+4.5_{2}−1.5_{2}$

CalcPowCalculate power

$-12y=(x+2)_{2}+20.25−2.25$

SubTermSubtract term

$-12y=(x+2)_{2}+18$

DivEqn$LHS/-12=RHS/-12$

$y=-12(x+2)_{2} $