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Given the directrix x=6 and the focus (3,-5) what is the vertex form of the equation of the parabola?

Given the directrix x=6 and the focus (3,-5) what is the vertex form of the equation-example-1
User Rosamund
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1 Answer

1 vote

Answer:


x=-(1)/(6)(y+5)^(2)+(9)/(2)

Step-by-step explanation:

Given a parabola such that:

• The directrix, x=6

,

• Focus = (3, -5)

The standard form of a left-right parabola is given as:


4p(x-h)=(y-k)^2

The focus is obtained using the formula:


\begin{gathered} Focus=(h+p,k) \\ \implies(h+p,k)=(3,-5) \\ \implies h+p=3\cdots(1) \end{gathered}

The directrix is obtained using the formula:


\begin{gathered} x=h-p \\ x=6 \\ \implies6=h-p\cdots(2) \end{gathered}

Solve equations 1 and 2 simultaneously:


\begin{gathered} \begin{equation*} h+p=3\cdots(1) \end{equation*} \\ h-p=6\operatorname{\cdots}(2) \\ Add \\ 2h=9 \\ h=4.5 \\ \text{ Solve for p} \\ 4.5+p=3 \\ p=3-4.5 \\ p=-1.5 \end{gathered}

Substitute into the standard form:


\begin{gathered} 4p(x-h)=(y-k)^2 \\ 4(-1.5)(x-4.5)=(y-(-5))^2 \\ -6(x-4.5)=(y+5)^2 \\ x-4.5=-(1)/(6)(y+5)^2 \\ x=-(1)/(6)(y+5)^2+(9)/(2) \end{gathered}

The vertex form of the parabola is:


x=-(1)/(6)(y+5)^(2)+(9)/(2)

User Afser
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