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User Gurushant
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1 Answer

19 votes
19 votes

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

The equation above is the standard form equation of a horizontal parabola and have the next caractheristics:

- The vertex is (h,k)

- If 4p > 0, it opens right

If 4p < 0, it opens left

- | p | is the distance from the vertex to the focus

- The directrix is vertical and the vertex is midway betweeen the focus and directrix

- The directrix is 2| p | units from the focus

For the given parabola:


\begin{gathered} (y+1)^2=12(x-3) \\ \\ k=-1 \\ h=3 \\ 4p=12 \\ p=(12)/(4)=3 \end{gathered}

Focus coordinates: It has the same y-coordinate that the vertex, the x-coorcdinate is |p| units to the right (as the parabola opens right) from the x-coordinate of the vertex:


\begin{gathered} \text{focus: } \\ (h+\lvert p\rvert,k) \\ \\ (3+3,-) \\ (6,-1) \end{gathered}

Directrix: it is a vertical line as follow (for a parabola that opens right)


\begin{gathered} \text{Directrix:} \\ x=h-\lvert p\rvert \\ \\ x=3-3 \\ x=0 \end{gathered}

The vertex of the parabola is (3, -1). The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x=0

User Adam Goucher
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