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Find an equation of the parabola with vertex at (-3, 1) and focus (-1, 1)
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@Directrix
\[ 4py=x^2 \]Has vertex \((0,0)\) and focus \((0,p)\)
Yes
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Translate the vertex: \[ 4p(y-1)=(x+3)^2 \]Now it has vertex \((-3, 1)\) and focus \((-3,p+1)\)
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I get (x+3)^2=4(y1)^2
(x+3)^2=4(y-1)^2 @wio
|dw:1364265229061:dw| Wait it is a horizontal parabola
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