document.write( "Question 982511: Parabola with vertex at (1, 3) and x = 5 as directrix
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Algebra.Com's Answer #603403 by josgarithmetic(39630) $\"\"$ $\"About$

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Knowing the vertex and directrix, you can determine the focus point. See 4 units to the left of x=1; this would be x=-3. The focus is (-3,3).\r
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\n" ); document.write( "\n" ); document.write( "If you already know the derived formula for a parabola based on the definition and the use of the Distance Formula, then you are ready to use it directly.\r
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\n" ); document.write( "\n" ); document.write( " $\"4px=y%5E2\"$ would be parabola with focus at (-p,0) for some number $\"p%3E0\"$ and vertex at the origin, (0,0). This is for a axis of symmetry being the x-axis.\r
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\n" ); document.write( "\n" ); document.write( "YOUR example is concave to the left, and also a horizontal parabola, and $\"abs%28p%29=4\"$ . Put all the data into the equation model, as translated from standard position: $\"highlight%28-4%2A4%2A%28x-1%29=%28y-3%29%5E2%29\"$ .
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\n" ); document.write( "You could also derive this from the known focus and given directrix according to the definition of a parabola. \n" ); document.write( "

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