document.write( "Question 1138078: Find the standard form of the equation of the parabola with a focus at (0, -3) and a directrix at y = 3. (5 points)
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\n" ); document.write( "y2 = -12x
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\n" ); document.write( "y2 = -3x
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\n" ); document.write( "y = negative 1 divided by 12 x2
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\n" ); document.write( "y = negative 1 divided by 3 x2
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Algebra.Com's Answer #755995 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "The standard form is \"%28x+-+h%29%5E2+=+4p+%28y+-+k%29\", where the focus is (\"h\", \"k+%2B+p\") and the directrix is \"y+=+k+-+p\".\r
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\n" ); document.write( "\n" ); document.write( "given: \r
\n" ); document.write( "\n" ); document.write( "a focus at (\"0\", \"-3\") =>(\"h\", \"k+%2B+p\")=(\"0\", \"-3\") \r
\n" ); document.write( "\n" ); document.write( "so, \"h=0\" and \"+k+%2B+p=-3\"=>\"k=-3-p\"....eq.1\r
\n" ); document.write( "\n" ); document.write( "a directrix at \"y+=+3\"=>\"k+-+p=3\"=>\"k=3%2Bp\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "from eq.1 and eq.2 we have
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\n" ); document.write( "\"-3-p=3%2Bp\".....solve for \"p\"
\n" ); document.write( "\"-3-3=p%2Bp\"
\n" ); document.write( "\"-6=2p\"
\n" ); document.write( "\"p=-3\"\r
\n" ); document.write( "\n" ); document.write( "go to \"k=3%2Bp\"....eq.2, plug in \"p\"\r
\n" ); document.write( "\n" ); document.write( "\"k=3-3\"\r
\n" ); document.write( "\n" ); document.write( "\"k=0\"\r
\n" ); document.write( "\n" ); document.write( "your equation is:\r
\n" ); document.write( "\n" ); document.write( "\"%28x-0%29%5E2+=+4%28-3%29+%28y+-+0%29\"\r
\n" ); document.write( "\n" ); document.write( "\"x%5E2+=+-12+y+\"\r
\n" ); document.write( "\n" ); document.write( "\"y=+-%281%2F12%29x%5E2++\" => your answer \r
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