document.write( "Question 752890: how do you write out an equation for a parabola with the vertex (0,0) and a focus (0, -1/12)? \r
\n" ); document.write( "\n" ); document.write( "i have tried to fit it in the equation (x-h)^2+(y-k)^2=r^2, but i got a completely off- answer... \r
\n" ); document.write( "\n" ); document.write( "I looked up the answer key online, and it said that the answer is \" y = -3x^2 \" , but i am NOT getting even close..
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Algebra.Com's Answer #458115 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
given:
\n" ); document.write( " the vertex (\"0\",\"0\") and a focus (\"0\",\"-1%2F12\")\r
\n" ); document.write( "\n" ); document.write( "The directed distance from the vertex to the focus is called \"p\", the \"parabolic constant\". If going from vertex to focus is an upward or rightward motion, then \"+p\" is positive. If downward or leftward, then \"p+\" is negative. Here \"p\" is negative, in fact,it is \"-1%2F12\" because going from the vertex (\"0\",\"0\") to the focus.\r
\n" ); document.write( "\n" ); document.write( "The value of \"p\" is \"-1%2F12\", since that is the directed distance from the vertex to the focus. Since the axis is vertical, the general equation is \r
\n" ); document.write( "\n" ); document.write( "\"y=+4px%5E2+\"\r
\n" ); document.write( "\n" ); document.write( " So your equation is :\r
\n" ); document.write( "\n" ); document.write( "\"y+=+4%28+-1%2F12%29x%5E2+\"\r
\n" ); document.write( "\n" ); document.write( "\"y+=-3x%5E2+\"\r
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\n" ); document.write( "\n" ); document.write( "\"+graph%28+600%2C600%2C+-5%2C+5%2C+-10%2C+5%2C+-3x%5E2%29%29+\" \r
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