document.write( "Question 617191:
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document.write( "(y+1)^2 = 12(x-1)\r
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document.write( "Graph, including the foci \n" );
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Algebra.Com's Answer #388204 by Edwin McCravy(20059)![]() ![]() You can put this solution on YOUR website! (y + 1)² = 12(x - 1)\r \n" ); document.write( "\n" ); document.write( "Graph, including the foci. \n" ); document.write( "---------------------------- \n" ); document.write( "We \n" ); document.write( "(y - k)² = 4p(x - h)\r \n" ); document.write( "\n" ); document.write( "has vertex (h,k). If p is positive it opens to the right. \n" ); document.write( "If p is negative it opens to the left. its fccus is inside \n" ); document.write( "the parabola on its axis on symmetry and |p| units from the \n" ); document.write( "vertex.And the latus rectum or focal chord is |4p| units long \n" ); document.write( "bisected by the focus.\r \n" ); document.write( "\n" ); document.write( "Comparing your equation to that, we get\r \n" ); document.write( "\n" ); document.write( "h=1, k= -1, 4p=12, so p=3\r \n" ); document.write( "\n" ); document.write( "So the graph opens to the right because p is positive\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "An the focus is a point inside the parabola |p| or 3 units \n" ); document.write( "to the right of the vertex. The vertex is (1,-1), so the \n" ); document.write( "focus is (4,-1) \r \n" ); document.write( "\n" ); document.write( " \r \n" ); document.write( "\n" ); document.write( " \r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |