document.write( "Question 808094: For the parabola find the vertex,focus and directrix
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Algebra.Com's Answer #486844 by Edwin McCravy(20055) $\"\"$ $\"About$
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document.write( "a) x² + 6x + 4y + 5 = 0\r\n" );
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document.write( "Since the squared variable is x², the parabola has a\r\n" );
document.write( "vertical axis of symmetry We get the equation in the \r\n" );
document.write( "standard form:\r\n" );
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document.write( "(x - h)² = 4p(y - k) \r\n" );
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document.write( "and the vertex will be (h,k) and the distance from\r\n" );
document.write( "the vertex to the focus and directrix will be |p|.\r\n" );
document.write( "If p > 0, the parabola will open upward, and if p < 0,\r\n" );
document.write( "the parabola will open downward:\r\n" );
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document.write( "x² + 6x + 4y + 5 = 0\r\n" );
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document.write( "Isolate x terms on the left:\r\n" );
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document.write( "x² + 6x = -4y - 5\r\n" );
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document.write( "1. Multiply coefficient of x by 1/2:  6· = 3 \r\n" );
document.write( "2. Square that resul:                 3² = 9\r\n" );
document.write( "3. Add that result to both sides of the equation: \r\n" );
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document.write( "x² + 6x + 9 = -4y - 5 + 9\r\n" );
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document.write( "Factor the left side as (x + 3)(x + 3) or (x + 3)²\r\n" );
document.write( "Combine number terms on the right\r\n" );
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document.write( "(x + 3)² = -4y + 4\r\n" );
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document.write( "Factor out the coefficient of y on the right\r\n" );
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document.write( "(x + 3)² = -4(y - 1)\r\n" );
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document.write( "Compare that to\r\n" );
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document.write( "(x - h)² = 4p(y - k)\r\n" );
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document.write( "-h = +3, so h = -3 \r\n" );
document.write( "-k = -1, so k = 1\r\n" );
document.write( "vertex = (h,k) = (-3,1)\r\n" );
document.write( "4p = -4, so p = -1\r\n" );
document.write( "p is negative so parabola opens downward:\r\n" );
document.write( "distance from vertex to focus = distance from vertex to directrix = \r\n" );
document.write( "|p| = |-1| = 1\r\n" );
document.write( "Since the parabola opens downward, the focus is the point which is\r\n" );
document.write( "|p| = 1 unit below the vertex at (-3,1-1) or (-3,0), and the directrix is\r\n" );
document.write( "the horizontal line 1 units above the vertex or y = 1+1 or y = 2\r\n" );
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document.write( "We plot the vertex (-3,1), the focus (-3,0) and the directrix y = 2:\r\n" );
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document.write( "We plot the \"latus rectum\" or \"focal chord\", a horizontal line\r\n" );
document.write( "4|p| or 4(1) or 4 units long bisected by the focus:\r\n" );
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document.write( "And sketch in the parabola:\r\n" );
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document.write( "b) y² + 6y - 8x - 31 = 0 \r\n" );
document.write( "\r\n" );
document.write( "Since the squared variable is y², the parabola has a\r\n" );
document.write( "horizontal axis of symmetry. We get the equation in the \r\n" );
document.write( "standard form:\r\n" );
document.write( "\r\n" );
document.write( "(y - k)² = 4p(x - h) \r\n" );
document.write( "\r\n" );
document.write( "and the vertex will be (h,k) and the distance from\r\n" );
document.write( "the vertex to the focus and directrix will be |p|.\r\n" );
document.write( "If p > 0, the parabola will open to the right, and \r\n" );
document.write( "if p < 0, the parabola will open to the left:\r\n" );
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document.write( "y² + 6y - 8x - 31 = 0 \r\n" );
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document.write( "Isolate y terms on the left:\r\n" );
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document.write( "y² + 6y = 8x + 31 \r\n" );
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document.write( "1. Multiply coefficient of y by 1/2:  6· = 3 \r\n" );
document.write( "2. Square that resul:                 3² = 9\r\n" );
document.write( "3. Add that result to both sides of the equation: \r\n" );
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document.write( "y² + 6y + 9 = 8x + 31 + 9\r\n" );
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document.write( "Factor the left side as (y + 3)(y + 3) or (y + 3)²\r\n" );
document.write( "Combine number terms on the right\r\n" );
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document.write( "(y + 3)² = 8x + 40\r\n" );
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document.write( "Factor out the coefficient of y on the right\r\n" );
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document.write( "(y + 3)² = 8(x - 5)\r\n" );
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document.write( "Compare that to\r\n" );
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document.write( "(y - k)² = 4p(x - h)\r\n" );
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document.write( "-k = +3, so k = -3 \r\n" );
document.write( "-h = -5, so h = 5\r\n" );
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document.write( "vertex = (h,k) = (5,-3)\r\n" );
document.write( "4p = 8, so p = 2\r\n" );
document.write( "p is positive so parabola opens to the right:\r\n" );
document.write( "distance from vertex to focus = distance from vertex to directrix = \r\n" );
document.write( "|p| = |2| = 2\r\n" );
document.write( "Since the parabola opens to the right, the focus is the point \r\n" );
document.write( "which is |p| = 2 unit to the right of the vertex at (5+2,-3) or \r\n" );
document.write( "(7,-3), and the directrix is the vertical line 2 units left of \r\n" );
document.write( "the vertex or x = 5-2 or x = 3\r\n" );
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document.write( "We plot the vertex (5,-3), the focus (7,-3) and the directrix x = 3:\r\n" );
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document.write( "We plot the \"latus rectum\" or \"focal chord\", a vertical line\r\n" );
document.write( "4|p| or 4(1) or 4 units long bisected by the focus:\r\n" );
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document.write( "And sketch in the parabola:\r\n" );
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document.write( "Edwin
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