document.write( "Question 283263: x^2+4x+4-4y^2+8y-8=0\r
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Algebra.Com's Answer #205641 by Edwin McCravy(20059) $\"\"$ $\"About$
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x^2+4x+4-4y^2+8y-8=0
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document.write( "Yes this is a hyperbola because the terms in \r\n" );
document.write( "and  have opposite signs.\r\n" );
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document.write( "How do I find the center, vertices and foci?\r\n" );
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document.write( "First get it in standard form, which is either\r\n" );
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document.write( " if the hyperbola opens right and left, \r\n" );
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document.write( "or\r\n" );
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document.write( " if the hyperbola opens upward and downward.\r\n" );
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document.write( "The first three terms will factor into a perfect square as they\r\n" );
document.write( "are.  So we do so:\r\n" );
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document.write( "Add 8 to both side to get the loose number off the left side:\r\n" );
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document.write( "  \r\n" );
document.write( "Factor out the coefficient of  out of the \r\n" );
document.write( "last two terms on the left. \r\n" );
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document.write( "Complete the square on  by multiplying\r\n" );
document.write( "the coefficient of y, which is -2 by  getting -1,\r\n" );
document.write( "and then squaring -1, getting +1.  And we add that inside the\r\n" );
document.write( "second parentheses.  However since there is a -4 in fromt of the\r\n" );
document.write( "second parentheses, adding -1 inside the parentheses amounts\r\n" );
document.write( "to adding  or -4 to the left side, so we must add -4 \r\n" );
document.write( "to the right side:\r\n" );
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document.write( "Factor the trinomial as the square of a binomial, and combine\r\n" );
document.write( "the numbers on the right:\r\n" );
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document.write( "Get a 1 on the right by dividing through by 4\r\n" );
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document.write( "Since the variable x comes first in the standard form, the\r\n" );
document.write( "hyperbola opens right and left.\r\n" );
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document.write( "So we compare that to:\r\n" );
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document.write( "The center is (-2,1).  \r\n" );
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document.write( "So we plot the center:\r\n" );
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document.write( " so , the semi-transverse axis is 2 unit\r\n" );
document.write( "long, so we draw the complete transverse axis right and left\r\n" );
document.write( "2 units from the center, that is, the tranverse axis is the \r\n" );
document.write( "horizontal green line below:\r\n" );
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document.write( "The vertices are the endpoints of the transverse axis,\r\n" );
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document.write( "so \r\n" );
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document.write( "the vertices are (-4,1) and (0,1)\r\n" );
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document.write( " so , the semi-conjugate axis is 1 unit\r\n" );
document.write( "long, so we draw the complete conjugate axis up and down\r\n" );
document.write( "1 units from the center, that is, the conjugate axis is the \r\n" );
document.write( "vertical green line below:\r\n" );
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document.write( "Now we draw in the defining rectangle\r\n" );
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document.write( "Now we can draw the asymptotes which are the extended diagonals\r\n" );
document.write( "of the defining rectangle:\r\n" );
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document.write( "and we can sketch in the hyperbola:\r\n" );
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document.write( "Finally we will find the two foci.  To do that,\r\n" );
document.write( "we find c, which is the distance from the center to each\r\n" );
document.write( "of the foci.  We use the hyperbola Pythagorean relation:\r\n" );
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document.write( "So the two foci are  units right and\r\n" );
document.write( "left of the center, and their coordinates are\r\n" );
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document.write( "(,1) and (,1)\r\n" );
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document.write( "or approximately the points:\r\n" );
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document.write( "(-4.2,1) and (.2,1)\r\n" );
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document.write( "The hyperbola alone is just this, if you erase all the guidelines:\r\n" );
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document.write( "Edwin
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