document.write( "Question 281903: Find the center, vertices, foci, and asymtotes of the hyperbola
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Algebra.Com's Answer #204759 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
what are the slopes of the asymptotes of the hyperbola with equation
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document.write( "First get it in standard form, which is either\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\" if the hyperbola opens right and left, \r\n" );
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document.write( "or\r\n" );
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document.write( "\"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\" if the hyperbola opens upward and downward.\r\n" );
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document.write( "\"4x%5E2-y%5E2-8x%2B2y-1=0\"\r\n" );
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document.write( "Get the loose number -1 off the left side\r\n" );
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document.write( "\"4x%5E2-y%5E2-8x%2B2y=1\"\r\n" );
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document.write( "Get the \"x\" term next to the \"x%5E2\" term.\r\n" );
document.write( "Get the \"y\" term next to the \"y%5E2\" term.\r\n" );
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document.write( "\"4x%5E2-8x-y%5E2%2B2y=1\"\r\n" );
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document.write( "Factor the coefficient of \"x%5E2\" out of the \r\n" );
document.write( "first two terms on the left. \r\n" );
document.write( "Farctor the coefficient of \"y%5E2\" out of the \r\n" );
document.write( "last two terms on the left. \r\n" );
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document.write( "\"4%28x%5E2-2x%29-1%28y%5E2-2y%29=1\"\r\n" );
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document.write( "Complete the square on \"%28x%5E2-2x%29\" by multiplying\r\n" );
document.write( "the coefficient of x, which is -2 by \"1%2F2\" getting -1,\r\n" );
document.write( "and then squaring -1, getting +1.  And we add that inside the\r\n" );
document.write( "first parentheses.  However since there is a 4 in fromt of the\r\n" );
document.write( "first parentheses, adding +1 inside the parentheses amounts\r\n" );
document.write( "to adding +4*1 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( "\"4%28x%5E2-2x%2Bred%281%29%29-1%28y%5E2-2y%29=1%2Bred%284%29\"\r\n" );
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document.write( "Complete the square on \"%28y%5E2-2y%29\" by multiplying\r\n" );
document.write( "the coefficient of y, which is -2 by \"1%2F2\" 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 -1 in fromt of the\r\n" );
document.write( "second parentheses, adding +1 inside the parentheses amounts\r\n" );
document.write( "to adding \"1%2A-1\" or -1 to the left side, so we must add -1 \r\n" );
document.write( "to the right side:\r\n" );
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document.write( "Factor the parentheses as squares of binomials, and combine\r\n" );
document.write( "the numbers on the right:\r\n" );
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document.write( "\"4%28x-1%29%5E2-%28y-1%29%5E2=4\"\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( "\"4%28x-1%29%5E2%2F4-%28y-1%29%5E2%2F4=4%2F4\"\r\n" );
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document.write( "\"%28x-1%29%5E2%2F1-%28y-1%29%5E2%2F4=1\"\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( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"\r\n" );
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document.write( "The center is at (1,1).  So we plot the center:\r\n" );
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document.write( "\"a%5E2=1\" so \"a=1\", the semi-transverse axis is 1 unit\r\n" );
document.write( "long, so we draw the complete transverse axis right and left\r\n" );
document.write( "1 unit 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 two vertices are the endpoints of the transverse axis, so the\r\n" );
document.write( "vertices are the two points \r\n" );
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document.write( "(0,1) and (2,1)\r\n" );
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document.write( "\"b%5E2=4\" so \"b=2\", the semi-conjugate axis is 2 units\r\n" );
document.write( "long, so we draw the complete conjugate axis up and down\r\n" );
document.write( "2 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( "The slopes of the asymptotes are \"%22%22%2B-b%2Fa+=+%22%22%2B-+2%2F1=%22%22%2B-+2\"\r\n" );
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document.write( "They go through the point which is the center of the hyperbola (1,1)\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
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document.write( "Using m=2\r\n" );
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document.write( "\"y-1=2%28x-1%29\"\r\n" );
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document.write( "\"y-1=2x-2\"\r\n" );
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document.write( "\"y=2x-1\"\r\n" );
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document.write( "Using m=-2\r\n" );
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document.write( "\"y-1=-2%28x-1%29\"\r\n" );
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document.write( "\"y-1=-2x%2B2\"\r\n" );
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document.write( "\"y=-2x%2B3\"\r\n" );
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document.write( "To find the foci. we find c using\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
document.write( "\"c%5E2=1%5E2%2B2%5E2\"\r\n" );
document.write( "\"c%5E2=1%2B4\"\r\n" );
document.write( "\"c%5E2=5\"\r\n" );
document.write( "\"c=%22%22%2B-sqrt%285%29\"\r\n" );
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document.write( "The foci are c units right and left of the center:\r\n" );
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document.write( "They are (1-\"sqrt%285%29\",1) and (1+\"sqrt%285%29\",1)\r\n" );
document.write( "-sqrt(5)\r\n" );
document.write( "I'll draw the foci in:\r\n" );
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document.write( "Edwin
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