document.write( "Question 699163: Find the coordinates of the center, the foci, and the vertices, and the equations and asymptotes of the graph of each equation. Then graph the equation.
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Algebra.Com's Answer #431239 by Edwin McCravy(20060) $\"\"$ $\"About$

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document.write( "4x²-y²-8x+20 = 0         |        x²-4y²-4x+24y-36 = 0\r\n" );
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document.write( "Rearrange the terms in this order x²,x,y²,y, and constants\r\n" );
document.write( "on the right:\r\n" );
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document.write( "   4x²-8x-y² = -20                      x²-4x-4y²+24y = 36 \r\n" );
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document.write( " 4(x²-2x)-y² = -20                   (x²-4x)-4(y²-6y) = 36\r\n" );
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document.write( "Complete the square in the parentheses. In the equation on\r\n" );
document.write( "the left since there is no y-term write y as (y-0)\r\n" );
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document.write( "4(x²-2x+1)-(y-0)² = -20+4         (x²-4x+4)-4(y²-6y+9) = 36+4=36\r\n" );
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document.write( "    4(x-1)²-(y-0)² = -16                (x-2)²-4(y-3)² = 4\r\n" );
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document.write( "Get 1 on the right by dividing through by the constant term:\r\n" );
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document.write( "               \r\n" );
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document.write( "               \r\n" );
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document.write( "Switch the terms on the left so that the positive term comes first:\r\n" );
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document.write( "               \r\n" );
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document.write( "The equation on the left has its y-term first and therefore represents a\r\n" );
document.write( "hyperbola that looks like this  whereas the equation on the right \r\n" );
document.write( "has its x-term first and therefore it represents a hyperbola that looks\r\n" );
document.write( "like this )(\r\n" );
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document.write( "               \r\n" );
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document.write( "  \r\n" );
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document.write( "For the hyperbola on the left:\r\n" );
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document.write( "center=(h,k)=(1,0)\r\n" );
document.write( "vertices=(h,k±a)=(1,4)&(1,-4)\r\n" );
document.write( "focus=(h,k±c)\r\n" );
document.write( "c²=a²+b²\r\n" );
document.write( "c²=16+4\r\n" );
document.write( "c²=20\r\n" );
document.write( "c=√20\r\n" );
document.write( "c=√4*5\r\n" );
document.write( "c=2√5\r\n" );
document.write( "focus=(h,k±c)=(1,±2√5)\r\n" );
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document.write( "asymptotes are the line through center (1,0)\r\n" );
document.write( "with slope  =  = ±2\r\n" );
document.write( "Use point slope form.\r\n" );
document.write( "They are y = 2x-2 and y = -2x+2 \r\n" );
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document.write( "For the hyperbola on the right:\r\n" );
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document.write( "center=(h,k)=(2,3)\r\n" );
document.write( "vertices=(h±a,k)=(4,3)&(0,3)\r\n" );
document.write( "focus=(h±c,k)\r\n" );
document.write( "c²=a²+b²\r\n" );
document.write( "c²=4+1\r\n" );
document.write( "c²=5\r\n" );
document.write( "c=2√5\r\n" );
document.write( "focus=(h±c,k)=(2±√5,3)\r\n" );
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document.write( "asymptotes are the line through center (2,3)\r\n" );
document.write( "with slope  = \r\n" );
document.write( "Use point slope form.\r\n" );
document.write( "They are y = x+2 and y = x+4 \r\n" );
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document.write( "Edwin

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