document.write( "Question 294406: what is the equation of a hyperbola with vertices (-5,3) (-1,3) and foci (-3-2sqrt(5), 3) and -3-2sqrt(5), 3) \n" ); document.write( "
Algebra.Com's Answer #212300 by Edwin McCravy(20059)\"\" \"About 
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what is the equation of a hyperbola with vertices (-5,3) (-1,3) and foci (\"-3-2sqrt%285%29\", 3) and (\"-3%2B2sqrt%285%29\", 3)
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document.write( "First we plot the vertices:\r\n" );
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document.write( "We see that the hyperbola opens right and left, that is, \r\n" );
document.write( "it looks something like this: )(\r\n" );
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document.write( "So we know its standard equation is this:\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( "We connect them to find the transverse axis:\r\n" );
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document.write( "We can see that the transverse axis is 4 units long, and since the\r\n" );
document.write( "transverse axis is \"2a\" units long, then \"2a=4\" and \"a=2\"\r\n" );
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document.write( "The center of the hyperbola is the midpoint of the transverse axis,\r\n" );
document.write( "and we can see that the midpoint of the transverse axis is (-3,3), so\r\n" );
document.write( "we have (h,k) = (-3,3) and that a=2. \r\n" );
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document.write( "We are given the foci  \r\n" );
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document.write( "foci (\"-3-2sqrt%285%29\",3) and (\"-3%2B2sqrt%285%29\", 3)\r\n" );
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document.write( "We plot the vertex and those two given foci:\r\n" );
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document.write( "The number of units from each of the foci to the center is the value c.\r\n" );
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document.write( "To find that distance we can subtract the x-coordinate of the center \r\n" );
document.write( "from the x-coordinate of the right focus, and get\r\n" );
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document.write( "\"c=%28-3%2B2swrt%285%29%29-%28-3%29=-3%2B2sqrt%285%29%2B3=2sqrt%285%29\"\r\n" );
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document.write( "Next we find b from the Pythagorean relationship common to all\r\n" );
document.write( "hyperbolas, which is\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
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document.write( "Substituting for c and a\r\n" );
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document.write( "\"%282sqrt%285%29%29%5E2=%282%29%5E2%2Bb%5E2\"\r\n" );
document.write( "\"4%2A5=4%2Bb%5E2\"\r\n" );
document.write( "\"20=4%2Bb%5E2\"\r\n" );
document.write( "\"16=b%5E2\"\r\n" );
document.write( "\"4=b\"\r\n" );
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document.write( "Next we draw in the conjugate axis which is 2b units\r\n" );
document.write( "or 8 units long with the center at its midpoint. That is,\r\n" );
document.write( "we draw a vertical line 4 units upward and 4 units downward\r\n" );
document.write( "from the center:\r\n" );
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document.write( "Next we draw the defining rectangle which has the transverse axis and\r\n" );
document.write( "the conjugate asis as perpendicular bisectors of its sides:\r\n" );
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document.write( "Next we draw the extended diagonals of the defining rectangle, which\r\n" );
document.write( "are the asymptotes of the hyperbola:\r\n" );
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document.write( "Finally we can sketch in the hyperbola:\r\n" );
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document.write( "But you only wanted the equation.  I could have answered that when we\r\n" );
document.write( "found b, but you will probably have to graph other hyperbolas, so I\r\n" );
document.write( "thought I would go ahead and complete the graph before giving the equation,\r\n" );
document.write( "which is gotten by substituting (h,k) = (-3,3), a=2, and b=4 in\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( "\"%28x-%28-3%29%29%5E2%2F2%5E2-%28y-3%29%5E2%2F4%5E2=1\"\r\n" );
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document.write( "\"%28x%2B3%29%5E2%2F4-%28y-3%29%5E2%2F16=1\"\r\n" );
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document.write( "Edwin

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