document.write( "Question 572756: Give the equation in standard form of the hyperbola with vertices (-4,2)
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Algebra.Com's Answer #368516 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Give the equation in standard form of the hyperbola with vertices (-4,2)\r\n" );
document.write( "and (1,2) and foci (-7,2) and (4,2). Give the center and the asymptotes. \r\n" );
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document.write( "In a message dated 6/24/2011 2:53:25 P.M. Eastern Daylight Time, AnlytcPhil@aol.com writes:\r\n" );
document.write( "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)\r\n" );
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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 the vertices to find the transverse axis:\r\n" );
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document.write( "We can see that the transverse axis is 5 units long, and since the\r\n" );
document.write( "transverse axis is 2a units long, then 2a=5 and a=\"5%2F2\"\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%2F2\",2), so\r\n" );
document.write( "we have (h,k) = (\"-3%2F2\",2).  So we plot the center:\r\n" );
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document.write( "To find a, we subtract the x-coordinate of \r\n" );
document.write( "the center from the x-coordinate of the right vertex, and get\r\n" );
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document.write( "1 - (\"-3%2F2\") = \"2%2F2\" + \"3%2F2\" = \"5%2F2\"\r\n" );
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document.write( "So a=\"5%2F2\". \r\n" );
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document.write( "We are given the foci (-7,2) and (4,2)\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 c, we 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 = 4 - (\"-3%2F2%29\" = \"8%2F2\" + \"3%2F2\" = \"11%2F2\"\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( "c² = a² + b²\r\n" );
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document.write( "Substituting for c and a\r\n" );
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document.write( "\"%2811%2F2%29%5E2\" = (\"5%2F2\")² + b²\r\n" );
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document.write( "\"121%2F4\" = \"25%2F4\" + b²\r\n" );
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document.write( "\"121%2F4\" - \"25%2F4\" = b²\r\n" );
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document.write( "\"96%2F4\" = b²\r\n" );
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document.write( "24 = b²\r\n" );
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document.write( "\"sqrt%2824%29\" = b\r\n" );
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document.write( "\"2sqrt%286%29\" = b\r\n" );
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document.write( "Now we can give the standard equation of the hyperbola, since we now\r\n" );
document.write( "know h, k, a, and b, a² and b²:\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%2F2%29%29%5E2%2F%285%2F2%29%5E2-%28y-2%29%5E2%2F24\" = 1\r\n" );
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document.write( "\"%28x%2B3%2F2%29%5E2%2F%2825%2F4%29-%28y-2%29%5E2%2F24\" = 1\r\n" );
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document.write( "Next we draw in the conjugate axis which is 2b units\r\n" );
document.write( "or \"2sqrt%286%29\" or about 4.9 units long with the center as its midpoint.\r\n" );
document.write( "That is, we draw a vertical line \"2sqrt%286%29\", about 2.45 units\r\n" );
document.write( "upward and the same number of units downward from the center:\r\n" );
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document.write( "Next we draw the defining 2a×2b rectangle which has the transverse axis \r\n" );
document.write( "and the conjugate axis as perpendicular bisectors of its sides:\r\n" );
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document.write( "Next we draw the extended diagonals of the defining rectangle:\r\n" );
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document.write( "We can now sketch in the hyperbola:\r\n" );
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document.write( "But we still have to find the equations of those two blue\r\n" );
document.write( "asymptotes.\r\n" );
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document.write( "We know one point they go through, namely the center (\"-3%2F2\",2)\r\n" );
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document.write( "We need to know the slopes of the two asymptotes. They are \r\n" );
document.write( "\"rise%2Frun\" = ±\"b%2Fa\" = ±\"2sqrt%286%29%2F%285%2F2%29\" = ±\"2sqrt%286%29\"·\"2%2F5\" = ±\"4sqrt%286%29%2F5\"\r\n" );
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document.write( "Now we use the point-slope form.\r\n" );
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document.write( "y - y1 = m(x - x1)\r\n" );
document.write( "y - 2 = ±\"4sqrt%286%29%2F5\"(x - (\"-5%2F2\"))\r\n" );
document.write( "y - 2 = ±\"4sqrt%286%29%2F5\"(x + \"5%2F2\")\r\n" );
document.write( "    y = 2 ± \"4sqrt%286%29%2F5\"(x + \"5%2F2\")\r\n" );
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document.write( "One asymptote has the equation with the positive slope,\r\n" );
document.write( "and the other has the equation with the negative slope.\r\n" );
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document.write( "Edwin
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