document.write( "Question 1167417: Find the equation of the hyperbola (in standard and general forms) which satisfies the conditions given.
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Algebra.Com's Answer #792175 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Instead of doing your problem for you, I'm doing one which is EXACTLY\r\n" );
document.write( "STEP-BY-STEP like yours.  All you have to do is use it as a model and\r\n" );
document.write( "follow it step-by-step.  Here's the problem that I'll solve for you: \r\n" );
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Find the equation of the hyperbola (in standard and general forms) which
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document.write( "We plot the center and foci to see whether the hyperbola is like \r\n" );
document.write( "this ) ( or like this: \r\n" );
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document.write( "The hyperbola is like this \") (\",so its equation is of this form:\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( "where (h,k) = the center = (4,5), \r\n" );
document.write( "and \"a\" = the distance from the center to the vertex, which is the\r\n" );
document.write( "distance from (4,5) to (9,5) which is 9-4=5\r\n" );
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document.write( "so we have this much of the \r\n" );
document.write( "equation:\r\n" );
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document.write( "\"%28x-2%29%5E2%2F5%5E2-%28y-3%29%5E2%2Fb%5E2=1\"\r\n" );
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document.write( "The distance from the center to either focus is the value c,\r\n" );
document.write( "and that is the distance from (4,5) to (4+√41,5), which is \r\n" );
document.write( "(4+√41)-(4) = √41\r\n" );
document.write( "and the Pythagorean relationship for all hyperbolas is\r\n" );
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document.write( "c2=a2+b2.  \r\n" );
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document.write( "\"%28sqrt%2841%29%29%5E2=%285%29%5E2%2Bb%5E2\"\r\n" );
document.write( "\"41=25%2Bb%5E2\"\r\n" );
document.write( "\"16=b%5E2\"\r\n" );
document.write( "\"4=b\"\r\n" );
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document.write( "\"%28x-4%29%5E2%2F5%5E2-%28y-5%29%5E2%2F4%5E2=1\"  <---that's the standard form.\r\n" );
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document.write( "That's the standard form equation. To draw the graph, we make a defining\r\n" );
document.write( "rectangle which is 2a=10 units wide and 2b=8 units high, with the center (4,5)\r\n" );
document.write( "as the center of the defining rectangle, and draw its extended diagonals,\r\n" );
document.write( "which are the asymptotes.\r\n" );
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document.write( "Now we can sketch in the hyperbola:\r\n" );
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document.write( "To get the general form we start with the standard form:\r\n" );
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document.write( "\"%28x-4%29%5E2%2F5%5E2-%28y-5%29%5E2%2F4%5E2=1\"\r\n" );
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document.write( "\"%28x-4%29%5E2%2F25%5E%22%22-%28y-5%29%5E2%2F16%5E%22%22=1\"\r\n" );
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document.write( "We multiply through by the LCD = (25)(16) = 400\r\n" );
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document.write( "\"16%28x-4%29%5E2-25%28y-5%29%5E2=400\"\r\n" );
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document.write( "\"16%28x-4%29%28x-4%29-25%28y-5%29%28y-5%29=400\"\r\n" );
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document.write( "\"16%28x%5E2-4x-4x%2B16%29-25%28y%5E2-5y-5y%2B25%29=400\"\r\n" );
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document.write( "\"16%28x%5E2-8x%2B16%29-25%28y%5E2-10y%2B25%29=400\"\r\n" );
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document.write( "\"16x%5E2-128x%2B256-25y%5E2%2B250y-625=400\"\r\n" );
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document.write( "\"16x%5E2-25y%5E2-128x%2B250y-369=400\"\r\n" );
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document.write( "\"16x%5E2-25y%5E2-128x%2B250y-769=0\"  <--that's the general form.\r\n" );
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document.write( "Now go do yours the exact same way, step by step.\r\n" );
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document.write( "You weren't asked to draw the graph, but it helps.\r\n" );
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document.write( "Edwin

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