document.write( "Question 1193003: Given the hyperbola with the equation y^2 -9x^2 =-9
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Algebra.Com's Answer #824949 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Instead of doing your homework for you, I'll change 9 to 16\r\n" );
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document.write( "\"y%5E2+-16x%5E2+=-16\"\r\n" );
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document.write( "You will do yours exactly like this, step by step:\r\n" );
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document.write( "The standard form of a hyperbola is either\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2\"\"%22%22=%22%22\"\"1\" for a hyperbopla like this \" )( \"\r\n" );
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document.write( "or  \r\n" );
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document.write( "\"%28y-k%29%5E2%2Fa%5E2%2B%28x-h%29%5E2%2Fb%5E2\"\"%22%22=%22%22\"\"1\" for a hyperbola like this \"  \"\r\n" );
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document.write( "Either way, we get 1 on the right side, by dividing through by -16\r\n" );
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document.write( "\"y%5E2%2F%28-16%29+-16x%5E2%2F%28-16%29+=%28-16%29%2F%28-16%29\"\r\n" );
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document.write( "Both terms on the left must have a denominator showing, so we put\r\n" );
document.write( "1 under the second term on the left:\r\n" );
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document.write( "\"-y%5E2%2F16%2Bx%5E2%2F1+=1\"\r\n" );
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document.write( "Reverse the terms on the left, so that the \"plus\" term is first:\r\n" );
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document.write( "\"x%5E2%2F1-y%5E2%2F16=1\"\r\n" );
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document.write( "We finish by rewriting x as (x-0) and y as (y-0)\r\n" );
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document.write( "\"%28x-0%29%5E2%2F1-%28y-0%29%5E2%2F16=1\"\r\n" );
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document.write( "Now that we have it in standard form, we see that it is a hyperbola\r\n" );
document.write( "that looks like this \" )( \".\r\n" );
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document.write( "Comparing it to \"%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "center = (h,k) = (0,0), a2 = 1, b2 = 16, so\r\n" );
document.write( "a = semi-transverse axis = √1 = 1 and \r\n" );
document.write( "b =semi-conjugate axis = √16 = 4\r\n" );
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document.write( "The defining rectangle has corners (h±a,k±b) = (0±1,0±4) or\r\n" );
document.write( "(1,4), (-1,4), (1,-4), (-1,-4), and the asymptotes are the\r\n" );
document.write( "extended diagonals of the defining rectangle:\r\n" );
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document.write( "    \r\n" );
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document.write( "The vertices are the ends of the transverse axis (1,0) and (-1,0),\r\n" );
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document.write( "the foci are just beyond the vertices.\r\n" );
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document.write( "To find the foci, we must find c by the Pythagorean relation for\r\n" );
document.write( "hyperbolas:\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
document.write( "\"c%5E2=1%5E2%2B4%5E2\"\r\n" );
document.write( "\"c%5E2=1%2B16\"\r\n" );
document.write( "\"c%5E2=17\"\r\n" );
document.write( "\"c=sqrt%2817%29\"\r\n" );
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document.write( "So the foci are \"%28matrix%281%2C3%2C1%2C%22%2C%22%2C+%22%22+%2B-+sqrt%2817%29%29%29\"\r\n" );
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document.write( "I'll draw them in:\r\n" );
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document.write( "We find the equation of the asymptote that leans to the right.\r\n" );
document.write( "It passes through the points (0,0) and (1,4).  It has slope\r\n" );
document.write( "\"m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29+=+%284-0%29%2F%281-0%29+=+4%2F1+=+4\".\r\n" );
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document.write( "It has equation\r\n" );
document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-0=4%28x-0%29\"\r\n" );
document.write( "\"y=4x\"\r\n" );
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document.write( "We find the equation of the asymptote that leans to the left.\r\n" );
document.write( "It passes through the points (0,0) and (-1,4).  It has slope\r\n" );
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document.write( "It has equation\r\n" );
document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-0=-4%28x-0%29\"\r\n" );
document.write( "\"y=-4x\"\r\n" );
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document.write( "Now do yours the same way, step-by-step.\r\n" );
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document.write( "Edwin
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