document.write( "Question 196562: May you please help me with the following hyperbola in standard form:\r
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Algebra.Com's Answer #147328 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
find the center, foci, the length of one of the two axes (transverse or conjugate) which is parallel to the y-axis, and the two asymptotes.\r
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document.write( "Rewrite that as:\r\n" );
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document.write( "\"%28%28x-1%29%5E2%29%2F10+-+%28%28y-1%29%5E2%29%2F1+=+1\"\r\n" );
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document.write( "and compare with\r\n" );
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document.write( "\"%28%28x-h%29%5E2%29%2Fa%5E2+-+%28%28y-k%29%5E2%29%2Fb%5E2+=+1\"\r\n" );
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document.write( "\"h+=+1\", \"k=1\", \r\n" );
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document.write( "\"a%5E2=10\", so \"a=sqrt%2810%29\"\r\n" );
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document.write( "\"b%5E2=1\", so \"b=1\"\r\n" );
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document.write( "The center (h,k) = (1,1)\r\n" );
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document.write( "We start out plotting the center C(h,k) = C(1,1)\r\n" );
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document.write( "Next we draw the left semi-transverse axis,\r\n" );
document.write( "which is a segment \"a=sqrt%2810%29\" units long horizontally \r\n" );
document.write( "left from the center.  This semi-transverse \r\n" );
document.write( "axis ends up at one of the two vertices (\"1-sqrt%2810%29\",1).\r\n" );
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document.write( "We'll call it V1(\"1-sqrt%2810%29\",1).:\r\n" );
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document.write( "Next we draw the right semi-transverse axis,\r\n" );
document.write( "which is a segment \"a=sqrt%2810%29\" units long horizontally \r\n" );
document.write( "right from the center. This other semi-transverse\r\n" );
document.write( "axis ends up at the other vertex (\"1%2Bsqrt%2810%29\",1).\r\n" );
document.write( "We'll call it V2(\"1%2Bsqrt%2810%29\",1).:\r\n" );
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document.write( "That's the whole transverse (\"trans\"=\"across\",\r\n" );
document.write( "\"verse\"=\"vertices\", the line going across from\r\n" );
document.write( "one vertex to the other. It is 2a in length,\r\n" );
document.write( "so the length of the transverse axis is \"2a=2sqrt%2810%29\"\r\n" );
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document.write( "Next we draw the upper semi-conjugate axis,\r\n" );
document.write( "which is a segment b=1 units long verically \r\n" );
document.write( "upward from the center.  This semi-conjugate\r\n" );
document.write( "axis ends up at (1,2).\r\n" );
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document.write( "Next we draw the lower semi-conjugate axis,\r\n" );
document.write( "which is a segment b=1 units long verically \r\n" );
document.write( "downward from the center.  This semi-conjugate\r\n" );
document.write( "axis ends up at (1,0). \r\n" );
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document.write( "That's the complete conjugate axis. It is 2b in length,\r\n" );
document.write( "so the length of the transverse axis is 2b=2(1)=2\r\n" );
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document.write( "Next we draw the defining rectangle which has the\r\n" );
document.write( "ends of the transverse and conjugate axes as midpoints\r\n" );
document.write( "of its sides:\r\n" );
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document.write( "Next we draw and extend the two diagonals of this defining\r\n" );
document.write( "rectangle:\r\n" );
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document.write( "Now we can sketch in the hyperbola:\r\n" );
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document.write( "Next we find the equations of the two asymptotes.\r\n" );
document.write( "Their slopes are ±\"b%2Fa\" or ±\"1%2Fsqrt%2810%29\"\r\n" );
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document.write( "The asymptote that has slope \"1%2Fsqrt%2810%29\" goes through the center\r\n" );
document.write( "C(1,1), so its equation is found using the point-slope\r\n" );
document.write( "formula:\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-%281%29=%281%2Fsqrt%2810%29%29%28x-1%29%29\"\r\n" );
document.write( "\"y-1=%281%2Fsqrt%2810%29%29%28x-1%29%29\"\r\n" );
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document.write( "Multiply through by \"sqrt%2810%29\"\r\n" );
document.write( "\"sqrt%2810%29y-sqrt%2810%29=x-1\"\r\n" );
document.write( "\"-x%2Bsqrt%2810%29y=sqrt%2810%29-1\"\r\n" );
document.write( "\"x-sqrt%2810%29y=-sqrt%2810%29%2B1\"\r\n" );
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document.write( "The asymptote that has slope \"-1%2Fsqrt%2810%29\" goes through the center\r\n" );
document.write( "C(1,1), so its equation is also found using the point-slope\r\n" );
document.write( "formula:\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-%281%29=%28-1%2Fsqrt%2810%29%29%28x-1%29%29\"\r\n" );
document.write( "\"y-1=%28-1%2Fsqrt%2810%29%29%28x-1%29%29\"\r\n" );
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document.write( "Multiply through by \"sqrt%2810%29\"\r\n" );
document.write( "\"sqrt%2810%29y-sqrt%2810%29=-1%28x-1%29\"\r\n" );
document.write( "\"sqrt%2810%29y-sqrt%2810%29=-x%2B1\"\r\n" );
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document.write( "\"x%2Bsqrt%2810%29y=sqrt%2810%29%2B1\"\r\n" );
document.write( "\"x-sqrt%2810%29y=-sqrt%2810%29%2B1\"\r\n" );
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document.write( "All that's left is to find the two foci.\r\n" );
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document.write( "The distance from the vertex, through the center\r\n" );
document.write( "to each foci is c units. We calculate c from\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
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document.write( "\"c%5E2=%28sqrt%2810%29%29%5E2%2B%281%29%5E2\"\r\n" );
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document.write( "\"c%5E2=%2810%2B1%29\"\r\n" );
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document.write( "\"c%5E2=11\"\r\n" );
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document.write( "\"c=sqrt%2811%29\"\r\n" );
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document.write( "So the left focus is \"sqrt%2811%29\" units\r\n" );
document.write( "left of the center (1,1), so the left\r\n" );
document.write( "focus is the point (\"1-sqrt%2811%29\",1).  That's\r\n" );
document.write( "just a little left of the vertex V1. And the right\r\n" );
document.write( "focus is \"sqrt%2811%29\" units right of the center\r\n" );
document.write( "(1,1), so the right focus is the point (\"1-sqrt%2811%29\",1).\r\n" );
document.write( "That's just a little right of the vertex V2. I won't\r\n" );
document.write( "bother to plot them.\r\n" );
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document.write( "Edwin
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