document.write( "Question 379778: Given: x^2 - y^2 = 8x -2y -13. Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please. \n" ); document.write( "
Algebra.Com's Answer #269660 by Edwin McCravy(20065) $\"\"$ $\"About$
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Given: $\"x%5E2+-+y%5E2+=+8x+-+2y+-+13\"$ . Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please
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document.write( "This is a hyperbola because the  and the  term have opposite\r\n" );
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document.write( "Get it like this:\r\n" );
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document.write( "Put parentheses around the first two terms:\r\n" );
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document.write( "Factor -1 out of the last two terms on the left\r\n" );
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document.write( "Take half of -8, get -4, square it get +16, add +16 inside the \r\n" );
document.write( "first parentheses and add +16 to the right side:\r\n" );
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document.write( "Take half of -2, get -1, square it get +1, add +1 inside the second\r\n" );
document.write( "parentheses, but because of the - in front of the second parentheses\r\n" );
document.write( "on the left we add -1 to the right side:\r\n" );
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document.write( "Factor the two parentheses as perfect squares, Combine the terms on the\r\n" );
document.write( "right.\r\n" );
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document.write( "Get a 1 on the right side by dividing every term through by 2\r\n" );
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document.write( "Compare that to\r\n" );
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document.write( "and since the term in x is the positive one, we know the hyperbola \r\n" );
document.write( "opens right and left. (As we know the x-axis goes right and left, a\r\n" );
document.write( "way to remember it).\r\n" );
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document.write( "Comparing further, center = (h,k) = (4,1)\r\n" );
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document.write( "We plot the center\r\n" );
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document.write( "Comparing further:\r\n" );
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document.write( " so \r\n" );
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document.write( "Length of semi-transverse axis = a = \r\n" );
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document.write( "So we draw the transverse axis  about 1.4 units\r\n" );
document.write( "to the right and to the left of the center:\r\n" );
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document.write( "The ends of the transverse axis are the vertices. We subtract \"a\" \r\n" );
document.write( "from \"h\" to get the x-coordinate of the left vertex, so the left\r\n" );
document.write( "vertex is \r\n" );
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document.write( "(4-,1).  \r\n" );
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document.write( "We add \"a\" to \"h\" to get the x-coordinate of the right vertex, so \r\n" );
document.write( "the right vertex is V(4+,1).  They have the same \r\n" );
document.write( "y-coordinate as the center. \r\n" );
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document.write( " so \r\n" );
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document.write( "Length of semi-conjugate axis = b = \r\n" );
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document.write( "So we draw the conjugate axis  about 1.4 units\r\n" );
document.write( "above and below the center:\r\n" );
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document.write( "Next we draw the defining rectangle around the two axes:\r\n" );
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document.write( "We draw the asymptotes by drawing the extended diagonals of the\r\n" );
document.write( "defining rectangle:\r\n" );
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document.write( "These asymptotes have slopes \r\n" );
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document.write( "The go through the center (4,1) so their equations are gotten using\r\n" );
document.write( "the point-slope formula:\r\n" );
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document.write( "Finding the equation of the asymptote with the positive slope\r\n" );
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document.write( "Finding the equation of the asymptote with the begative slope\r\n" );
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document.write( "Finally we sketch in the hyperbola:\r\n" );
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document.write( "Edwin
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