document.write( "Question 1143793: what the slope of the tangent line of the curve x^2-xy+y^2=3 at a point (1,2) \n" ); document.write( "
Algebra.Com's Answer #764694 by ikleyn(52776)\"\" \"About 
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document.write( "To answer the question, find the derivative y' = \"%28dy%29%2F%28dx%29\".\r\n" );
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document.write( "For it, differentiate the given equation. You will get\r\n" );
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document.write( "    2x - y - xy' + 2y*y' = 0,\r\n" );
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document.write( "which implies\r\n" );
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document.write( "    2x - y = xy' - 2y*y',    or\r\n" );
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document.write( "    2x - y = y'*(x - 2y)\r\n" );
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document.write( "    y' = \"%282x-y%29%2F%28x-2y%29\".\r\n" );
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document.write( "Now substitute the coordinates of the given point  x= 1, y= 2  into the last formula to get\r\n" );
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document.write( "    y' = \"%282%2A1-2%29%2F%281+-+2%2A2%29\" = 0.\r\n" );
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document.write( "ANSWER.  The slope of the tangent line of the curve x^2-xy+y^2=3 at the point (1,2) is equal to 0 (zero, ZERO).\r\n" );
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