document.write( "Question 1082518: An airplane is flying along the hyperbolic
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Algebra.Com's Answer #696597 by ikleyn(52794) $\"\"$ $\"About$

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document.write( "1.  Our curve is specific !  If (x,y) is the point on the curve then\r\n" );
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document.write( "    2y^2 - x^2 = 8,  which implies  y^2 = .   (1)\r\n" );
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document.write( "2.  The distance from ANY point (x,y) on the coordinate plane to the point (3,0) is\r\n" );
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document.write( "     = .    (2)\r\n" );
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document.write( "    Now, if the point lies on the hyperbola, you have (1), and you can substitute this expression for  into the formula (2).\r\n" );
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document.write( "    You will get\r\n" );
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document.write( "     =  +  =  +  = .   (3)\r\n" );
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document.write( "    So, you need to minimize (3). In other words, you need to find the value of \"x\" which minimizes this quadratic function.\r\n" );
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document.write( "4.  The quadratic function of the general form q(x) =  achieves the maximum at x = .\r\n" );
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document.write( "    In your case this \"x\" is x = -  =  = 2.\r\n" );
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document.write( "    So, we found the value of \"x\". It is x= 2.\r\n" );
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document.write( "    Then the corresponding value of \"y\" on your curve is \r\n" );
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document.write( "          = 8  ---->   = 8 + 4 = 12  ---->   =  = 6.\r\n" );
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document.write( "     Thus your \"closest\" point on the curve is  (x,y) = (,).\r\n" );
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document.write( "5.  Now you can find that minimal distance:\r\n" );
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document.write( "     =  =  = 1 + 6 = 7.\r\n" );
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document.write( "    Hence, the minimal distance itself is   = 2.646 (approximately).\r\n" );
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