Algebra.Com's Answer #814765 by ikleyn(52787)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Find the locus of point if point moves in such way that the difference of its distance from two points (8, 0) and (-8, 0) always remains 4?
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document.write( "If the difference of distances of the point from two given points is a constant, then (it is a known fact) the locus is a HYPERBOLA.\r\n" );
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document.write( "The problem is to write an equation if this hyperbola.\r\n" );
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document.write( "There are two ways to do it.\r\n" );
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document.write( " One way is long and tortured: it is to write literally what is given\r\n" );
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document.write( "  -  = 4,\r\n" );
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document.write( " then to isolate square roots and square both sides;\r\n" );
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document.write( " then to isolate square root again and square both sides AGAIN;\r\n" );
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document.write( " then simplify, and at the end you will get the sought equation.\r\n" );
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document.write( " For the general case, this way is shown in many standard textbooks.\r\n" );
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document.write( " You can read it from the lesson\r\n" );
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document.write( " - Hyperbola focal property \r\n" );
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document.write( " in this site.\r\n" );
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document.write( "Another way is shorter; but it assumes that you just have some preliminaries knowledge on the subject.\r\n" );
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document.write( " So, the focuses of the hyperbola are (-8,0) and (8,0); thus the transverse axis is x-axis and the hyperbola is centered at (0,0);\r\n" );
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document.write( " one vertex is (x,0), the other vertex is (-x,0);\r\n" );
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document.write( " for the vertex (x,0) the difference of distances from the focuses gives this equation\r\n" );
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document.write( " (8-x) - (x-(-8)) = 4 ====> 8-x - (x+8) = 4 ====> 8 - x - x - 8 = 4 ====> -2x = 4 ====> x = -2.\r\n" );
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document.write( " So, the vertices are (-2,0) and (2,0).\r\n" );
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document.write( " Thus the transverse semi-axis is a =  = 2 units long,\r\n" );
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document.write( " and for the imaginary semi-axis \"b\" we have  +  =  , where c = 8 is half the distance between the focuses;\r\n" );
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document.write( " so  + =  , or = 64 - 4 = 60, b =  .\r\n" );
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document.write( " Thus the canonic equation of the hyperbola is\r\n" );
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document.write( "  -  = 1.\r\n" );
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document.write( "Solved.\r
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document.write( "For the preliminaries knowledge on hyperbolas, see the lessons\r
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document.write( " - Hyperbola definition, canonical equation, characteristic points and elements \r
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document.write( " - Hyperbola focal property \r
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document.write( " - Tangent lines and normal vectors to a hyperbola \r
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document.write( " - Optical property of a hyperbola.\r
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document.write( "in this site.\r
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