Algebra.Com's Answer #648719 by ikleyn(52778)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "find the equation of the tangent and normal of the curve x^2-2y^2+4=0, perpendicular to 2x-y+7=0
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document.write( "You, probably, know how to write an equation of the straight line perpendicular to 2x-y+7 = 0. \r\n" );
document.write( "It is 2y+x+c = 0. It is the entire family of parallel lines depending on the coefficient \"c\". \r\n" );
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document.write( "Now, to get the tangent line to the given hyperbola, you need to find the coefficient \"c\" (\"the line\") from the system\r\n" );
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document.write( " =  (1)\r\n" );
document.write( "x + 2y + c = 0 (2)\r\n" );
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document.write( "under the condition that the system has one and only one solution.\r\n" );
document.write( "In this way you will get the tangent line.\r\n" );
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document.write( "How to solve the system (1), (2)?\r\n" );
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document.write( "Very simple. Express x from (2) and substitute it into (1).\r\n" );
document.write( "You will get a single quadratic equation for y.\r\n" );
document.write( "The condition that it has one and only one real solution is vanishing its discriminant d = 0.\r\n" );
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document.write( "So, express that \"d\" via \"c\" and find \"c\" as a root of the equation d = 0.\r\n" );
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document.write( "For inspiration, see the lesson\r\n" );
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document.write( "Tangent lines and normal vectors to a hyperbola. \r\n" );
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document.write( "By the way, the curve x^2-2y^2+4=0 is a hyperbola.\r\n" );
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document.write( "Good luck! \r
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