document.write( "Question 252347: If P is the any point on the hyperbola whose axis are equal,prove that SP.SP'=CP^2. please explain it completely. \n" ); document.write( "
Algebra.Com's Answer #184342 by Edwin McCravy(20060)\"\" \"About 
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If P is the any point on the hyperbola whose axis are equal,prove that \"SP%2AS%27P=CP%5E2\", where S and S' are the foci, and C is the center.
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document.write( "Let the tranverse axis be along the x-axis and the conjugate axis be\r\n" );
document.write( "along the y-axis, with the center C at (0,0), the origin.\r\n" );
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document.write( "Let both axes be 2, so that both the semi-tranverse axis, \"a\", and \r\n" );
document.write( "semi-conjugate axis, \"b\", are 1 each.   Then the equation of the \r\n" );
document.write( "hyperbola, which is \"x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1\", becomes simply \"x%5E2-y%5E2=1\".\r\n" );
document.write( "In a hyperbola, \"c%5E2=a%5E2%2Bb%5E2\", so \"c%5E2=1%5E2%2B1%5E2=1%2B1=2\", therefore\r\n" );
document.write( "\"c+=+sqrt%282%29\", where \"c\" is the distance from the center to the focus.\r\n" );
document.write( "Therefore S and S' are the points (\"%22%22%2B-sqrt%282%29\",0).  \r\n" );
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document.write( "[Do not confuse the center \"C(0,0)\" with the value of \"c\", the distance \r\n" );
document.write( "from the center to each focus.]\r\n" );
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document.write( "Let P(x,y) be any arbitrary point on the hyperbola:\r\n" );
document.write( "The blue line is CP:\r\n" );
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document.write( "The graph is:\r\n" );
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document.write( "Using the distance formula to find SP and SP' in terms of x:\r\n" );
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document.write( "Since the equation of the hyperbola is \"x%5E2-y%5E2=1\", then \"y%5E2=x%5E2-1\",\r\n" );
document.write( "so substituting we get:\r\n" );
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document.write( "\"SP+=+sqrt%28x%5E2-2x%2Asqrt%282%29%2B2%2Bx%5E2-1%29=+sqrt%282x%5E2-2x%2Asqrt%282%29%2B1%29\"\r\n" );
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document.write( "Similarly,\r\n" );
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document.write( "As before, since the equation of the hyperbola is \"x%5E2-y%5E2=1\",\r\n" );
document.write( "then \"y%5E2=x%5E2-1\", so substituting we get:\r\n" );
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document.write( "So \r\n" );
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document.write( "Multiplying under the radicals:\r\n" );
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document.write( "Next we use the distance formuls to find CP where C is the origin (0,0).\r\n" );
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document.write( "\"CP+=+sqrt%28%28x-0%29%5E2%2B%28y-0%29%5E2%29=sqrt%28x%5E2%2By%5E2%29\"\r\n" );
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document.write( "Since the equation of the hyperbola is \"x%5E2-y%5E2=1\", then \"y%5E2=x%5E2-1\",\r\n" );
document.write( "so substituting we get\r\n" );
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document.write( "\"CP+=sqrt%28x%5E2%2By%5E2%29=sqrt%28x%5E2%2Bx%5E2-1%29=+sqrt%282x%5E2-1%29\"\r\n" );
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document.write( "so \"CP%5E2+=+%28sqrt%282x%5E2-1%29%29%5E2+=+abs%282x%5E2-1%29\"\r\n" );
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document.write( "Therefore \"SP%2A%22SP%27%22+=+CP%5E2\", because both equal \"abs%282x%5E2-1%29\" \r\n" );
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document.write( "Edwin
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