document.write( "Question 1152643: Find the value/s of U so that the graph of the equation of Ux2 + y^2 − 2Ux = 0 is a hyperbola.\r
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Algebra.Com's Answer #854695 by KMST(5398)\"\" \"About 
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I believe the equation was meant to be \"Ux%5E2+%2B+y%5E2+-+2Ux+=+0\" .
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\n" ); document.write( "FACTS ABOUT HYPERBOLAS:
\n" ); document.write( "The equation for a hyperbola centered at (h,k) can be written as
\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=%22+%22%2B-+1\" for some pair (a,b) of positive numbers.
\n" ); document.write( "For \"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=-1\" <--> \"%28y-k%29%5E2%2Fb%5E2-%28x-h%29%5E2%2Fa%5E2=1\" <--> \"%28y-k%29%5E2%2Fb%5E2=%28x-h%29%5E2%2Fa%5E2%2B1\" , you see that it must be \"%28y-k%29%5E2%2Fb%5E2%3E=1\" --> \"abs%28y-k%29%3E=b\" , so the graph will have an upper branch and a lower branch, like this:
\n" ); document.write( " The red and green lines are the asymptotes, with slopes \"b%2Fa\" and \"-b%2Fa\" .
\n" ); document.write( "In the other hand, for \"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\" <--> \"%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2%2B1\" , so it must be \"%28x-h%29%5E2%2Fa%5E2%3E=1\" <--> \"abs%28x-h%29%3E=a\" , so thee is a left branch and a tight branch to the graph, that looks like this:
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\n" ); document.write( "SOLVING THE PROBLEM:
\n" ); document.write( "Adding \"U\" to both sides of \"Ux%5E2+%2B+y%5E2+-+2Ux+=+0\" , we get
\n" ); document.write( "\"Ux%5E2-2Ux%2BU%2By%5E2=U\" --> \"U%28x%5E2-2x%2B1%29%2By%5E2=U\" --> \"U%28x-1%29%5E2%2By%5E2=U\" --> \"highlight%28%28x-1%29%5E2%2By%5E2%2FU=1%29\"
\n" ); document.write( "That equation represents a hyperbola for any negative value of \"U\" .
\n" ); document.write( "If allowed to choose a value for \"U\" to calculate distances, I would choose \"U=-1\" .
\n" ); document.write( "That makes the equation \"%28x-1%29%5E2-y%5E2=1\" , matching the general equation \"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\" with \"system%28h=1%2Ck=0%2Ca=1%2Cb=1%29\".
\n" ); document.write( "The equation represents a hyperbola with center (1,0),
\n" ); document.write( "vertices \"V%280%2C0%29\" and \"W%282%2C0%29\" , asymptotes \"y=%22+%22%2B-+%28x-1%29\" , focal distance \"c=sqrt%28a%5E2%2Bb%5E2%29=sqrt%282%29\" , and foci \"F\" and \"%22F%27%22\" at \"%22%28%22\"\"1+%2B-+sqrt%282%29\"\"%22%2C+0%29%22\" .
\n" ); document.write( "The distance between the foci is \"2sqrt%282%29\" .
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\n" ); document.write( "If a generic \"U\" is expected, we can say \"U=-b%5E2\" for any \"b%3E0\"
\n" ); document.write( "Then we have \"%28x-1%29%5E2-y%5E2%2Fb%5E2=1\" .
\n" ); document.write( "That equation represents a hyperbola with center (1,0),
\n" ); document.write( "vertices \"V%280%2C0%29\" and \"W%282%2C0%29\" , asymptotes \"y=%22+%22%2B-+b%28x-1%29\" , focal distance \"c=sqrt%281%2Bb%5E2%29\" , and foci at \"%22%28%22\"\"1+%2B-+sqrt%281%2Bb%5E2%29\"\"%22%2C+0%29%22\" .
\n" ); document.write( "The distance between the foci is \"2sqrt%281%2Bb%5E2%29\" .
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