document.write( "Question 1039199: A hyperbola has a vertical transverse axis of length 14 and asymptotes of
\n" ); document.write( "y=9/8x + 1 and y= -9/8x - 9. Find the center of the hyperbola, its focal length, and its eccentricity. \r
\n" ); document.write( "\n" ); document.write( "Can you please show me the work to get the center mainly?
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Algebra.Com's Answer #653970 by KMST(5328)\"\" \"About 
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The center is the intersection of the asymptotes.
\n" ); document.write( "That is the point that is a solution to \"system%28y=%289%2F8%29x%2B1%2Cy=-%289%2F8%29x-9%29\" .
\n" ); document.write( "You can solve that system of equations by substitution.
\n" ); document.write( "For example, you could start by substituting the expression \"%289%2F8%29x%2B1\" for \"y\" in \"y=-%289%2F8%29x-9\" .
\n" ); document.write( "\"system%28y=%289%2F8%29x%2B1%2Cy=-%289%2F8%29x-9%29\" --> \"system%28y=%289%2F8%29x%2B1%2C%289%2F8%29x%2B1=-%289%2F8%29x-9%29\" --> \"system%28y=%289%2F8%29x%2B1%2C%289%2F8%29x%2B%289%2F8%29x=-9-1%29\" --> \"system%28y=%289%2F8%29x%2B1%2C%289%2F4%29x=-10%29\" --> \"system%28y=%289%2F8%29x%2B1%2Cx=-10%284%2F9%29%29\" --> \"system%28y=%289%2F8%29x%2B1%2Cx=-40%2F9%29\" ,
\n" ); document.write( "and then, you could substitute the value found for one variable into one of the equations, to find the value of the other variable:
\n" ); document.write( "\"system%28y=%289%2F8%29x%2B1%2Cx=-40%2F9%29\" --> \"system%28y=%289%2F8%29%28-40%2F9%29%2B1%2Cx=-40%2F9%29\" --> \"system%28y=%289%2F8%29%28-40%2F9%29%2B1%2Cx=-40%2F9%29\" --> \"system%28y=-5%2B1%2Cx=-40%2F9%29\" --> \"highlight%28system%28y=-4%2Cx=-40%2F9%29%29\"
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\n" ); document.write( "The asymptotes slopes give you the \"b%2Fa=9%2F8\" ratio of conjugate to transverse axes.
\n" ); document.write( "The transverse axis is the segment joining the vertices of the hyperbola.
\n" ); document.write( "If the length is \"14\" , the distance from the center to each vertex must be
\n" ); document.write( "\"a=14%2F2=7\" .
\n" ); document.write( "From there you can find \"b\" , \"c=sqrt%28b%5E2%2Ba%5E2%29\" , and the eccentricity: \"c%2Fa\" .
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