document.write( "Question 576840: determine the equation of the hyperbola whose asymptotes are y=±2x and which passes through (5/2, 3) \n" ); document.write( "
Algebra.Com's Answer #370681 by lwsshak3(11628)\"\" \"About 
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determine the equation of the hyperbola whose asymptotes are y=±2x and which passes through (5/2, 3)
\n" ); document.write( "**
\n" ); document.write( "given asymptotes show that center of given hyperbola is at (0,0) because y-intercepts of straight line asymptotes=0
\n" ); document.write( "center:(0,0)
\n" ); document.write( "..
\n" ); document.write( "Given point thru which hyperbola passes is below the asymptote with the positive slope of 2; therefore, hyperbola has a horizontal transverse axis,and it follows that the slope is ±b/a
\n" ); document.write( "b/a=2
\n" ); document.write( "b=2a
\n" ); document.write( "..
\n" ); document.write( "Equation:
\n" ); document.write( "x^2/a^2-y^2/b^2=1
\n" ); document.write( "x^2/a^2-y^2/4a^2=1
\n" ); document.write( "LCD:4a^2
\n" ); document.write( "4(x^2)-y^2=4a^2
\n" ); document.write( "plug in coordinates of given point (5/2,3)
\n" ); document.write( "4(5/2)^2-3^2=4a^2
\n" ); document.write( "4(25/4)-9=4a^2
\n" ); document.write( "25-9=4a^2
\n" ); document.write( "16=4a^2
\n" ); document.write( "a^2=4
\n" ); document.write( "a=2
\n" ); document.write( "a^2=4
\n" ); document.write( "b=2a=4
\n" ); document.write( "b^2=16
\n" ); document.write( "..
\n" ); document.write( "Equation of hyperbola:
\n" ); document.write( "x^2/4-y^2/16=1\r
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