document.write( "Question 598440: Direction: Write an equation for each hyperbola. \r
\n" ); document.write( "\n" ); document.write( "vertices: (1,1) and (1,-3)
\n" ); document.write( "foci: {1, -1 +or- sqrt(5)}\r
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\n" ); document.write( "\n" ); document.write( "I would like to say please and thank you for helping me, in advance.(:
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Algebra.Com's Answer #378587 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
Direction: Write an equation for each hyperbola.
\n" ); document.write( "vertices: (1,1) and (1,-3)
\n" ); document.write( "foci: {1, -1 +or- sqrt(5)}
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\n" ); document.write( "This is a hyperbola with vertical transverse axis.
\n" ); document.write( "Its standard form of equation: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center.
\n" ); document.write( "For given hyperbola:
\n" ); document.write( "x-coordinate of center=1
\n" ); document.write( "y-coordinate of center=(1-3)/2=-2/2=-1 (by midpoint formula)
\n" ); document.write( "center: (1,-1)
\n" ); document.write( "length of vertical transverse axis=4 (-3 to 1)=2a
\n" ); document.write( "a=2
\n" ); document.write( "a^2=4
\n" ); document.write( "Given coordinates of Foci: (1,-1±√5)=(1,-1±c)
\n" ); document.write( "c=√5
\n" ); document.write( "c^2=5
\n" ); document.write( "c^2=a^2+b^2
\n" ); document.write( "b^2=c^2-a^2=5-4=1
\n" ); document.write( "Equation:
\n" ); document.write( "(y+1)^2/4-(x-1)^2/1=1
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