through geometry and proportions, Omar Khayyam found that the solution to that problem was the same as an intersection of a parabola and a circle!
6
Expression 7: left parenthesis, "x" minus StartFraction, "b" Over 2 "a" , EndFraction , right parenthesis squared plus "y" squared equals left parenthesis, StartFraction, "b" Over 2 "a" , EndFraction , right parenthesis squaredx−b2a2+y2=b2a2
7
Expression 8: "y" equals StartFraction, "x" squared Over StartRoot, "a" , EndRoot , EndFractiony=x2a
8
can you show a solution of the red system is also a solution of the blue?