Suggested: if (x-iy)1/3=a-ib then show that x/a+y/b=4(a2-b2) - x/a+y/b=a+b x/a2+y/b2=2 - x/a+y/b=2 ax-by=a2-b2 - x+y=a+b ax-by=a2-b2 - x/a-y/b=0 ax+by=a2+b2 - z=ax+by+a2+b2 - b/ax+a/by=a2+b2 and x+y=2ab - ax-by=a2+b2 x+y=2a - a3+b3/a2+2ab+b2 by a2-b2/a-b - (a-b)x+(a+b)y=a2-2ab-b2 (a+b)(x+y)=a2+b2 - a/x-b/y=0 ab2/x+a2b/y=a2+b2 - find the area enclosed by the ellipse x2/a2+y2/b2=1 - find the area of the ellipse x2/a2+y2/b2=1 by double integration - find the area of the region bounded by the ellipse x2/a2+y2/b2=1 - by=a2+b2 Browse related:
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