Question

积分：Ax^2+2Bxy+Cy^2=1 (AC-B>0) 围成面积

Answer 1

这题可以用旋转坐标系的办法

设再原坐标系中:

x=rcos(p+m)

y=rsin(p+m)

m为常量

坐标系旋转m,在新坐标系中,相应点变为:

x1=rcosp

y1=rsinp

所以:x=rcos(p+m)=rcospcosm-rsinpsinm=cosm*x1-sinm*y1

y=rsin(p+m)=rsinpcosm+rcospsinm=cosm*y1+sinm*x1

代入Ax^2+2Bxy+Cy^2=1,得:

(A(cosm)^2+Bsin2m+C(sinm)^2)x1^2+(A(sinm)^2-Bsin2m+C(cosm)^2)y1^2

+(-Asin2m+2Bcos2m+Csin2m)x1y1=1

令:-Asin2m+2Bcos2m+Csin2m=0

则:ctg2m=(A-C)/(2B),m=(1/2)arcctg((A-C)/(2B))

令:A(cosm)^2+Bsin2m+C(sinm)^2=1/a^2

则:a^2=1/(A(cosm)^2+Bsin2m+C(sinm)^2)

令:A(sinm)^2-Bsin2m+C(cosm)^2=1/b^2

则:b^2=1/(A(sinm)^2-Bsin2m+C(cosm)^2)

则:x1^2/a^2+y1^2/b^2=1

因此,在新坐标系中,方程已经是标准的椭圆方程

面积=pi*a