跪求解答一道英文数学题,Derive the formulas for A' and C' so that the eq

x=x'cosu-y'sinu
y=x'sinu+y'cosu
Ax^2+Bxy+Cy^2
=A(x'^2cosu^2+y'^2sinu^2-2x'y'sinucosu)
+B(x'y'(cosu^2-sinu^2)+(x'^2-y'^2)sinucosu)
+C(x'^2sinu^2+y'^2cosu^2+2x'y'sinucosu)
=(Acosu^2+Bsin2u/2+Csinu^2)x'^2
+(-Asin2u+Bcos2u+Csin2u)x'y'
+(Asinu^2-Bsin2u/2+Ccosu^2)y'^2
-Asin2u+Bcos2u+Csin2u=0
Bcos2u=(A-C)sin2u
B/(A-C)=tan2u
(cos2u)^2=1/1+tan2u^2=1/1+B^2/(A-C)^2=(A-C)^2/(B^2+(A-C)^2)
(sin2u)^2=B^2/(B^2+(A-C)^2)
A'=Acosu^2+Csinu^2+Bsin2u/2
=A(1+cos2u)/2+C(1-cos2u)/2+Bsin2u^2/2
=(1/2)[A+C+(A-C)cos2u+Bsin2u^2]
=(1/2)[(A+C)+ [(A-C)^2+B^2]/√[B^2+(A-C)^2]] 或 =(1/2)[(A+C)-[(A-C)^2+B^2]/√[B^2+(A-C)^2]]
=(1/2)[(A+C)+ √(B^2+(A-C)^2] =(1/2)[(A+C)-√(B^2+(A-C)^2)]