双曲线标准方程推导（详细过程）√(x+c)^2+y^2-√(x-c)^2+y^2=2a

Question

双曲线标准方程推导（详细过程）
√(x+c)^2+y^2-√(x-c)^2+y^2=2a

Answer 1

√[(x+c)^2+y^2]-√[(x-c)^2+y^2]=2a √[(x+c)^2+y^2]={2a-√[(x-c)^2+y^2]}√[(x+c)^2+y^2]^2={2a-√[(x-c)^2+y^2]}^2(x+c)^2+y^2=4a^2-4a√[(x-c)^2+y^2]+(x-c)^2+y^22cx=4a^2-4a√[(x-c)^2+y^2]-2cx4a√[(x-c)^2+y^2]=4a^2-4cxa√[(x-c)^2+y^2]=a^2-cxa^2*[(x-c)^2+y^2]=(a^2-cx)^2a^2x^2-2ca^2x+a^2c^2+a^2y^2=a^4-2a^2cx+c^2x^2(a^2-c^2)x^2+a^2y^2=a^4-a^2c^2(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)x^2/a^2+y^2/(a^2-c^2)=1x^2/a^2-y^2/(c^2-a^2)=1设：c^2-a^2=b^2,则有：x^2/a^2-y^2/b^2=1,为双曲线的标准方程.