问一道数学题：化简：根号下[(x+c)2+y2]+根号下[(x-c)2+y2]=2a速求!

Question

问一道数学题：化简：根号下[(x+c)2+y2]+根号下[(x-c)2+y2]=2a 速求!

Answer 1

根号[(x+c)2+y2]+根号[(x-c)2+y2]=2a (x+c)^2+y^2+(x-c)^2+y^2+2根号{[(x+c)2+y2][(x-c)2+y2]}=4a^2 2x^2+2y^2+2c^2+2根号{[(x+c)2+y2][(x-c)2+y2]}=4a^2 根号{[(x+c)2+y2][(x-c)2+y2]}=2a^2-x^2-y^2-c^2 [(x+c)2+y2][(x-c)2+y2]=(2a^2-x^2-y^2-c^2)^2 (x^2+c^2+y^2)^2-(2cx)^2=4a^4-4a^2(x^2+c^2+y^2)+(x^2+c^2+y^2)^2 4a^4-4a^2(x^2+c^2+y^2)+4c^2x^2=0 (c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2) 设c^2-a^2=b^2 x^2/a^2-y^2/b^2=1