Question

求到定点F（c,0)与到定直线l:x=a^2/c距离之比是c/a(c/a>1)的点M的轨迹

Answer 1

设M(x,y),c/a>1 c>a

MF=√[(x-c)^2+y^2],点M到直线L的距离=[x-a^2/c]

√[(x-c)^2+y^2]/[x-a^2/c]=c/a [(x-c)^2+y^2]/(x-a^2/c)^2=c^2/a^2

x^2-2cx+c^2+y^2=(c^2/a^2)(x^2-2a^2x/c+a^4/c^2)=c^2x^2/a^2-2cx+a^2

x^2+c^2+y^2=c^2x^2/a^2+a^2 a^2x^2+a^2c^2+a^2y^2=c^2x^2+a^4

(c^2-a^2)x^2-a^y^2=a^2(c^2-a^2) 设c^2-a^2=b^2

b^2x^2-a^2y^2=a^2b^2 两边同除a^2b^2

x^2/a^2-y^2/b^2=1