椭圆x^2/a^2+y^2/b^2=1,(a>b>0);右焦点为F,直线l方程为x=a2/c,求证椭圆上任意点P到F的距离与到l的距离之比为c/a

Question

椭圆x^2/a^2+y^2/b^2=1,(a>b>0); 右焦点为F,直线l方程为x=a2/c,求证椭圆上任意点P到F的距离与到l的距离之比为c/a

Answer 1

d1=√[(x-c)^2+y^2],d2=a^2/c-x,a^2=b^2+c^2
x^2/a^2+y^2/b^2=1 ====>y^2=b^2-(b^2/a^2)*x^2
则d1/d2={√[(x-c)^2+y^2]}/(a^2/c-x)
={√[x^2-2cx+c^2+a^2-c^2-(1-c^2/a^2))*x^2]}/(a^2/c-x)
={√[-2c^2x+ca^2+(c^2/a^2)*cx^2]}/(a^2-cx)
=(c/a)*{√[c^2x^2-2ca^2x+a^4]}/(a^2-cx)
=(c/a)*{√[(cx-a^2)^2]}/(a^2-cx)
=(c/a)*{1}=c/a

其实x=a^2/c就是此椭圆的右准线,椭圆第一定律就是这样的,第二定律是：某点到两不同定点的距离之和为定值.