Question

过双曲线x^2/9-y^2/16=1的右焦点F作倾斜角为π/4的弦AB

Answer 1

1

令坐标A(x1,y1),B(x2,y2)

则有:

x1^2/9-y1^2/16=1;

x2^2/9-y2^2/16=1;

两式相减得:

(x1+x2)(x1-x2)/9=(y1+y2)(y1-y2)/16;

而由题意有:

斜率(y1-y2)/(x1-x2)=arctan(π/4)=1;

则:(x1+x2)=[(y1+y2)/16]·1=(y1+y2)/16

令x0=(x1+x2)/2; y0=(y1+y2)/2;

则 y0=16x0.

由弦AB过右焦点F(5,0)可知直线AB方程为y=x-5;

则有:y0=x0-5;

与y0=16x0联立解得:x0=-1/3; y0=-16/3.

则AB的中点F的距离=√[(-1/3-5)^2 +(-16/3-0)^2]=80√2/7.

2

（x1,y1）B(x2,y2) y2