Question

过Q（1,1）做双曲线(x^2)/4-(y^2)/2 = 的弦MN,使Q为MN的中点 ,求MN方程和弦长

Answer 1

用点差法的过程是这样:

记M(x1,y1)N(x2,y2) 所在直线y-1=k(x-1)

(x1+x2)/2=1

(y1+y2)/2=1

(x1^2)/4-(y1^2)/2=1

(x2^2)/4-(y2^2)/2=1

相减得(x1+x2)(x1-x2)/4-(y1+y2)(y1-y2)/2=0

k=(y1-y2)/(x1-x2)=2(x1+x2)/4(y1+y2)=1/2

∴直线方程y=(x+1)/2

y=(x+1)/2

(x^2)/4-(y^2)/2=1

联立得x^2-2x-9=0

x1+x2=2

x1x2=-9

|MN|=√[1+(1/2)^2]√[(x1+x2)^2-4x1x2]

=5√2/2