题目

直线l与圆x²+y²=2相切与点p(p不在坐标轴上),l与双曲线x²-y²/2=1
相交与不同的两点A,B.求证OA⊥OB

提问时间：2021-03-25

答案

设P(m,n) ,mn≠0
P在圆x^2+y^2=2上,m²+n²=2
那么过P点圆的切线与OP垂直,
斜率k=-m/n
∴L:y-n=-m/n(x-m)
即y=-m/nx+(m²+n²)/n
y=-m/nx+2/n
联立方程组：
{y=(-mx+2)/n
{x²-y²/2=2
==>
x²-(2-mx)²/(2n²)=1
==>
(2n²-m²)x²+4mx-4-2n²=0
∵n²=2-m²
∴方程即
(4-3m²)x²+4mx+2m²-8=0
需4-3m²≠0
△=16m²+4(3m²-4)(2m²-8)>0
设A(x1,y1),B(x2,y2)
那么x1+x2=-4m/(4-3m²)
x1x2=(2m²-8)/(4-3m²)
∴OA·OB
=x1x2+y1y2
=x1x2+(2-mx1)(2-mx2)/n²
=x1x2+[4-2m(x1+x2)+m²x1x2]/n²
=[(m²+n²)x1x2-2m(x1+x2)+4]/n²
分子(m²+n²)x1x2-2m(x1+x2)+4
=2(2m²-8)/(4-3m²)+2m*4m/(4-3m²)+4
=(12m²-16)/(4-3m²)+4
=-4+4=0
∴OA·OB=0
即OA⊥OB