设直线y=k(x-p)+q①与椭圆x^2+y^2/2=1②相切,则
把①代入②,2x^2+(kx+q-kp)^2=2,
整理得(2+k^2)x^2+2k(q-kp)x+(q-kp)^2-2=0,
△/4=k^2(q-kp)^2-(2+k^2)[(q-kp)^2-2]
=4+2k^2-2(q-kp)^2=0,
∴2+k^2-(q-kp)^2=0,③
以-1/k代k,得2+1/k^2-(q+p/k)^2=0,
∴2k^2+1-(kq+p)^2=0,④
③+④,3(k^2+1)-(k^2+1)(p^2+q^2)=0,
∴p^2+q^2=3,
即互相垂直的切线的交点P(p,q)在圆x^2+y^2=3上.