f(x)=√x=g(x)=a/x
所以,a≥0
且,x=a^(2/3)
即,交點為(a^(2/3),a^(1/3))
又,f'(x)=(1/2)[1/√x]；g'(x)=-a/x^2
已知在交點處的切線互相垂直,所以：f'(a^(2/3))=-1/g'(a^(2/3))
===> (1/2)*a^(-1/3)=a^(1/3)
===> a^(2/3)=1/2
===> a=√2/4
交點坐標為(1/2,√2/2)
切線斜率分別為：√2/2,-√2
所以切線分別為：2√2x-4y+√2=0；√2x+y-√2=0