Let the coordinates of A, B and C be `(ct_(1),(c)/(t_(1))),(ct_(2),(c)/(t_(2))) and (ct_(3),(c)/(t_(3)))`, respectively.
`"Slope of CA"=((c)/(t_(3))-(c)/(t_(1)))/(ct_(3)-ct_(1))=-(1)/(t_(3)t_(1))`
`"Slope of CB"=-(1)/(t_(2)t_(3))`
Given that `CA _|_ CB`.
`therefore" "(-(1)/(t_(1)t_(3)))xx(-(1)/(t_(2)t_(2)))=-1`
`rArr" "(-(1)/(t_(3)^(2)))xx(-(1)/(t_(1)t_(2)))=-1`
`"Slope of AB"=(1)/(t_(1)t_(2))`
Also, differentiating the equation of curve, we get
`(dy)/(dx)=-(y)/(x)=-(c//t)/(ct)=-(1)/(t^(2))`
`therefore" Slope of tangent to curve at point C"=-(1)/(t_(3)^(2))`
Hence, form Eq. (1), normal at C is parallel to AB.