If $L$ is the normal drawn to the parabola $y^2=8 x$ at the point $t=\frac{1}{\sqrt{2}}$, then the foot of the perpendicular drawn from the focus of the parabola on to the normal L is

• A

  $(3,2)$
• B

  $(5, \sqrt{2})$
• C

  $(0, \sqrt{2})$
• Correct answer
  D

  $(3, \sqrt{2})$

Hence, the correct answer is option: 'D'.