Given, equation of curve is

x = 4y²

Differentiate w.r.t. x, we get

      1 = 8y * (dy/dx)

=> dy/dx = 1/8y

Let the point at tangent is P(a₁ , b₁ )

So, (dy/dx)_P = 1/8b₁

Now, equation of tangent is

=> y - b₁ = (dy/dx)_P * (x - a₁ )

=> y - b₁ = (1/8b₁ ) * (x - a₁ )

Again equation of normal at point Q(a₂ , b₂ ) is

=> y - b₂ = -1/(dy/dx)_Q * (x - a₂ )

=> y - b₂ = -1/(1/8b₂ ) * (x - a₂ )

=> y - b₂ = -8b₂  * (x - a₂ )