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The point on the parabola is $(1,2)$. Implicit differentiation gives $2ydy/dx=4$, so the tangent line has slope one there. Its perpendicular normal line to a plane curve therefore has slope $-1$. The required line is

The derivative of the parabola is $dy/dx=2+2x$, giving tangent line slope $4$ at $(1,3)$. A normal line to a plane curve has the negative reciprocal slope, since its direction is orthogonal to a tangent vector. HenceThe point $(1,3)$ satisfies the equation, and the product of the two slopes is $-1$.

At $x=3$, the point on the parabola is $(3,8)$ and the tangent line has slope $f^{\prime }(3)=6$. The normal line to a plane curve therefore has slope $-1/6$. Its equation isThus the requested results are

The normal line to a plane curve is perpendicular to the tangent line, so its slope is the negative reciprocal $1/\sqrt{6}$. The required equation isThe normal direction also agrees with the gradient $(4x,2y)$ of the defining level function.