Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 9

The positive horizontal coordinate is $x=\sqrt{3/2}=\sqrt{6}/2$. By implicit differentiation,At the selected point this gives $dy/dx=-\sqrt{6}$.

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.