Mathematics - Application of Derivative Question with Solution | TestHub

$$P=\left(y^2,y\right)$$ Perpendicular distance from $P$ to $x-y+1=0$ is $\frac{|y^2-y+1|}{\sqrt{2}}$ $$\begin{array}{rl}& {\mathrm{y}}^2-\mathrm{y}+1>0\forall \mathrm{y}\in \mathrm{R}\\ & \therefore \text{ Coefficient }{\mathrm{y}}^2>0\\ & \therefore \text{ Min value }=\frac{1}{\sqrt{2}}\left(\frac{4\mathrm{ac}-{\mathrm{b}}^2}{4\mathrm{a}}\right)=\frac{3}{4\sqrt{2}}\end{array}$$