Question

Find the point on the curvey y²=2x which is at a minimum distance from the point (1,4).

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{a}\mathrm{point}\left(\mathrm{x},\mathrm{y}\right)\mathrm{on}\mathrm{the}\mathrm{curve}{y}^2=2x\mathrm{is}\mathrm{nearest}\mathrm{to}\mathrm{the}\mathrm{point}\left(1,4\right).\mathrm{Then}, \\ {y}^2=2x \\ \Rightarrow x=\frac{y^2}{2}...\left(1\right) \\ {d}^2={\left(x-1\right)}^2+{\left(y-4\right)}^2\left[\mathrm{Using}\mathrm{distance}\mathrm{formula}\right] \\ \mathrm{Now}, \\ Z=d^2={\left(x-1\right)}^2+{\left(y-4\right)}^2 \\ \Rightarrow Z={\left(\frac{y^2}{2}-1\right)}^2+{\left(y-4\right)}^2\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow Z=\frac{y^4}{4}+1-y^2+y^2+16-8y \\ \Rightarrow \frac{dZ}{dy}=y^3-8 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dy}=0 \\ \Rightarrow y^3-8=0 \\ \Rightarrow y^3=8 \\ \Rightarrow y=2 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ x=2 \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{y}^2}=3y^2 \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=12>0 \\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{nearest}\mathrm{point}\mathrm{is}\left(2,2\right). \end{gathered}$$