Question

Find the point on the curve y2=4x which is nearest to the point (2, $-$8).

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{point}\left(x,y\right)\mathrm{be}\mathrm{the}\mathrm{nearest}\mathrm{to}\mathrm{the}\mathrm{point}\left(2,-8\right).\mathrm{Then}, \\ {y}^2=4x \\ \Rightarrow x=\frac{y^2}{4}...\left(1\right) \\ {d}^2={\left(x-2\right)}^2+{\left(y+8\right)}^2\left[\mathrm{Using}\mathrm{distance}\mathrm{formula}\right] \\ \mathrm{Now}, \\ Z=d^2={\left(x-2\right)}^2+{\left(y+8\right)}^2 \\ \Rightarrow Z={\left(\frac{y^2}{4}-2\right)}^2+{\left(y+8\right)}^2\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow Z=\frac{y^4}{16}+4-y^2+y^2+64+16y \\ \Rightarrow \frac{dZ}{dy}=\frac{4y^3}{16}+16 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dy}=0 \\ \Rightarrow \frac{4y^3}{16}+16=0 \\ \Rightarrow \frac{4y^3}{16}=-16 \\ \Rightarrow y^3=-64 \\ \Rightarrow y=-4 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ x=4 \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{y}^2}=\frac{12y^2}{16} \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=12>0 \\ \mathrm{So},\mathrm{the}\mathrm{nearest}\mathrm{point}\mathrm{is}\left(4,-4\right). \end{gathered}$$