Question

Determine the points on the curve x2 = 4y which are nearest to the point (0,5).

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{point}\left(x,y\right)\mathrm{on}\mathrm{the}\mathrm{curve}{x}^2=4y\mathrm{be}\mathrm{nearest}\mathrm{to}\left(0,5\right).\mathrm{Then}, \\ {x}^2=4y \\ \Rightarrow y=\frac{x^2}{4}...\left(1\right) \\ \mathrm{Also}, \\ {d}^2={\left(x\right)}^2+{\left(y-5\right)}^2\left[\mathrm{Using}\mathrm{distance}\mathrm{formula}\right] \\ \mathrm{Now}, \\ Z=d^2={\left(x\right)}^2+{\left(y-5\right)}^2 \\ \Rightarrow Z={\left(x\right)}^2+{\left(\frac{x^2}{4}-5\right)}^2\left[\mathrm{Using}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow Z=x^2+\frac{x^4}{16}+25-\frac{5x^2}{2} \\ \Rightarrow \frac{dZ}{dy}=2x+\frac{4x^3}{16}-5x \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dy}=0 \\ \Rightarrow 2x+\frac{4x^3}{16}-5x=0 \\ \Rightarrow \frac{4x^3}{16}=3x \\ \Rightarrow x^3=12x \\ \Rightarrow x^2=12 \\ \Rightarrow x=\pm 2\sqrt{3} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ y=3 \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{y}^2}=2+\frac{12x^2}{16}-5 \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=9-3=6>0 \\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{nearest}\mathrm{point}\mathrm{is}\left(\pm 2\sqrt{3},3\right). \end{gathered}$$