Question

Find the point on the parabolas x² = 2y which is closest to the point (0,5).

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{required}\mathrm{point}\mathrm{be}\left(x,y\right).\mathrm{Then}, \\ {x}^2=2y \\ \Rightarrow y=\frac{x^2}{2}...\left(1\right) \\ \mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{points}\left(x,y\right)\mathrm{and}\left(0,5\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ {d}^2={\left(x\right)}^2+{\left(y-5\right)}^2 \\ \mathrm{Now}, \\ {d}^2=Z \\ \Rightarrow Z={\left(x\right)}^2+{\left(\frac{x^2}{2}-5\right)}^2 \\ \Rightarrow Z=x^2+\frac{x^4}{4}+25-5x^2 \\ \Rightarrow \frac{dZ}{dy}=2x+x^3-10x \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dy}=0 \\ \Rightarrow x^3-8x=0 \\ \Rightarrow x^2=8 \\ \Rightarrow x=\pm 2\sqrt{2} \\ \text{Substituting the value of}\mathit{\text{x}}\text{in eq.}\left(\text{1}\right)\text{, we get} \\ y=4 \\ \frac{d^{2}Z}{d{y}^2}=3x^2-8 \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=24-8=16>0 \\ \text{So, the nearest point is}\left(\pm 2\sqrt{2},4\right). \end{gathered}$$