#### Question

Find the point on the parabolas x^{2} = 2y which is closest to the point (0,5) ?

#### Solution

\[\text { Let the required point be } \left( x, y \right) . \text { Then }, \]

\[ x^2 = 2y\]

\[ \Rightarrow y = \frac{x^2}{2} . . . \left( 1 \right)\]

\[\text { The distance between points } \left( x, y \right) \text { and } \left( 0, 5 \right) \text { is given by }\]

\[ d^2 = \left( x \right)^2 + \left( y - 5 \right)^2 \]

\[\text { Now,} \]

\[ d^2 = Z\]

\[ \Rightarrow Z = \left( x \right)^2 + \left( \frac{x^2}{2} - 5 \right)^2 \]

\[ \Rightarrow Z = x^2 + \frac{x^4}{4} + 25 - 5 x^2 \]

\[ \Rightarrow \frac{dZ}{dy} = 2x + x^3 - 10x\]

\[\text { For maximum or a minimum values of Z, we must have }\]

\[\frac{dZ}{dy} = 0\]

\[ \Rightarrow x^3 - 8x = 0\]

\[ \Rightarrow x^2 = 8\]

\[ \Rightarrow x = \pm 2\sqrt{2}\]

\[\text { Substituting the value ofxin eq }.\left( 1 \right), \text { we get }\]

\[y = 4\]

\[\frac{d^2 Z}{d y^2} = 3 x^2 - 8\]

\[ \Rightarrow \frac{d^2 Z}{d y^2} = 24 - 8 = 16 > 0\]

\[\text { So, the nearest point is }\left( \pm 2\sqrt{2}, 4 \right) .\]