#### Question

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

#### Solution

\[\text { Suppose a point }\left( x, y \right)\text { on the curve } y^2 = 2x \text { is nearest to the point }\left( 1, 4 \right) . \text { Then }, \]

\[ y^2 = 2x\]

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

\[ d^2 = \left( x - 1 \right)^2 + \left( y - 4 \right)^2 ...................\left[\text { Using distance formula } \right]\]

\[\text { Now,} \]

\[Z = d^2 = \left( x - 1 \right)^2 + \left( y - 4 \right)^2 \]

\[ \Rightarrow Z = \left( \frac{y^2}{2} - 1 \right)^2 + \left( y - 4 \right)^2 .......................\left[ \text { From eq. } \left( 1 \right) \right]\]

\[ \Rightarrow Z = \frac{y^4}{4} + 1 - y^2 + y^2 + 16 - 8y\]

\[ \Rightarrow \frac{dZ}{dy} = y^3 - 8\]

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

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

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

\[ \Rightarrow y^3 = 8\]

\[ \Rightarrow y = 2\]

\[\text { Substituting the value of y in } \left( 1 \right),\text { we get }\]

\[x = 2\]

\[\text { Now,} \]

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

\[ \Rightarrow \frac{d^2 Z}{d y^2} = 12 > 0\]

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