#### Question

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

#### Solution

\[\text {Let point }\left( x, y \right)\text { be the nearest to the point } \left( 2, - 8 \right) . \text { Then }, \]

\[ y^2 = 4x\]

\[ \Rightarrow x = \frac{y^2}{4} . . . \left( 1 \right)\]

\[ d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2 \left[ \text { Using distance formula } \right]\]

\[\text { Now,} \]

\[Z = d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2 \]

\[ \Rightarrow Z = \left( \frac{y^2}{4} - 2 \right)^2 + \left( y + 8 \right)^2 \left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow Z = \frac{y^4}{16} + 4 - y^2 + y^2 + 64 + 16y\]

\[ \Rightarrow \frac{dZ}{dy} = \frac{4 y^3}{16} + 16\]

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

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

\[ \Rightarrow \frac{4 y^3}{16} + 16 = 0\]

\[ \Rightarrow \frac{4 y^3}{16} = - 16\]

\[ \Rightarrow y^3 = - 64\]

\[ \Rightarrow y = - 4\]

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

\[x = 4\]

\[\text { Now, }\]

\[\frac{d^2 Z}{d y^2} = \frac{12 y^2}{16}\]

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

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