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Question
Find the point on the curve x2 = 8y which is nearest to the point (2, 4) ?
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Solution
\[\text { Let }\left( x, y \right) \text {be nearest to the point } \left( 2, 4 \right) . \text { Then }, \]
\[ x^2 = 8y\]
\[ \Rightarrow y = \frac{x^2}{8} ............\left( 1 \right)\]
\[ d^2 = \left( x - 2 \right)^2 + \left( y - 4 \right)^2 ................\left[ \text {Using distance formula} \right]\]
\[\text { Now,} \]
\[Z = d^2 = \left( x - 2 \right)^2 + \left( y - 4 \right)^2 \]
\[ \Rightarrow Z = \left( x - 2 \right)^2 + \left( \frac{x^2}{8} - 4 \right)^2 .............\left[\text {From eq. } \left( 1 \right) \right]\]
\[ \Rightarrow Z = x^2 + 4 - 4x + \frac{x^4}{64} + 16 - x^2 \]
\[ \Rightarrow \frac{dZ}{dy} = - 4 + \frac{4 x^3}{64}\]
\[\text {For maximum or minimum values of Z, we must have }\]
\[\frac{dZ}{dy} = 0\]
\[ \Rightarrow - 4 + \frac{4 x^3}{64} = 0\]
\[ \Rightarrow \frac{x^3}{16} = 4\]
\[ \Rightarrow x^3 = 64\]
\[ \Rightarrow x = 4\]
\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]
\[y = 2\]
\[\text { Now,} \]
\[\frac{d^2 Z}{d y^2} = \frac{12 x^2}{64}\]
\[ \Rightarrow \frac{d^2 Z}{d y^2} = 3 > 0\]
\[\text { So, the nearest point is } \left( 4, 2 \right) .\]
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