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प्रश्न
Find the point on the curve y2 = 4x which is nearest to the point (2,\[-\] 8).
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उत्तर
\[\text {Let point }\left( x, y \right)\text { be the nearest to the point } \left( 2, - 8 \right) . \text { Then }, \]
\[ y^2 = 4x\]
\[ \Rightarrow x = \frac{y^2}{4} . . . \left( 1 \right)\]
\[ d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2 ............\left[ \text {Using distance formula } \right]\]
\[\text { Now,} \]
\[Z = d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2 \]
\[ \Rightarrow Z = \left( \frac{y^2}{4} - 2 \right)^2 + \left( y + 8 \right)^2 ................\left[ \text { From eq. }\left( 1 \right) \right]\]
\[ \Rightarrow Z = \frac{y^4}{16} + 4 - y^2 + y^2 + 64 + 16y\]
\[ \Rightarrow \frac{dZ}{dy} = \frac{4 y^3}{16} + 16\]
\[\text { For maximum or minimum values of Z, we must have }\]
\[\frac{dZ}{dy} = 0\]
\[ \Rightarrow \frac{4 y^3}{16} + 16 = 0\]
\[ \Rightarrow \frac{4 y^3}{16} = - 16\]
\[ \Rightarrow y^3 = - 64\]
\[ \Rightarrow y = - 4\]
\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]
\[x = 4\]
\[\text { Now, }\]
\[\frac{d^2 Z}{d y^2} = \frac{12 y^2}{16}\]
\[ \Rightarrow \frac{d^2 Z}{d y^2} = 12 > 0\]
\[\text { So, the nearest point is }\left( 4, - 4 \right) .\]
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