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प्रश्न
Find the point on the parabolas x2 = 2y which is closest to the point (0,5) ?
उत्तर
\[\text { Let the required point be } \left( x, y \right) . \text { Then }, \]
\[ x^2 = 2y\]
\[ \Rightarrow y = \frac{x^2}{2} .............\left( 1 \right)\]
\[\text { The distance between points } \left( x, y \right) \text { and } \left( 0, 5 \right) \text { is given by }\]
\[ d^2 = \left( x \right)^2 + \left( y - 5 \right)^2 \]
\[\text { Now,} \]
\[ d^2 = Z\]
\[ \Rightarrow Z = \left( x \right)^2 + \left( \frac{x^2}{2} - 5 \right)^2 \]
\[ \Rightarrow Z = x^2 + \frac{x^4}{4} + 25 - 5 x^2 \]
\[ \Rightarrow \frac{dZ}{dy} = 2x + x^3 - 10x\]
\[\text {For maximum or a minimum values of Z, we must have }\]
\[\frac{dZ}{dy} = 0\]
\[ \Rightarrow x^3 - 8x = 0\]
\[ \Rightarrow x^2 = 8\]
\[ \Rightarrow x = \pm 2\sqrt{2}\]
\[\text { Substituting the value of x in eq. }\left( 1 \right), \text { we get }\]
\[y = 4\]
\[\frac{d^2 Z}{d y^2} = 3 x^2 - 8\]
\[ \Rightarrow \frac{d^2 Z}{d y^2} = 24 - 8 = 16 > 0\]
\[\text { So, the nearest point is }\left( \pm 2\sqrt{2}, 4 \right) .\]
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