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Find the Point on the Parabolas X2 = 2y Which is Closest to the Point (0,5) ? - Mathematics

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प्रश्न

Find the point on the parabolas x2 = 2y which is closest to the point (0,5) ?

बेरीज

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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पाठ 18: Maxima and Minima - Exercise 18.5 [पृष्ठ ७४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 18 Maxima and Minima
Exercise 18.5 | Q 32 | पृष्ठ ७४

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