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Solution for Determine the Points on the Curve X2 = 4y Which Are Nearest to the Point (0,5) ? - CBSE (Science) Class 12 - Mathematics

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Question

Determine the points on the curve x2 = 4y which are nearest to the point (0,5) ?

Solution

\[\text { Let the point } \left( x, y \right) \text { on the curve} x^2 = 4y \text { be nearest to } \left( 0, 5 \right) . \text { Then }, \]

\[ x^2 = 4y\]

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

\[\text { Also }, \]

\[ d^2 = \left( x \right)^2 + \left( y - 5 \right)^2 \left[\text {  Using distance formula } \right]\]

\[\text { Now,} \]

\[Z = d^2 = \left( x \right)^2 + \left( y - 5 \right)^2 \]

\[ \Rightarrow Z = \left( x \right)^2 + \left( \frac{x^2}{4} - 5 \right)^2 \left[ \text { Using eq }. \left( 1 \right) \right]\]

\[ \Rightarrow Z = x^2 + \frac{x^4}{16} + 25 - \frac{5 x^2}{2}\]

\[ \Rightarrow \frac{dZ}{dy} = 2x + \frac{4 x^3}{16} - 5x\]

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

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

\[ \Rightarrow 2x + \frac{4 x^3}{16} - 5x = 0\]

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

\[ \Rightarrow x^3 = 12x\]

\[ \Rightarrow x^2 = 12\]

\[ \Rightarrow x = \pm 2\sqrt{3}\]

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

\[y = 3\]

\[\text { Now,} \]

\[\frac{d^2 Z}{d y^2} = 2 + \frac{12 x^2}{16} - 5\]

\[ \Rightarrow \frac{d^2 Z}{d y^2} = 9 - 3 = 6 > 0\]

\[\text { So, the required nearest point is } \left( \pm 2\sqrt{3}, 3 \right) .\]

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Solution Determine the Points on the Curve X2 = 4y Which Are Nearest to the Point (0,5) ? Concept: Graph of Maxima and Minima.
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