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The Point on the Curve Y2 = 4x Which is Nearest To, the Point (2,1) is (A) 1 , 2 √ 2 (B) (1,2) (C) (1, − 2) (D) ( − 2,1) - CBSE (Science) Class 12 - Mathematics

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Question

The point on the curve y2 = 4x which is nearest to, the point (2,1) is _______________ .

  • \[1, 2\sqrt{2}\]

  • (1, 2)

  • (1, -2)

  • ( -2,1)

Solution

\[\left( 1, 2 \right)\]

 

\[\text { Let the required point be } \left( x, y \right) . \text { Then }, \]

\[ y^2 = 4x\]

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

\[\text { Now,} \]

\[d = \sqrt{\left( x - 2 \right)^2 + \left( y - 1 \right)^2}\]

\[\text { Squaring both sides, we get }\]

\[ \Rightarrow d^2 = \left( x - 2 \right)^2 + \left( y - 1 \right)^2 \]

\[ \Rightarrow d^2 = \left( \frac{y^2}{4} - 2 \right)^2 + \left( y - 1 \right)^2 \]

\[ \Rightarrow d^2 = \frac{y^4}{16} + 4 - y^2 + y^2 + 1 - 2y ..............\left[ \text{From eq. }\left( 1 \right) \right]\]

\[\text { Now }, \]

\[Z = d^2 = \frac{y^4}{16} + 4 - y^2 + y^2 + 1 - 2y\]

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

\[ \Rightarrow \frac{dZ}{dy} = \frac{y^3}{4} - 2\]

\[ \Rightarrow \frac{y^3}{4} - 2 = 0\]

\[ \Rightarrow y^3 = 8\]

\[ \Rightarrow y = 2\]

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

\[x = 1\]

\[\text { Now,} \]

\[\frac{d^2 Z}{d y^2} = \frac{3 y^2}{4}\]

\[ \Rightarrow \frac{d^2 Z}{d y^2} = \frac{3 \left( 2 \right)^2}{4} = 3 > 0\]

\[\text { So, the nearest point is } \left( 1, 2 \right) . \]

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Solution The Point on the Curve Y2 = 4x Which is Nearest To, the Point (2,1) is (A) 1 , 2 √ 2 (B) (1,2) (C) (1, − 2) (D) ( − 2,1) Concept: Graph of Maxima and Minima.
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