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Question
The point on the curve y2 = 4x which is nearest to, the point (2,1) is _______________ .
Options
\[1, 2\sqrt{2}\]
(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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