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

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\left( 1, 2 \right) \text { Let the required point be } \left( x, y \right) . \text { Then }, y^2 = 4x \Rightarrow x = y^$\frac{2}{4}$ ............. \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( y^$\frac{2}{4}$ - 2 \right)^2 + \left( y - 1 \right)^2 \Rightarrow d^2 = y^$\frac{4}{16}$ + 4 - y^2 + y^2 + 1 - 2y ..............\left[ \text{From eq. }\left( 1 \right) \right] \text { Now }, Z = d^2 = y^$\frac{4}{16}$ + 4 - y^2 + y^2 + 1 - 2y \Rightarrow dZ/dy = y^$\frac{3}{4}$ - 2y + 2y - 2 \Rightarrow dZ/dy = y^$\frac{3}{4}$ - 2 \Rightarrow y^$\frac{3}{4}$ - 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,} d^2 Z/d y^2 = 3 y^$\frac{2}{4}$ \Rightarrow 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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