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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

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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