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# Find the Point on the Curvey Y2=2x Which is at a Minimum Distance from the Point (1,4). - CBSE (Science) Class 12 - Mathematics

#### Question

Find the point on the curvey y2 = 2x which is at a minimum distance from the point (1, 4).

#### Solution

$\text { Suppose a point }\left( x, y \right)\text { on the curve } y^2 = 2x \text { is nearest to the point }\left( 1, 4 \right) . \text { Then },$

$y^2 = 2x$

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

$d^2 = \left( x - 1 \right)^2 + \left( y - 4 \right)^2 ...................\left[\text { Using distance formula } \right]$

$\text { Now,}$

$Z = d^2 = \left( x - 1 \right)^2 + \left( y - 4 \right)^2$

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

$\Rightarrow Z = \frac{y^4}{4} + 1 - y^2 + y^2 + 16 - 8y$

$\Rightarrow \frac{dZ}{dy} = y^3 - 8$

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

$\frac{dZ}{dy} = 0$

$\Rightarrow y^3 - 8 = 0$

$\Rightarrow y^3 = 8$

$\Rightarrow y = 2$

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

$x = 2$

$\text { Now,}$

$\frac{d^2 Z}{d y^2} = 3 y^2$

$\Rightarrow \frac{d^2 Z}{d y^2} = 12 > 0$

$\text { So, the required nearest point is } \left( 2, 2 \right) .$

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Solution Find the Point on the Curvey Y2=2x Which is at a Minimum Distance from the Point (1,4). Concept: Graph of Maxima and Minima.
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