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# Find the Point on the Curve X2=8y Which is Nearest to the Point (2,4) ? - Mathematics

#### Question

Find the point on the curve x2 = 8y which is nearest to the point (2, 4) ?

#### Solution

$\text { Let }\left( x, y \right) \text {be nearest to the point } \left( 2, 4 \right) . \text { Then },$

$x^2 = 8y$

$\Rightarrow y = \frac{x^2}{8} ............\left( 1 \right)$

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

$\text { Now,}$

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

$\Rightarrow Z = \left( x - 2 \right)^2 + \left( \frac{x^2}{8} - 4 \right)^2 .............\left[\text {From eq. } \left( 1 \right) \right]$

$\Rightarrow Z = x^2 + 4 - 4x + \frac{x^4}{64} + 16 - x^2$

$\Rightarrow \frac{dZ}{dy} = - 4 + \frac{4 x^3}{64}$

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

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

$\Rightarrow - 4 + \frac{4 x^3}{64} = 0$

$\Rightarrow \frac{x^3}{16} = 4$

$\Rightarrow x^3 = 64$

$\Rightarrow x = 4$

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

$y = 2$

$\text { Now,}$

$\frac{d^2 Z}{d y^2} = \frac{12 x^2}{64}$

$\Rightarrow \frac{d^2 Z}{d y^2} = 3 > 0$

$\text { So, the nearest point is } \left( 4, 2 \right) .$

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