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# Solution for Determine the Points on the Curve X2 = 4y Which Are Nearest to the Point (0,5) ? - CBSE (Science) Class 12 - Mathematics

#### Question

Determine the points on the curve x2 = 4y which are nearest to the point (0,5) ?

#### Solution

$\text { Let the point } \left( x, y \right) \text { on the curve} x^2 = 4y \text { be nearest to } \left( 0, 5 \right) . \text { Then },$

$x^2 = 4y$

$\Rightarrow y = \frac{x^2}{4} . . . \left( 1 \right)$

$\text { Also },$

$d^2 = \left( x \right)^2 + \left( y - 5 \right)^2 \left[\text { Using distance formula } \right]$

$\text { Now,}$

$Z = d^2 = \left( x \right)^2 + \left( y - 5 \right)^2$

$\Rightarrow Z = \left( x \right)^2 + \left( \frac{x^2}{4} - 5 \right)^2 \left[ \text { Using eq }. \left( 1 \right) \right]$

$\Rightarrow Z = x^2 + \frac{x^4}{16} + 25 - \frac{5 x^2}{2}$

$\Rightarrow \frac{dZ}{dy} = 2x + \frac{4 x^3}{16} - 5x$

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

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

$\Rightarrow 2x + \frac{4 x^3}{16} - 5x = 0$

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

$\Rightarrow x^3 = 12x$

$\Rightarrow x^2 = 12$

$\Rightarrow x = \pm 2\sqrt{3}$

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

$y = 3$

$\text { Now,}$

$\frac{d^2 Z}{d y^2} = 2 + \frac{12 x^2}{16} - 5$

$\Rightarrow \frac{d^2 Z}{d y^2} = 9 - 3 = 6 > 0$

$\text { So, the required nearest point is } \left( \pm 2\sqrt{3}, 3 \right) .$

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Solution Determine the Points on the Curve X2 = 4y Which Are Nearest to the Point (0,5) ? Concept: Graph of Maxima and Minima.
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