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Find the Point on the Parabolas X2 = 2y Which is Closest to the Point (0,5) ? - CBSE (Arts) Class 12 - Mathematics

Question

Find the point on the parabolas x2 = 2y which is closest to the point (0,5) ?

Solution

$\text { Let the required point be } \left( x, y \right) . \text { Then },$

$x^2 = 2y$

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

$\text { The distance between points } \left( x, y \right) \text { and } \left( 0, 5 \right) \text { is given by }$

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

$\text { Now,}$

$d^2 = Z$

$\Rightarrow Z = \left( x \right)^2 + \left( \frac{x^2}{2} - 5 \right)^2$

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

$\Rightarrow \frac{dZ}{dy} = 2x + x^3 - 10x$

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

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

$\Rightarrow x^3 - 8x = 0$

$\Rightarrow x^2 = 8$

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

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

$y = 4$

$\frac{d^2 Z}{d y^2} = 3 x^2 - 8$

$\Rightarrow \frac{d^2 Z}{d y^2} = 24 - 8 = 16 > 0$

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

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Solution Find the Point on the Parabolas X2 = 2y Which is Closest to the Point (0,5) ? Concept: Graph of Maxima and Minima.
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