PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Find the Point on the Curve Y2=4x Which is Nearest to the Point (2, − 8). - PUC Karnataka Science Class 12 - Mathematics

#### Question

Find the point on the curve y2 = 4x which is nearest to the point (2,$-$ 8).

#### Solution

$\text {Let point }\left( x, y \right)\text { be the nearest to the point } \left( 2, - 8 \right) . \text { Then },$

$y^2 = 4x$

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

$d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2 ............\left[ \text {Using distance formula } \right]$

$\text { Now,}$

$Z = d^2 = \left( x - 2 \right)^2 + \left( y + 8 \right)^2$

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

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

$\Rightarrow \frac{dZ}{dy} = \frac{4 y^3}{16} + 16$

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

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

$\Rightarrow \frac{4 y^3}{16} + 16 = 0$

$\Rightarrow \frac{4 y^3}{16} = - 16$

$\Rightarrow y^3 = - 64$

$\Rightarrow y = - 4$

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

$x = 4$

$\text { Now, }$

$\frac{d^2 Z}{d y^2} = \frac{12 y^2}{16}$

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

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

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Solution Find the Point on the Curve Y2=4x Which is Nearest to the Point (2, − 8). Concept: Graph of Maxima and Minima.
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