Application of Derivatives — Class 12 Maths Solution

Option A is correct

Let $Z$ be the square of distance of the point $P(x,y)$ on ${x^2} = 2y$ from the point $A(0,\;5)$ , then $Z = |PA{|^2}$

$\Rightarrow Z = {(x - 0)^2} + {(y - 5)^2} = 2y + {(y - 5)^2} = {y^2} - 8y + 25$
$\Rightarrow \cfrac{{dZ}}{{dy}} = 2y - 8$

Now, $\cfrac{{dZ}}{{dy}} = 0 \Rightarrow 2y - 8 = 0 \Rightarrow y = 4$
Therefore, ${x^2} = 2(4) = 8$[From ${x^2} = 2y$]

$\Rightarrow x = 2\sqrt 2$
${\left( {\cfrac{{{d^2}Z}}{{d{y^2}}}} \right)_{v = 4}} = 2 > 0$

Hence we can say that $Z$ is minimum at $(2\sqrt 2 ,4)$

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