The point $(0, 3)$ is nearest to the curve $x^{2} = 2 y$ at

(2 \sqrt{2}, 0)

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(0, 0)

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(2, 2)

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none of these

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Solution

The correct option is D $(2, 2)$
Let $(a, b)$ is the point on the parabola from which its distance from $(0, 3)$ is minimum
$\Rightarrow a^{2} = 2 b$
Assume $d$ is the distance
$d^{2} = a^{2} + (b - 3)^{2} = 2 b + (b - 3)^{2}$
For minimum value of $d$
$\frac{d (d^{2})}{d b^{2}} = 0 = 2 + 2 (b - 3)$
$\Rightarrow b = 2$
and $a^{2} = 2 b = 4 \Rightarrow a = \pm 2$
Hence require points on the parabola are $(2, - 2)$ and $(2, 2)$

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The point (0,3) is nearest to the curve x2=2y at

The correct option is D (2,2)Let (a,b) is the point on the parabola from which its distance from (0,3) is minimum⇒a2=2b Assume d is the distanced2=a2+(b−3)2=2b+(b−3)2For minimum value of dd(d2)db2=0=2+2(b−3)⇒b=2and a2=2b=4⇒a=±2Hence require points on the parabola are (2,−2) and (2,2)

The point $(0, 3)$ is nearest to the curve $x^{2} = 2 y$ at