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Q.

The minimum distance of a point on the curve $y = x^{2} - 4$ from the origin is

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a

$\frac{\sqrt{15}}{2}$

b

$\frac{\sqrt{19}}{2}$

c

$\sqrt{\frac{15}{2}}$

d

$\sqrt{\frac{19}{2}}$

answer is A.

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Detailed Solution

$y = x^{2} - 4, p (x, y)$
$f (y) = o p^{2} = x^{2} + y^{2} = y + 4 + y^{2}$
$f^{1} (y) = 0 \Rightarrow 1 + 2 y = 0 \Rightarrow y = \frac{- 1}{2}$
$\Rightarrow x^{2} = \frac{- 1}{2} + 4 = \frac{7}{2}$
$o p = \sqrt{x^{2} + y^{2}} = \frac{\sqrt{15}}{2}$

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