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**Hint:**Assume the variable point on the given curve. Now find its distance from the origin. By using the equation of the curve, simplify the distance of the variable point from the origin. Now differentiate the equation and equate it to $0$ to get the minimum distance between the variable point and the origin.

**Complete step-by-step answer:**

Let us assume the point $(h,k)$ such that it lies on the curve $y = {x^2} - 4$. Since $(h,k)$ lies on the curve, therefore replacing $x$ by $h$ and $y$ by $k$, we get

$k = {h^2} - 4 - - - - - (1)$

Distance between the points $({x_1},{y_1})$ and $({x_2},{y_2})$

Distance formula $ = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} $

Distance between $(h,k)$ and $(0,0)$

$D$$ = \sqrt {{{(h - 0)}^2} + {{(k - 0)}^2}} $

${D^2} = {h^2} + {k^2}$

From (1), $k = {h^2} - 4$

${h^2} = k + 4$

${D^2} = {k^2} + k + 4$

As stated in the question, we have to find the minimum distance between the point on the curve and the origin.

Hence, for finding the minimum distance we need to differentiate the equation and equate it to $0$.

The value of $k$ for which the equation will satisfy will be the $y$ coordinate of that point.

By differentiating the equation,

$2k + 1 = 0$

$k = \dfrac{{ - 1}}{2}$

Substituting in (1),

$\dfrac{{ - 1}}{2} = {h^2} - 4$

$h = \pm \sqrt {\dfrac{7}{2}} $

$x - $coordinate of the point is $ \pm \sqrt {\dfrac{7}{2}} $

$y - $coordinate of the point is $ - \dfrac{1}{2}$

Therefore minimum distance is given by substituting coordinates value in equation ${D^2} = {h^2} + {k^2}$ we get,

$ = \sqrt {\dfrac{7}{2} + \dfrac{1}{4}} = \dfrac{{\sqrt {15} }}{2}$

**So, the correct answer is “Option A”.**

**Note:**An important step in this question is the formation of the equation (1).Students should remember that for finding maximum or minimum point we have to differentiate the equation and equate it to 0.And also should remember the distance between two points formula i.e $ = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} $ for solving these types of questions.

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