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

A variable point on the given parabola can be taken as $\left(\alpha , {\alpha }^2-4\right)$



Distance from origin $D= \sqrt{{\alpha }^2+{\left({\alpha }^2-4\right)}^2}$

$\Rightarrow$ $D^2={\alpha }^2+{\alpha }^4+16-8{\alpha }^2$

$={\alpha }^4-7{\alpha }^2+16$

$\frac{d{D}^2}{d\alpha }=4{\alpha }^3-14 \alpha =0$

$\Rightarrow$ $2\alpha \left(2{\alpha }^2-7\right)=0$

$\Rightarrow$$\alpha =0, \sqrt{\frac{7}{2}}, -\sqrt{\frac{7}{2}}$



$\therefore D_{min }^2 at {\alpha }^2=\frac{7}{2}$

$D_{min}^2=\frac{49}{4}-\frac{49}{2}+16$

$= -\frac{49}{4}+16=\frac{15}{4}$

$D_{min}=\frac{\sqrt{15}}{2}$