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 $(\alpha , {\alpha }^{2}-4)$(α,α2−4)

Distance from origin $D= \sqrt{{\alpha }^{2}+{({\alpha }^{2}-4)}^{2}}$D=α2+(α2−4)2​

$\Rightarrow$⇒ ${D}^{2}={\alpha }^{2}+{\alpha }^{4}+16-8{\alpha }^{2}$D2=α2+α4+16−8α2

$$={\alpha }^{4}-7{\alpha }^{2}+16$$=α4−7α2+16

$$\frac{d{D}^{2}}{d\alpha }=4{\alpha }^{3}-14 \alpha =0$$dαdD2​=4α3−14α=0

$\Rightarrow$⇒ $2\alpha (2{\alpha }^{2}-7)=0$2α(2α2−7)=0

$\Rightarrow$⇒$\alpha =0, \sqrt{\frac{7}{2}}, -\sqrt{\frac{7}{2}}$α=0,27​​,−27​​

$$\therefore {D}_{min }^{2} at {\alpha }^{2}=\frac{7}{2}$$∴Dmin2​atα2=27​

$${D}_{min}^{2}=\frac{49}{4}-\frac{49}{2}+16$$Dmin2​=449​−249​+16

$$= -\frac{49}{4}+16=\frac{15}{4}$$=−449​+16=415​

$${D}_{min}=\frac{\sqrt{15}}{2}$$Dmin​=215​​