Question

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

• A. $\frac{\sqrt{19}}{2}$
• B. $\sqrt{\frac{15}{2}}$
• C. $\frac{\sqrt{15}}{2}$
• D. $\sqrt{\frac{19}{2}}$

Answer 1

Answer: (C)

$D=\sqrt{{\alpha }^2+{\left({\alpha }^2-4\right)}^2}$
$D^2={\alpha }^2+{\alpha }^4+16-8{\alpha }^2$
$={\alpha }^4-7{\alpha }^2+16$
$\frac{d{D}^2}{d\alpha }=4{\alpha }^3-14\alpha =0$
$2\alpha \left(2{\alpha }^2-7\right)=0$
${\alpha }^2=\frac{7}{2}$
$D^2=\frac{49}{4}-\frac{49}{2}+16=-\frac{49}{2}+16=\frac{15}{4}$
$D=\frac{\sqrt{15}}{2}$