Question

The least distance of the point $Q(0,-2)$ from the point $P(x,y)$ where $y=\frac{16}{\sqrt{3}{x}^3}-2$ and $x>0$ is

• A. $3\sqrt{3}$
• B. $4\sqrt{3}$
• C. $\frac{4}{3}$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

Answer: (D)

$PQ=\sqrt{x^2+{\left(\frac{16}{\sqrt{3}{x}^3}\right)}^2}$
$\Rightarrow PQ=\sqrt{x^2+\frac{256}{3x^6}}$
$\text{ A.M. }\ge \text{ G.M. }$
$\frac{\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{256}{3x^6}}{4}\ge {\left(\frac{x^6\cdot 256}{3^4\cdot x^6}\right)}^{\frac{1}{4}}.$
$\Rightarrow P{Q}_{\min }=\frac{4}{\sqrt{3}}$