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)

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