Question

The minimum distance from the origin to a point on the curve $x^{2/3}+y^{2/3}=a^{2/3}(a>0)$ is?

• A. a
• B. $\frac{a}{2}$
• C. $\frac{a}{\sqrt{8}}$
• D. $a^{2/3}$

Answer 1

Answer: (C)

$\frac{a}{\sqrt{8}}$

$x^{2/3}+y^{2/3}=a^{2/3}$

$\frac{2}{3}{x}^{-1/3}\frac{dx}{dy}+\frac{2}{3}{y}^{-1/3}=0$

$-\frac{dx}{dy}=\frac{x^{1/3}}{y^{1/3}}$

Distance from origin to the the curve at $(x,y)$

$\sqrt{x^2+y^2}$

Differentiating it wrt y, we get

$\frac{1}{2\sqrt{x^2+y^2}}[2x\frac{dx}{dy}+2y]$

Equating it to zero, we get

$-\frac{dx}{dy}=\frac{y}{x}$

$y^{4/3}=x^{4/3}$ or $x=y$

$2x^{2/3}=a^{2/3}$

$x=\frac{a}{\sqrt{8}}$