Question

The point on the curve $x{y}^2=1$ which is nearest to the origin is

• A. (1, 1)
• B. (1, -1)
• C. $(4,\frac{1}{2})$
• D. $(\frac{1}{2^{\frac{1}{3}}},2^{\frac{1}{6}})$

Answer 1

The correct option is D $(\frac{1}{2^{\frac{1}{3}}},2^{\frac{1}{6}})$

Let point on curve be $(\frac{1}{y^2},y)$
So distance $=\sqrt{\frac{1}{y^4}+y^2}$
$\frac{d\left(d\right)}{dy}=\frac{1}{2}\left(\frac{-4y^{-5}+2y}{\sqrt{\frac{1}{y^4}+y^2}}\right)$
$y^6=2$
$y=2^{\frac{1}{6}}$
$x=\frac{1}{2^{\frac{1}{3}}}$