Question

The points on the curve $x{y}^2=1$ which are nearest to the origin, are

• A. $[{\left(\frac{1}{2}\right)}^{1/3},\pm {\left(\frac{1}{2}\right)}^{-1/6}]$
• B. $[{\left(\frac{1}{2}\right)}^{1/3},2^{-1/6}]$
• C. $[2^{1/3},\pm {\left(\frac{1}{2}\right)}^{-1/6}]$
• D. None of these

Answer 1

The correct option is A $[{\left(\frac{1}{2}\right)}^{1/3},\pm {\left(\frac{1}{2}\right)}^{-1/6}]$

Any point on the curve $x{y}^2=1$ is of form $P(\frac{1}{t^2},t)$

Let distance of $P$ from the origin $=a$

$a=\sqrt{{\left(\frac{1}{t^2}\right)}^2+t^2}a=\sqrt{\frac{1}{t^4}+t^2}$

At minimum distance $\frac{da}{dt}=0$

$\Rightarrow \frac{d}{dt}\left(\sqrt{\frac{1}{t^4}+t^2}\right)=0\Rightarrow \frac{\frac{-4}{t^5}+2t}{2\sqrt{\frac{1}{t^4}+t^2}}=0\frac{-4}{t^5}+2t=0\Rightarrow t^6=2\Rightarrow t=\pm 2^{1/6}\Rightarrow t^2=2^{1/3}\Rightarrow \frac{1}{t^2}=\frac{1}{2^{1/3}}={\left(2\right)}^{-1/3}$

So, the points on the curve nearest to origin are $(2^{-1/3},\pm 2^{1/6})$ or $({\left(\frac{1}{2}\right)}^{1/3},\pm {\left(\frac{1}{2}\right)}^{-1/6})$.

So, option A is correct.