Question

The shortest distance between line $y-x=1$ and curve $x=y^2$

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{3\sqrt{2}}{8}$
• C. $\frac{8}{3\sqrt{2}}$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{3\sqrt{2}}{8}$

Let $(a^2,a)$ be the point of shortest distance on $x=y^2$

The distance between $(a^2,a)$ and line $x-y+1=0$ is given by

$D=\frac{a^2-a+1}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}[{(a-\frac{1}{2})}^2+\frac{3}{4}]$

It attains a minimum value at $a=\frac{1}{2}$

$\therefore D_{min}=\frac{1}{\sqrt{2}}\times \frac{3}{4}=\frac{3}{4\sqrt{2}}$

$\therefore D_{min}=\frac{3}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{8}$