Question

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

• A. $\frac{3\sqrt{2}}{5}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{3\sqrt{2}}{8}$
• D. $\frac{2\sqrt{3}}{8}$

Answer 1

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

let the point on parabola be $(t,t^2)$

distance of point to line $y-x-1=0$ is

$p=\frac{t^2-t+1}{\sqrt{(1{)}^2+(-1{)}^2}}=\frac{t^2-t+1}{\sqrt{2}}$

given p is minimum

so $\frac{dp}{dt}=2t-1=0$

$=>t=\frac{1}{2}$

now $\frac{d^{2}p}{d{t}^2}=2>0$

so pi minimum at $t=\frac{1}{2}$

therefore $p=\frac{(\frac{1}{2}{)}^2-\frac{1}{2}+1}{\sqrt{2}}=\frac{3\sqrt{2}}{8}$