Question

Find the shortest distance between the line $x-y+1=0$ and the curve $y^2=x$.

Answer 1

Line is $x-y+1=0$ and curve $y^2=x$

On the line and curve you can take any point that should satisfy the them.

So for line if $x=t-1$ and $y=t$, $A(t-1,t)$ then they are satisfying the line.

For curve $x=t^2$ , $y=t$, they are satisfying the curve $B(t^2,t)$

So using distance formula, we have

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

$=t^2-t+1$

Let $f\left(t\right)=t^2-t+1$

$f^{\prime }\left(t\right)=2t-1$

For critical point ,$f^{\prime }\left(t\right)=0$

So,

$2t-1=0$

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

For minima ,$f^{\prime \prime }\left(t\right)>0$

So, $f^{\prime \prime }\left(t\right)=2>0$

So at $t=\frac{1}{2}$, minima occurs

Hence the minimum distance $=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$