Shortest distance

Question

Minimum distance between the curves $y^2=x-1$ and $x^2=y-1$
I just know that to go for common normal for such kind of problems but I am unable to go with it . It would be helpful if i can get exactly how to approach by this method. Any other methods are also welcomed as it can set an example for further persons who have same kind of difficulty.

Answer 1

The parametric equation of $y=x^2+1$ is $(t, t^2+1)$ and of $x=y^2+1$ is $(u^2+1, u)$. The distance between any two parametric points is then, $$d(t,u)=\sqrt{(t-(u^2+1))^2+(t^2+1-u)^2}.$$ Now take partial derivatives w.r.t $u$ and $t$ and set equal to zero to obtain, $${-1+3t+2t^3-2tu-u^2=0\\1+t^2-3u+2tu-2u^3=0}~.$$ Solving this pair of simultaneous equations gives real solutions $u=t=1/2$. Hence

$$d(1/2,1/2)=\sqrt{9/16+9/16}=\frac{3\sqrt{2}}{4}.$$

Answer 2

Hint the curves are symmetric about $y=x $. Thus for shortest distance the normal should be perpendicular to this line. Hence slope of normal =-1. The equation of normal for parabolas is $y=mx-2am-am^3$ we have $a=1/4,m=-1$. Find the point of intersection of this line with parabolas . Can you continue from here?