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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.

Ted Shifrin
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palash jain
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2 Answers2

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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

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

pshmath0
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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?

Archis Welankar
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  • I can continue from here , we were lucky enough to have symmetric equation but if this is not then how to approach it . Any link to solved examples or any solved problem are heartedly invited – palash jain May 17 '17 at 16:26
  • Find the equation of common normal get the points of intersection with each curves and then distance formula. – Archis Welankar May 17 '17 at 16:49
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    @palashjain I would hazard a guess that the problem is deliberately set up with this symmetry to encourage you to take advantage of it to simplify the solution. – amd May 18 '17 at 00:45
  • @amd ty so much – palash jain May 18 '17 at 10:32