Alternate methods for finding the shortest distance between these two curves

Question

There are two curves $2y^2=2x-1$ and $2x^2=2y-1$ Find the shortest distance between these two curves.

I'm well aware of the traditional method of solving these type of questions. The concept is that the minimum distance is along the common normal. And then we find $y'$ and the points and then we proceed accordingly. I want to know if there is some other method of solving this particular question.

A method that uses minimum calculus is appreciated. That being said, it doesn't mean that calculus based methods are not welcomed. I'm posting my attempt below as answer.

Answer 1

A couple solutions.

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The curves are mirror images of each other about the line $y=x$. They are concave in a direction away from $y=x$. This suggest their point of closest approach to each other is the point of closest approach to that line. The point of closest approach has the same tangent as the line $y=x$, i.e. $1$.

$2y^2=2x-1\implies 4yy'=2$. If we swap in $y'=1$, then $4y=2\implies y=1/2\implies x=3/2$.

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Alternative parameter approach

$2y^2=2x-1. 2u^2=2v-1$.

$y=t. x=t^2+1/2. u=s. v=s^2+1/2$

$d^2=(x-u)^2+(y-v)^2$

$(t^2+1/2-s)^2+(s^2+1/2-t)^2=D^2$

$\partial(D^2)/\partial t= 4t(t^2+1/2-s)-2(s^2+1/2-t)=0$

$\partial (D^2)/\partial s= -2(t^2+1/2-s)+4s(s^2+1/2-t)=0$

$4t^3+4t-4ts-2s^2-1=0$

$2s^2+4ts-4t^3-4t+1=0$

$-2t^2-1+4s^3+4s-4st=0$

$2t^2+4st-4s^3-4s+1=0$

$(s^2-t^2)+2(s^3-t^3)+2(s-t)=0$

$(s-t)[(s+t)+2(s^2+t^2+st)+2]=0$

$(s-t)[s^2+t^2+2st + s^2+s+t^2+t+2]=(s-t)[(s+t)^2+(s+1/2)^2+(t+1/2)^2+3/2]=0$

It follows that $s=t$ is the only solution over the reals.

$d^2=2(t^2+1/2-t)^2\implies d=\sqrt{2}(t^2+1/2-t)$ which has a critical point at $t=1/2\implies d=1/2\sqrt{2}$

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Having trouble getting this to work.

Lagrange Multipliers.

$d^2=(x-u)^2+(y-v)^2. 2y^2-2x+1=0. 2u^2-2v+1=0.$

$2(x-u)=-2\lambda_1$

$2(y-v)=4y\lambda_1$

$-2(x-u)=4u\lambda_2$

$-2(y-v)=-2\lambda_2$

That's 6 equations in 6 unknowns, so I think it's solvable. Finding the ratio of $(y-v)/(x-u)$ implies $4uy=1$. By symmetry, it is expected $u=y$, so this says $u=\pm 1/2$. Try the two candidates to get the answer.

Answer 2

We can assign parametric coordinates to the curves. Let them be $\left(a^2+\frac12,a\right)$ and $\left(b,b^2+\frac12\right)$

Distance between any two points will be $$\sqrt{\left(a^2+\frac12-b\right)^2+\left(b^2+\frac12-a\right)^2}$$ Now using some inequalities we can say (it's the quadratic mean inequality) $$\sqrt{\left(a^2+\frac12-b\right)^2+\left(b^2+\frac12-a\right)^2}\ge\frac{a^2+b^2-a-b+1}{\sqrt{2}}=\frac{\left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\frac12}{\sqrt2}$$ which has a minimum value equal to $\frac{1}{2\sqrt2}$ Hence the minimum distance between the two curves is $$\boxed{\frac{1}{2\sqrt2}}$$