Two Passing Parabolas

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Shortest distance will be the common normal. By symmetry it's perpendicular to y = x (has slope -1)

Hence the corresponding tangents must have slope = 1.

Differentiating we get $4yy' = 2$ or $y' = \frac{1}{2y} = 1$ giving $y = \frac{1}{2}$ and $x = \frac{3}{4}$ . The other point will be mirror image about y = x i.e. $(\frac{1}{2},\frac{3}{4})$ Giving the distance sought to be $\frac{1}{\sqrt{8}} = 0.353553. . .$

Looking at the equation shows they are inverse of each other, thus, symmetric about $x=y$ . Hence, we have to look for the shortest distance between line $y=x$ and any one of parabola and double it up. Let's choose $2x^2=2y-1$

$y'=2x$

For minimum distance, $y'$ must be equal to slope of line $y=x$ which is obviously 1.

$2x=1\Rightarrow x=\dfrac{1}{2}$

$y=\dfrac{2x^2+1}{2}=\dfrac{3}{4}$

Distance of point $\left (\dfrac{1}{2},\dfrac{3}{4}\right )$ from line $y=x$ is $\left |\dfrac{\dfrac{3}{4}-\dfrac{1}{2}}{\sqrt{1^2+1^2}}\right | =\dfrac{1}{4\sqrt{2}}$

Therefore distance between parabola is $2×\dfrac{1}{4\sqrt{2}}=\dfrac{1}{2\sqrt{2}}=0.3535$

for equation y^2....use parametric coordinates as T square divided by 4 + 1/2 , T divided by 2 (T is a parameter). for x^2... use parametric coordinates as T divided by 2, T square divided by 4 +1/2(T is a parameter). Now, minimize the distance between these 2 pairs of coordinates x,y using calculus. you will obtain T=1. Now, put this value of T in the distance formula, you will get the answer 3.5 2^{34}