the shortest distance between the parabola two x square is equal to 2x minas 1 and 2 x square is equal to 2y minus 1
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap x,yx,y to get to the second parabola. They are mirror images with respect to line x=yx=y. Required point should have this slope y′=1y′=1 for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating 2yy′=1,2y=1,2yy′=1,2y=1, and the x,yx,y coordinates are
(54,12)
(54,12)
and the other point of tangency is again swapped to
(12,54);
(12,54);
Now use distance formula between the tangent points to get
d=32–√4.