Find the minimum distance between the curves y^2 = x-1 and x^2 = y-1
Please tell the procedure, how to solve this?
Thanks
For this particular problem, if you draw a diagram you'll see that the two parabolas are symmetric about the line y=x. So the shortest distance between them will be between the two points where the tangents are parallel to that line.
The points on the parabolas where the tangents have gradient 1 are $\displaystyle \bigl(\tfrac12,\tfrac54\bigr)$ on $\displaystyle x^2=y-1$, and $\displaystyle \bigl(\tfrac54,\tfrac12\bigr)$ on $\displaystyle y^2=x-1$. The distance between these points is $\displaystyle \boxed{\tfrac34\sqrt2}$.
In general, if you can't use that sort of symmetry to simplify the problem, you would have to find the equation of the normal at a general point on one curve. Then find the point(s), if any, where that line meets the other curve, determine the distance along the normal to the closest point on the other curve, and finally minimise that distance.
Edit. For the general case, Mr F's method (see next comment) is much better than my laborious procedure of using normals.
The problem can be done using calculus. The start would be to let (a, a^2 + 1) be a point on x^2 = y-1 and (b^2 + 1, b) be a point on y^2 = x-1. Substitute these points into the formula for the distance between two points to get an expression for the distance D in terms of a and b. You require the values of a and b so that the distance is a minimum.
However ..... here's a hint for an easier way: note that the two curves are inverses of each other and are therefore reflections of each other in the line y = x.
Edit: I was very slow. Beaten by Opalg.