Question

Find the shortest distance between the line and the curve?

`y-x=1` and `x=y^2`

Answer 1

`(3sqrt(2))/8`

Explanation:

Here's a method that does not use differentiation.

Given:

`{ (y - x = 1), (x = y^2) :}`

The graphs of these equations look something like this:
graph{(y-x-1)(x-y^2) = 0 [-5, 5, -2.5, 2.5]}

Let's find a line parallel to `y-x=1` which just touches the parabola.

Given a system of equations:

`{ (y - x = k), (x = y^2) :}`

we want to find the value of `k` which yields exactly one real solution.

Substituting `x = y-k` into the second equation, we get:

`y - k = y^2`

That is:

`y^2 - y + k = 0`

This is a quadratic in standard form:

`ay^2+by+c = 0`

with `a=1`, `b=-1` and `c = k`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-1)^2-4(1)(k) = 1-4k`

So this quadratic has exactly one root when `Delta = 0` and hence `k = 1/4`

So the parabola `x = y^2` is touched by the line `y-x = 1/4`

graph{(y-x-1)(y-x-1/4)(x-0.02-y^2) = 0 [-2.5, 2.5, -1.25, 1.25]}

Since the lines are diagonal, the distance between them is:

`(1-1/4)sqrt(2)/2 = (3sqrt(2))/8`