# Find the shortest distance between the line and the curve?

## $y - x = 1$ and $x = {y}^{2}$

Sep 22, 2017

$\frac{3 \sqrt{2}}{8}$

#### Explanation:

Here's a method that does not use differentiation.

Given:

$\left\{\begin{matrix}y - x = 1 \\ x = {y}^{2}\end{matrix}\right.$

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:

$\left\{\begin{matrix}y - x = k \\ x = {y}^{2}\end{matrix}\right.$

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:

$a {y}^{2} + b y + c = 0$

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

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(1\right) \left(k\right) = 1 - 4 k$

So this quadratic has exactly one root when $\Delta = 0$ and hence $k = \frac{1}{4}$

So the parabola $x = {y}^{2}$ is touched by the line $y - x = \frac{1}{4}$

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:

$\left(1 - \frac{1}{4}\right) \frac{\sqrt{2}}{2} = \frac{3 \sqrt{2}}{8}$