Find min distance from y = cosh x to y = x

Question

The full question is this: Point P on curve y = cosh x is such that its perpendicular distance from the line y = x is a minimum. Show P's coordinates are (ln(1 + root 2), root 2).

I am completely at a loss as to what to do. I cannot find any example online for a similar question for a hyperbolic function, and the similar questions I have found aren't that well explained. This question is from an old textbook from an old specification for Further Maths. So it seems I may never learn the information I need to solve this, or I just haven't learnt it yet. All examples online include a parabola, which seems a lot simpler than a hyperbolic. Any help appreciated.

Answer 1

A general trend I noticed (not sure if it is correct) is that the minimum distance between a curve and a line (no intersection) occurs when the gradient of the curve equals the gradient of the line. So $y=\cosh x=\frac {e^x}2+\frac 1{2e^x}$.

Taking the gradients of $y=\cosh x$ and $y=x:$
$\therefore 1 = \frac {e^x}2-\frac 1{2e^x}$

Let $e^x=z$. Then we are looking for the positive solutions of $1=\frac z2-\frac 1{2z}\iff2z=z^2-1\iff z^2-2z-1=0$, that is $z=1+\sqrt 2$ or $z=1-\sqrt 2$, which is negative.

From here $x=\ln (1+\sqrt 2)$ and

$$\begin{align}y&=\frac {e^x}2+\frac 1{2e^x}=\frac {1+\sqrt 2}2+\frac 1{2+\sqrt 2}\\ &=\frac 12 +\frac {\sqrt 2}2-\frac 12+\frac {\sqrt 2}2\\ &=\sqrt 2. \end{align}$$

So $P=\bigl(\ln (1+\sqrt 2), \sqrt 2\bigr)$.

Does this work for you? I am a year $11$ student so it probably should but just ask if you have any queries.