Problem 1594

1. Given the curve $\frac{y}{c} = \cosh \left(\frac{x}{c}\right)$, we can rewrite it as:
$$y = c \cosh \left(\frac{x}{c}\right) = \frac{c}{2} \left(e^{\frac{x}{c}} + e^{-\frac{x}{c}}\right)$$

2. The points $P(a, b)$ and $Q(0, c)$ lie on this curve. Therefore, for point $P(a, b)$:
$$b = \frac{c}{2} \left(e^{\frac{a}{c}} + e^{-\frac{a}{c}}\right)$$

3. To find the slope of the normal at point $P$, we first need the derivative of $y$ with respect to $x$:
$$y' = \frac{d}{dx} \left( \frac{c}{2} \left(e^{\frac{x}{c}} + e^{-\frac{x}{c}}\right) \right) = \frac{1}{2} \left(e^{\frac{x}{c}} - e^{-\frac{x}{c}}\right)$$
At $x = a$:
$$y'(a) = \frac{1}{2} \left(e^{\frac{a}{c}} - e^{-\frac{a}{c}}\right)$$

4. The slope of the normal at $P$ is the negative reciprocal of the slope of the tangent:
$$\text{slope of normal} = -\frac{1}{y'(a)} = -\frac{2}{e^{\frac{a}{c}} - e^{-\frac{a}{c}}}$$

5. The equation of the line through $Q(0, c)$ parallel to the normal at $P$ is:
$$y - c = -\frac{2}{e^{\frac{a}{c}} - e^{-\frac{a}{c}}} \cdot x$$
Simplifying, we get:
$$y = -\frac{2}{e^{\frac{a}{c}} - e^{-\frac{a}{c}}} \cdot x + c$$

6. To find the x-intercept $R$, set $y = 0$:
$$0 = -\frac{2}{e^{\frac{a}{c}} - e^{-\frac{a}{c}}} \cdot x + c$$
Solving for $x$:
$$\frac{2}{e^{\frac{a}{c}} - e^{-\frac{a}{c}}} \cdot x = c$$
$$x = \frac{c (e^{\frac{a}{c}} - e^{-\frac{a}{c}})}{2}$$
Therefore, the coordinates of $R$ are:
$$R \left( \frac{c}{2} (e^{\frac{a}{c}} - e^{-\frac{a}{c}}), 0 \right)$$

7. The distance $QR$ is calculated as:
$$QR = \sqrt{\left( \frac{c}{2} (e^{\frac{a}{c}} - e^{-\frac{a}{c}}) - 0 \right)^2 + (0 - c)^2}$$
$$QR = \sqrt{\left( \frac{c}{2} (e^{\frac{a}{c}} - e^{-\frac{a}{c}}) \right)^2 + c^2}$$
$$QR = \sqrt{\frac{c^2}{4} (e^{\frac{a}{c}} - e^{-\frac{a}{c}})^2 + c^2}$$
$$QR = \sqrt{\frac{c^2}{4} (e^{\frac{a}{c}} - e^{-\frac{a}{c}})^2 + \frac{4c^2}{4}}$$
$$QR = \sqrt{\frac{c^2}{4} (e^{\frac{a}{c}} - e^{-\frac{a}{c}})^2 + \frac{4c^2}{4}}$$
$$QR = \sqrt{\frac{c^2}{4} \left( (e^{\frac{a}{c}} - e^{-\frac{a}{c}})^2 + 4 \right)}$$
Using the identity $(e^{\frac{a}{c}} - e^{-\frac{a}{c}})^2 + 4 = (e^{\frac{a}{c}} + e^{-\frac{a}{c}})^2$:
$$QR = \sqrt{\frac{c^2}{4} (e^{\frac{a}{c}} + e^{-\frac{a}{c}})^2}$$
$$QR = \frac{c}{2} (e^{\frac{a}{c}} + e^{-\frac{a}{c}})$$
$$QR = b$$

$\blacksquare$

The final answer is $\boxed{ QR = b }$