How can I find a curve based on its tangent lines? Let's say for some curve its tangent lines at every point have a property that the length of a segment within the first quarter $[0;+\infty)^2$ is exactly $C>0$.
How can such a curve be defined analytically? Maybe even in terms of $y=f(x)$.

Now about the tangent lines themselves. Noticeably for all $k<0$ the tangent line $y = kx - \frac{Ck}{\sqrt{k^2 + 1}}$ seems to fit the description precisely because the length of its segment within the first quarter is equal to $C$. But what to do next?
 A: Probably not the most precise argument, but here's one way to go about it: let's take the tangent lines to have the form
$$y = kx - \frac{Ck}{\sqrt{k^2 + 1}}$$
for $k < 0$, as I mentioned in a comment. Two of these tangent lines with slopes $k$ and $k'$ intersect at the point with $x$-coordinate given by
$$kx_\text{int} - \frac{Ck}{\sqrt{k^2 + 1}} = k'x_\text{int} - \frac{Ck'}{\sqrt{k'^2 + 1}}$$
or
$$x_\text{int} = \frac{C}{(k' - k)}\biggl(\frac{k'}{\sqrt{k'^2 + 1}} - \frac{k}{\sqrt{k^2 + 1}}\biggr)$$
If the difference between these slopes is very small, $k' = k + \epsilon$, this expression simplifies to
$$x_\text{int} = \frac{C}{(k^2 + 1)^{3/2}} + O(\epsilon)$$
which tells us the $x$-coordinate of the point where two lines of similar slope intersect.
It stands to reason that if the curve we're looking for, $y = f(x)$, is not too "crazy", the points at which it is tangent to two lines converge to the intersection of those two lines as the difference between the slopes of the lines goes to zero. So in the $\epsilon\to 0$ limit, we get an expression for the slope at the convergence point:
$$f'\Biggl(\frac{C}{(k^2 + 1)^{3/2}}\Biggr) = k$$
or equivalently,
$$f'(x_\text{int}) = -\sqrt{\biggl(\frac{C}{x_\text{int}}\biggr)^{2/3}-1}$$
As achille hui pointed out in the comments, the relevant solution to this differential equation is the function
$$f(x) = C\Biggl[1 - \biggl(\frac{x}{C}\biggr)^{2/3}\Biggr]^{3/2}$$
which is called an astroid.

