Setting up a differential equation in a geometry problem

Question

I excuse myself in advance if the answer to this problem seems blatantly simple, yet I seem to be hitting a brick wall and not getting anywhere, and it seems to me to be a relatively simple problem.

The problem:

Find all curves in the first quadrant so that the following is true: the triangle, bounded by the tangent line to the curves in point T, the x-axis and the line OT has a constant surface area of p (where p is a positive integer, naturally). The point O marks the origin. This must be true for every point T (meaning the choice of T is arbitrary).

Attempt at the solution:

The idea is of course to set up a differential equation, since we have $f'$ as the slope of the tangent line. The equation of the tangent line can be derived as follows:

$y-f(a)=f'(a)(x-a)$,

where I marked $y$ as the tangent, $f$ as the function we are seeking and $a$ the point $x$ in which we determined the tangent to the function.

Now, marking the sides of the triangle as:

$|OX|$ - the side laying of the x-axis

$|OT|$ - the side from the origin to the point T

$|TX|$ - the tangent line to the x-axis

Now, I reasoned to obtain the expressions for these sides as follows:

$|OX|=a-\frac{f(a)}{f'(a)}$ (by setting $y$ to $0$)

$|OT|=\sqrt{(a-0)^2+(f(a)-0)^2}=\sqrt{a^2+f(a)^2}$ (Euclidian distance)

$|TX|=\sqrt{(a-(a-\frac{f(a)}{f'(a)}))^2+(f(a)-0)^2}=\sqrt{(\frac{f(a)}{f'(a)})^2+f(a)^2}=\frac{f(a)\sqrt{1+f'(a)^2}}{f'(a)}$

and the generalise $a->x$, $f(a)->y$ and $f'(a)->y'$, thus obtaining:

$|OX|=x-\frac{y}{y'}=a$

$|OT|=\sqrt{x^2+y^2}=b$

$|TX|=\frac{y\sqrt{1+y'^2}}{y'}=c$

We may then connect the sides of the triangle with its surface by Heron's formula:

$S_{triangle}=p=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$

However all this seems a bit cumbersome and rather than just plunging myself into all this algebra I spent my time thinking about more elegant ways. I thought about $S=rs$, where $r$ is the radius of the incircle, but then I just translated the problem into finding r, for which I had no idea.

Another idea was trying to express the differential equation through integral application as follows:

$S_{triangle}=p=\int_0^{f(a)}dy\int_{\frac{ay}{f(a)}}^{a+\frac{y-f(a)}{f'(a)}}dx$

I have obtained the above integral like this:

1. First we integrate from $0$ to $f(a)$ by $y$ to avoid having to integrate twice.
2. We integrate by $x$ from the line $|OT|$ to the line $|TX|$; the equation for $|OT|$ is simply $y=\frac{f(a)}{a}x$ and I have already written the tangent line above.

So, my question here is, if the above reasonings and methods are correct (or so to speak, which, if any, is correct). I am only concerned here with setting up the correct differential equation; solving that is not a problem. I am particularly uncomfortable at the moment by the substitution $a->x$, $f(a)->y$ and $f'(a)->y'$ as it seems kind of forced and I am wondering if it is the right thing to do.

I am grateful for all the answers and possibly (indeed probably) better solutions. Many thanks!

Answer 1

The height of the triangle is $f(x)$, the base is $\Big(x-\frac{f(x)}{f'(x)}\Big)$, thus $$2p=f(x)\Big(x-\frac{f(x)}{f'(x)}\Big)$$ $$2pf'=xff'-f^2$$ The solution is $$f(x)=\frac{x\pm\sqrt{x^{2}-4cp}}{2c}$$

Answer 2

Write your curve in the form $$y\mapsto\bigl(g(y),y\bigr)$$ with an unknown function $x=g(y)$, assumed decreasing for the moment. Then we obtain the condition $${1\over2} y\bigl(g(y)-y g'(y)\bigr)=p\ ,\tag{1}$$ which is an inhomogeneous linear differential equation for $g$. The corresponding homogeneous equation $g(y)-y g'(y)=0$ has the solutions $g_{\rm hom}(y)=C y$, and the "variation of the constant" Ansatz then leads to $$g(y)=c_0 y+{p\over y}\ .$$ This is only the general solution of $(1)$. One now has to do some qualitative geometric analysis in order to check to which extent this is the solution of the original problem.