Prove that $\frac{1}{x}$ is the only function that satisfies the following condition For all values of t, the area bounded by the tangent line at the point (t,f(t)), the x-axis and the y-axis is constant.
I saw this from a YouTube video by blackpenredpen.
For my proof attempt, I first showed that this condition holds true for f(x)=$\frac{1}{x}$
Equation of the tangent line:
$y-f(t)=f'(t)(x-t)$
$y-\frac{1}{t}=-\frac{1}{t^2}(x-t)$
let x-intercept be at the point $(a,0)$ and y-intercept by at the point $(0,b)$
At $y=0:$
$-\frac{1}{t}=-\frac{1}{t^2}(a-t)$
$a=2t$
At $x=0:$
$b-\frac{1}{t}=-\frac{1}{t^2}(-t)$
$b=\frac{2}{t}$
If $A(t)$ is the function that area of the triangle formed by the tangent line and the two axis:
$A(t)=\frac{1}{2}ab$
$A(t)=\frac{1}{2}(2t)(\frac{2}{t})$
$A(t)=2$
Next, I tried to find an expression for area for a general function f(x):
Again, the equation of the tangent line is:
$y-f(t)=f'(t)(x-t)$
At $y=0:$
$-f(t)=f'(t)(a-t)$
$a=t-\frac{f(t)}{f'(t)}$
At $x=0:$
$b-f(t)=f'(t)(-t)$
$b=f(t)-f'(t)t$
Then:
$A(t)=\frac{1}{2}ab$
$A(t)=tf(t)+\frac{1}{2}(t^2f'(t)-\frac{f^2(t)}{f'(t)})$
Therefore, functions must be chosen such that the above expression is constant when evaluated at all values of t.
However, I am unsure on how to show that $\frac{1}{x}$ is the only function that satisfies the above condition.
 A: Suppose that $f'(t)\neq 0$, otherwise area is either infinite or zero. Note that $b=-a f'(t)$. $$A(t)={\rm const}\Rightarrow ab={\rm const}\Rightarrow a^2 f'(t)={\rm const} \Rightarrow b^2={\rm const}\cdot f'(t)$$
$$\left(\frac{b^2}{f'}\right)'=0 \Rightarrow \frac{2b b'}{f'}-\frac{b^2}{f'^2}f''=0 \Rightarrow b(2b'f'-bf'')=0$$
$$b=f-tf'\Rightarrow b'=-tf'' \Rightarrow (f-tf')(-2tf''f'-(f-tf')f'')=0\Rightarrow$$ $$(f-tf')f''(f+tf')=0$$
Note that $f-tf'=0 \Rightarrow f''=0$, so we can consider this case as part of more general case $f''=0$.
$$f''=0 \Rightarrow f(t)=kt+l$$
$$f+tf'=0 \Rightarrow f(t)=\frac{C}{t}$$
$f(t)$ can also be piecewise combination of these functions, for example
$$
f(t)=
\begin{cases}
20-20t,\ t <\frac12 \\
\frac5t,\ \frac12 \leq t \leq 1 \\
10-5t,\ t > 1
\end{cases}
$$
A: Another problem which is connected with this question and excludes straight line segments. Let find function $f(t)$ such that every straight line passing through points $(a,0)$ and $\left(0,\frac{2A}{a}\right)$  is tangent of $f(t)$ plot. Here $A\neq 0$ is constant value and $a\neq 0$ is free variable, used to  parameterize set of straight lines.
Let consider two lines: first passing through $(a,0)$ and $\left(0,\frac{2A}{a}\right)$ and second passing through $(a_1,0)$ and $\left(0,\frac{2A}{a_1}\right)$, where $a_1^2\neq a^2$. One can compute common point of these lines:
$$\frac{2A}{a}-\frac{2A}{a^2}t=\frac{2A}{a_1}-\frac{2A}{a_1^2}t\Rightarrow aa_1^2-ta_1^2=a_1a^2-ta^2\Rightarrow t=\frac{aa_1}{a+a_1} \Rightarrow\\ \frac{2A}{a}-\frac{2A}{a^2}t=\frac{2A(a-t)}{a^2}=\frac{2A}{a+a_1}$$
When $a_1\to a$ two tangents will get intersect in tangent point, therefore $$f\left(\lim_{a_1\to a}\frac{aa_1}{a+a_1}\right)=\lim_{a_1\to a}\frac{2A}{a+a_1} \Rightarrow f\left(\frac{a}2\right)=\frac{A}{a}$$
Let $\frac{a}{2}=t$, then $f(t)=\frac{A}{2t}$.
