Question: Find the functions for which the area of the triangle made by the two axis and the tangent any point on the function is independent of the coordinate of the point.
Now I approached the problem like this.
Let the function be $f (x) $. Hence the slope of the tangent at a point $(h,f (h)) $ is $f'(h)$. Hence the equation of the tangent at that point is \begin{align*}
y-f (h)=f'(h)(x-h)& \implies -xf'(h)+y=f (h)-hf'(h)\\
& \implies \frac{x}{\dfrac{hf'(h)-f (h)}{f'(h)}}+\frac{y}{f (h)-hf'(h)}=1
\end{align*}
Hence he intercepts of the tangent are $\left(\dfrac{hf'(h)-f (h)}{f'(h)},0\right) $, $ \left(0, f (h)-hf'(h)\right) $.
Hence the area of the triangle is $$\frac{1}{2}\cdot\dfrac{ f (h)-hf'(h)}{f'(h)}\times( f (h)-hf'(h))=\dfrac{( f (h)-hf'(h))^2}{2f'(h)}$$
According to the question $\dfrac{( f (h)-hf'(h))^2}{2f'(h)}$ is independent of $h $ and $f(h) $. Which means it is constant (Let $k $). Hence we have to solve the differential equation $$\dfrac{( f (h)-hf'(h))^2}{2f'(h)}=k$$
Now I cant move further. Also am I going in the correct way?
Best Answer
The sign of your area is wrong. One has $$A=-{(f(h)-h f'(h))^2\over 2 f'(h)}$$ (note that $f'(h)<0$ in our problem), so that we obtain, using other letters, the ODE $$(y- x y')^2=-2 A y'\ .$$ This can be written as $$y=x y'+\sqrt{-2 A y'}\ ,$$ with $+$ at the square root, since we want $y(0)>0$. This is a Clairaut differential equation. Its solutions are the hypotenuses of the triangles we constructed, plus the envelope of this family of lines. The envelope is the curve we are after. Computation gives the equation $2xy=A$ for the envelope.