The equation of the curve whose tangent makes a constant area with the axes 
My turn:
Let the equation of the curve is $$y=f(x)$$
Then the slope of the tangent is $$\frac{dy}{dx}$$
The equation of the tangent at some point $x_o$ Is 
$$y= f'(x_o) x + b $$ , $$b = f(x_o) -x_o f'(x_o)$$
Now the area of the triangle is $$A = 0.5 \frac{x_o f'(x_o) -f(x_o)}{f'(x_o)}\times (f(x_o) -x_o f'(x_o))$$
Now i think i need to put A as a constant an integrate the last equation with respect to $x_o$ to get the equation of th as a e curve but i am confused with how can i consider $x_o$ as a variable even though it is constant ?
 A: It is difficult problem
The intercets on $x$ and $y$ axes by the tangent at the point $(x,y)$ are $\Delta x=x-ydx/dy$ and $\Delta y=y-xdy/dx.$ Then the area of the triangle with thes intercepts is
$$|\frac{1}{2} (x-ydx/dy)(y-xdy/dx)|=A^2~~~(1)$$
Let us denote $\frac{dy}{dx}=p$ and let $B^2p>0$
Then we have $$x^2p^2+y^2-2xyp=B^2 p \implies |y-xp|= B\sqrt{p}~~~(2)$$
$$y=xp+B\sqrt{p} \implies y=x\frac{dy}{dx}+B\sqrt{\frac{dy}{dx}}~~~(3)$$
D.w.r.t.x to get $$y'=xy''+y'+ \frac{B}{2}\sqrt{\frac{dx}{dy}} y'' \implies y'' \left [x+\frac{B}{2}\sqrt{\frac{dx}{dy}} \right]=0~~~~(4)$$
$y''=0$ gives trivial solution $(y'=C$) which can be put in (3) to get the general solution of the first order ODE (3) which will be family of lines. For a fixed non-trivial solution of (3),  we take
$$\frac{B}{2}\sqrt{\frac{dx}{dy}}=-x \implies  dy= \frac{B^2}{4x^2}dx \implies \int dy=\frac{B^2}{4} \int \frac{dx}{x^2}~~~~(5)$$ 
Finally, we get $$xy=-\frac{B^2}{4}~~~~(6),$$ the rectangular hyperbola. Note that this (6) is the singular solution of the Clairaut equaion (3). Here $B$ or $A$ is a fixed given constant.
Notte: it is rather simple to show the converse that the  itercepts of the tangents to the curve $xy=D^2$ make a triangle of constant area.
A: I would write $$A=\frac{1}{2}\frac{(y_0-f'(x_0)x_0)^2}{|f'(x_0)|}$$
It is $$A=\frac{1}{2}\frac{(f(x_0)-x0_0f'(x_0))^2}{|f'(x_0)|}$$
