Find a curve with tangents making a constant area with coordinate axes Find a curve for which each of its tangent lines forms with the coordinate axes a triangle of constant area $a^2$.
The answer is supposed to be $2xy = a^2$, but I haven't figured out how to arrive at the answer.
I'm having trouble with visualizing what is really needed as part of the area. Is it the whole thing under the tangent line or just the the triangle between tangent and subtangent?
I tried adding the components for each side (subtangent + subnormal, etc.) but it got way too messy quick. I was expecting some cancellations, but it just got harder to deal with.
Thank you in advance.
 A: Let the curve be $y=f(x)$. Let $P(x,y)$ be the point of tangency. Then using the equation of the tangent line we get that the line will intercept the axes at $Q\left(0,\,y-xy'\right)$ and $R\left(\frac{xy'-y}{y'}, 0\right)$.

The area of the $\triangle QOR$ (where $O$ is the origin) is given to be a constant $a^2$. So $$\frac{1}{2}(y-xy')(\frac{xy'-y}{y'})=a^2 \implies (xy'-y)^2=-2a^2y'.$$
Can you solve this differential equation?
A: This should help you visualize the situation:

For a given value of $x$, what value of $y$ gives you a triangle area of $a^2$?
This gives you the slope (derivative) of the function.
Can you finish the rest?
A: $$\dfrac{y-y_1}{x-x_1}=y'$$
Put $(x=0,y=0)$ separately and get the intercepts and take their area product say equal to $c^2$ and drop the suffixes.
$$(xy'-y)^2+2c^2y'=0 \tag1$$
In $p$ discriminant method put $ y'=p$ and partially differentiate w.r.t. $p$ to get $c$  the envelope:
$$ c= \dfrac{y-a^2/x}{x^2} \tag2$$
Plug into 1) and simplify
$$ 2 xy= 3 c^2 $$
$$ x y = C_1^2 \tag3 $$
