Find a curve such that the surface of the triangle bounded by the line going through the tangent point and perpendicular to the x-axis and the tangent line to the graph is equal to $a^2$.
I didn't understand the question at first glance, and I found related answers like this one – but this question also assumes the triangle is bounded by a perpendicular line to the $x$ axis.
This is what I got so far, but I don't think I got the DE entirely correct. I think I need some help with interpreting the question.
The tangent line to the curve at any point $x$ is given by:
$y-xy^{\prime}=0$
This line intersects the $x$ axis at some point $x_{0}$ s.t. $y(x_0)=0$.
Then, for any $x>x_{0}$, a $\perp$ line of height $y\left(x\right)$ intersects the tangent line.
So the triangle is defined by:
$A\left(x_{0},0\right)$
$B\left(x,0\right)$
$C\left(x,y-xy^{\prime}\right)$
Thus, the area of $\triangle ABC$ is given by:
$\frac{1}{2}\left(x-x_{0}\right)\left(y-xy^{\prime}\right) =a^{2}$
Best Answer
Let the point of tangency on the curve $y=g(x)$ be $(X,g(X))$.
Equation of tangent is $\frac{y-g(X)}{x-X}=g'(X)$.
$x-$intercept of tangent is $X-g(X)/g'(X)$.
The area of the triangle in question is $\frac12\times|X-(X-g(X)/g'(X))|\times|g(X)|=a^2$.
This gives the ODE$$|g'|=g^2/2a^2$$
Since we want to find any one solution, we let $g'\ge0$. Thus$$\int\frac{dg}{g^2}=-\frac1g=\frac x{2a^2}+c$$You can assume $g'\le 0$ and you would get $1/g=x/2a^2+c_1$ which is also admissible.