Find curve that has tangent and normal lines that create a triangle

Find the implicit equation of the curve that goes through the point (3, 1) and whose tangent and normal lines always form with the x axis a triangle whose area is equal o the slope of the tangent line. Assume y` > 0 and y > 0.

Hint: integral( sqrt(a^2 - u^2) / u du = sqrt(a^2-u^2) - a*ln | [a+sqrt(a^2-u^2)] / u | + C
(sorry, I don't know how to use the math writer yet)

This is a question from an introductory differential equations class. I have absolutely no idea how to do this! I haven't really gotten anywhere yet. This is what I've done:

let f(x) denote the curve we're looking for. Then the tangent line will have equation:
y_t = df/dx * x + C
Normal line will have equation y_n = -1/(df/dx) * x + k

Together they will form a triangle with area = df/dx, at any point on f(x). I wanted to find an expression for area in terms of df/dx, simplify it, and solve the resulting differential equation, but I can't figure out a DE for the area! I'm getting very frustrated, as we've never been shown a question like this in lecture, and I can't find any examples in my textbook.

Help would be very much appreciated!

I would assign some x-values if I were you. Suppose is the point at which you're taking the tangent and normal lines (which we will eventually need to be able to move, so just think of it as "temporarily constant"), let be the x-intercept of the tangent line, and let be the x-intercept of the normal line. Because you have assumed and you know that Why?

Then you want to set



Your goal is to eliminate and How can you do that?