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

Oct 2011
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!


MHF Hall of Honor
Jun 2010
I would assign some x-values if I were you. Suppose \(\displaystyle x_{0}\) 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 \(\displaystyle x_{1}\) be the x-intercept of the tangent line, and let \(\displaystyle x_{2}\) be the x-intercept of the normal line. Because you have assumed \(\displaystyle y'>0\) and \(\displaystyle y>0,\) you know that \(\displaystyle x_{2}>x_{1}.\) Why?

Then you want to set

\(\displaystyle \frac{y(x_{0})(x_{2}-x_{1})}{2}=y'(x_{0}).\)

Your goal is to eliminate \(\displaystyle x_{1}\) and \(\displaystyle x_{2}.\) How can you do that?