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

#### Shamako

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!

#### Ackbeet

MHF Hall of Honor
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?