1. ## Orthogonal trajectories

Show that the orthogonal trajectories for the family of curves $\displaystyle y = ae^{-x}$ (a constant) are a family of parabolas.

Attempt:

$\displaystyle y = ae^{-x}$

$\displaystyle \frac{dy}{dx} = -ae^{-x}$

$\displaystyle \frac{dy}{dx} = \frac{1}{ae^{-x}}$

$\displaystyle \int dy = \int \frac{1}{ae^{-x}}dx$

$\displaystyle y = \frac{1}{a}\int e^{x}dx$

$\displaystyle y = \frac{1}{a}e^{x} + C$

$\displaystyle \frac{1}{2}y^2 = x +C$

I cant seem to see where I went wrong though.

2. Welll, maybe you could just change this part:

$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{ae^{-x}}$

$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{y}$

$\displaystyle \int y \ dy = \int \ dx$

$\displaystyle \frac12 y^2 = x + C$

From what you did though.. maybe you can do this:

$\displaystyle y = \frac{1}{a}e^{x} + C$

$\displaystyle y = \frac{1}{ae^{-x}} + C$

$\displaystyle y = \frac{1}{y} + C$

$\displaystyle y^2 = 1 + yC$

$\displaystyle y^2 = 1 + ae^{-x}C$

But then, it gets complicated for me... the constant becomes another constant...