# problem of differential equations

Considering a family of curves $k(x,y,\lambda)=0$ defined in a domain $\omega$ of $R^2$ with $\lambda$ real, I have to calculate the differential equation of the curves intersect those under a constant angle $\alpha \in (o,\pi/2)$

Can you give a hint about how to do it? because I dont even know how to start it

• Shout "$o$" in "$\alpha \in (o,\pi/2)$" be a zero? – Vedran Šego Oct 2 '13 at 11:57
• Vedran Šego: yes is a zero! Sorry – user98227 Oct 2 '13 at 12:03

Let $\alpha(t)=(x(t),y(t))$ be a curve such as we are looking for.
Let $\lambda(t)$ be such that $k(x(t),y(t),\lambda(t))=0$.
At the point $\alpha(t)$, the tangent of $\alpha$ is $\alpha'(t)=(x'(t),y'(t))$ and the normal to the curve $k(x,y,\lambda)=0$ is $$\left(\frac{\partial{k}}{\partial x}(x(t),y(t),\lambda(t)),\frac{\partial{k}}{\partial y}(x(t),y(t),\lambda(t))\right)$$ Then...
• Hint 2: For convenience, you can suppose $\alpha$ is parametrized with arc-lenght paramenter ( $||\alpha'(t)||=1$) . – Pocho la pantera Oct 2 '13 at 14:54