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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

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

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...

May you continue from here?

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  • $\begingroup$ Hint 2: For convenience, you can suppose $\alpha$ is parametrized with arc-lenght paramenter ( $||\alpha'(t)||=1$) . $\endgroup$ – Pocho la pantera Oct 2 '13 at 14:54

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