# orthogonal trajectories

• October 21st 2008, 06:04 PM
desdemoniagirl
orthogonal trajectories
hello, my first time and im soo excited to have some help, been beeting my head against the wall on this question,
1) Show that both of these families are orthogonal trajectories and graph on the same axis...
x^2+y^2=ax and x^2+y^2=by
thanks in advance to whoever can help, my other tutor was very lost.
Jo
• October 21st 2008, 08:58 PM
mr fantastic
Quote:

Originally Posted by desdemoniagirl
hello, my first time and im soo excited to have some help, been beeting my head against the wall on this question,
1) Show that both of these families are orthogonal trajectories and graph on the same axis...
x^2+y^2=ax and x^2+y^2=by
thanks in advance to whoever can help, my other tutor was very lost.
Jo

Differentiate each relation implicitly to get dy/dx.

First curve:

$2x + 2y \frac{dy}{dx} = a \Rightarrow \frac{dy}{dx} = \frac{a-2x}{2y}$.

But $x^2 + y^2 = ax \Rightarrow a = \frac{x^2 + y^2}{x}$.

Therefore $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

In a similar way you can show that for the second curve, $\frac{dy}{dx} = \frac{2x}{b - 2y} = \frac{2xy}{x^2 - y^2}$.

What do you notice about the product of the two derivatives ....?

By the way, I hope you realise that the two curves are circles .......

First curve: $x^2 - ax + y^2 = 0 \Rightarrow \left(x - \frac{a}{2}\right)^2 + y^2 = \left( \frac{a}{2} \right)^2$.

Second curve: Left for you.
• October 21st 2008, 10:03 PM
desdemoniagirl
thank you so very much, help like this is sooo greatly appreciated.(Rofl)