<con...@lawyer.com> wrote: >I'm showing that a given family of curves are >orthogonal trajectories of each other.
>I have the two curves: >x^2 + y^2 = ax >x^2 + y^2 = by
>It is said that the two circles intersect >at the origin. How is this evident >other than by plugging in 0?
Plugging in (0, 0) for (x, y) and seeing that the result is 0 in both cases seems about as evident a method as one could wish for! What worries you about it? (Something to do with orthogonality?)
In article <6fc9fdd6-dead-4bb1-9f3d-0c749553c...@8g2000hse.googlegroups.com>,
conrad <con...@lawyer.com> wrote: > I'm showing that a given family of curves are > orthogonal trajectories of each other.
> I have the two curves: > x^2 + y^2 = ax > x^2 + y^2 = by
> It is said that the two circles intersect > at the origin. How is this evident > other than by plugging in 0?
Complete the square in x^2 + y^2 = ax to see you have a circle of radius __ centered at __. This will immediately tell you the circle goes thru the origin, as well as what the tangent line at the origin has to be. Same with the other circle.
Alternately, you could down a couple of pints of Guinness and then see what occurs to you.
>> I'm showing that a given family of curves are >> orthogonal trajectories of each other.
>> I have the two curves: >> x^2 + y^2 = ax >> x^2 + y^2 = by
>> It is said that the two circles intersect >> at the origin. How is this evident >> other than by plugging in 0?
> Plugging in (0, 0) for (x, y) and seeing that the result is 0 in > both cases seems about as evident a method as one could wish for! > What worries you about it? (Something to do with orthogonality?)
