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Finding the eccentricity of an ellipse when a line joining the foot of the perpendiculars from a point of a known ellipse (having eccentricity e) at 2 perpendicular lines(example the x and y axes ;they are not necessarily the axes of the known ellipse) is a normal to the unknown ellipse -

Can anyone give a hint on how to go about this one? Is using the condition of normality sufficient to approach this? Or is there a better way?

Thanks a lot.

user232216
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  • Sounds like an initial value problem. I'd use the slope to find the curve and then from there calculate the eccentricity. Not sure if there is some formula for the relation between the derivative at a point and the eccentricity. Seems like there should be. Is this what you're asking for? Do you have a specific example we can work on? – John Molokach Jul 25 '15 at 12:00
  • @JohnMolokach If you want a specific example,you can choose the perpendicular axes as x and y axes. Let the known ellipse be of standard form : like x^2/a^2 + y^2/b^2 = 1. I think that the unknown ellipse also has the same eccentricity in this case. Thanks again for looking into this. – user232216 Jul 25 '15 at 12:20
  • https://en.m.wikipedia.org/wiki/Eccentricity_(mathematics) Look under 'Values' and you'll find $e$ in terms of $a$ and $b.$ Of course the axes lengths are $2a$ and $2b.$ This should be true regardless of the rotation of the axes. – John Molokach Jul 25 '15 at 13:20
  • @JohnMolokach Are you referring to the axes of the ellipse? – user232216 Jul 25 '15 at 13:25
  • Yes. Is there something else I should use? – John Molokach Jul 25 '15 at 15:51

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