Angry Badger Member Oct 13, 2008 #1 Hi, I'm stuck on this! Have had several attempts but can't arrive at the given solution. I've no propensity for maths! **broken link removed** Any help gratefully appreciated.
Hi, I'm stuck on this! Have had several attempts but can't arrive at the given solution. I've no propensity for maths! **broken link removed** Any help gratefully appreciated.
M MrAl Well-Known Member Most Helpful Member Oct 13, 2008 #2 Hi, That's an ellipse. p^2=x^2/(1-y^2/q^2) so p1=sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1 p2=-sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1 BTW, p is the distance along the x axis from the center to each edge of the ellipse. Last edited: Oct 13, 2008
Hi, That's an ellipse. p^2=x^2/(1-y^2/q^2) so p1=sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1 p2=-sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1 BTW, p is the distance along the x axis from the center to each edge of the ellipse.
Angry Badger Member Oct 13, 2008 #3 Hello Mr Al, Don't know if your still around. Thanks for your prompt reply earlier. Is this how you arrived at your solution? **broken link removed** That was one of the solutions that I arrived at but the solution in the text book is: **broken link removed** Actually, in the text book solution the x is X (capitalised). Must be a typo. Is the solution correct? Thanks again.
Hello Mr Al, Don't know if your still around. Thanks for your prompt reply earlier. Is this how you arrived at your solution? **broken link removed** That was one of the solutions that I arrived at but the solution in the text book is: **broken link removed** Actually, in the text book solution the x is X (capitalised). Must be a typo. Is the solution correct? Thanks again.
M MrAl Well-Known Member Most Helpful Member Oct 13, 2008 #4 Hi again, Oh ok, well that requires one little extra step then... starting again with: p^2=x^2/(1-y^2/q^2) multiply the top and bottom of the right side by q^2 to get: p^2=x^2*q^2/(q^2-y^2) Now when we take the square root we can perform this on the top but not on the bottom so we get: p=x*q/sqrt(q^2-y^2) Ok now?
Hi again, Oh ok, well that requires one little extra step then... starting again with: p^2=x^2/(1-y^2/q^2) multiply the top and bottom of the right side by q^2 to get: p^2=x^2*q^2/(q^2-y^2) Now when we take the square root we can perform this on the top but not on the bottom so we get: p=x*q/sqrt(q^2-y^2) Ok now?
Angry Badger Member Oct 13, 2008 #5 Hi, Thanks. I can follow your working, I'd never have 'seen' the way to proceed.