This problem bothers me a bit: Find eccentricity of ellipse if distance between its foci is arithmetical average of length of semi major and semi minor axis. Well I know that e=c/a and c^2 = a^2 - b^2. 2c=a+b, should I just plugin inside e=c/a for c? Is there something I am missing here?
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It's basic use of formulae. – StubbornAtom Sep 01 '16 at 17:39
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Did I set up formulas correctly? Is it just simple algebra from here? What is the solution to this? – Sep 01 '16 at 17:42
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$e=\sqrt {1-\frac{b^2}{a^2}}$ and $2ae=\frac{a+b}{2}$ for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. – StubbornAtom Sep 01 '16 at 17:46
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So it appears my solution was correct. Thank you for your answer! I realize it is a simple elementary problem, but I got confused if I was missing some information since solution was straightforward. – Sep 01 '16 at 17:49
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$4a^2-4b^2=4c^2= (a+b)^2=a^2+2ab+b^2.$ So $3a^2-5b^2-2ab=0.$ Dividing thru by $a^2$ and putting $b/a=e,$ we have $3-5e^2-2e=0.$ With one positive root, $e=3/5.$
DanielWainfleet
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