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Let the ellipse $x^2+2y^2+2xy=1$ Find the eccentricity of this ellipse

Ben Grossmann
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Nikos127
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    What are your thoughts on the problem? What have you tried? – Ben Grossmann Aug 17 '20 at 09:21
  • on a side note, there is a general formula for eccentricity for any 2 degree curve available on Math.S.E. – Anindya Prithvi Aug 17 '20 at 09:23
  • There is a formula for the eccentricity given here. If this is not the kind of solution you want, then it would help if you explained how you are meant to solve the problem. If this is from a class, then how have similar problems been solved? – Ben Grossmann Aug 17 '20 at 09:27
  • @BenGrossmann If i had encountered the question by the OP, it would have been https://wikimedia.org/api/rest_v1/media/math/render/svg/7243bc5d4717aaf1723f4046ea15c1f83a943b32 because this ellipse is rotated, same wiki page, under the heading Eccentricity in terms of coefficients – Anindya Prithvi Aug 17 '20 at 09:32
  • https://math.stackexchange.com/questions/3775650/coordinates-of-focus-of-parabola/3775661#3775661 – lab bhattacharjee Aug 17 '20 at 09:35
  • This isnt from class, this is a part of an excercise from calculus – Nikos127 Aug 17 '20 at 09:42
  • Can you rotate the ellipse, to put it into standard form? – Angina Seng Aug 17 '20 at 10:06

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$$x^2+2y^2+2xy=1$$ Let $x=r \cos t, y= r\sin t$, then $$r^2=\frac{1}{\cos^2 t+ 2\sin^2 t+\sin 2t}= \frac{1}{1+\sin^2 t+\sin 2t}=\frac{2}{2+1-\cos 2t+\sin 2t}$$ $$\implies r^2=\frac{2}{3-\sqrt{2}(\sin (t-\pi/4)}$$ So $$r_{min}=\sqrt{\frac{2}{3+\sqrt{2}}}. r_{max}==\sqrt{\frac{2}{3-\sqrt{2}}}$$ So the eccentricity $$e=\sqrt{1-\frac{r_{min}^2}{r_{max}^2}}$$ $$=\sqrt{1-\frac{3-\sqrt{2}}{3+\sqrt{2}}}=\sqrt{\frac{2\sqrt{2}}{3+\sqrt{2}}}=0.8004...$$

Z Ahmed
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