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The latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the same as latus rectum of a parabola $y^2=4cx$ . Find eccentricity of the ellipse .

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I'm not sure I understand: by definition, that ellipse's eccentricity is $$\frac1{\max(a,b)}\sqrt{|a^2-b^2|}$$ so what the latus rectum's business is here? – DonAntonio May 25 '13 at 9:53
up vote 1 down vote accepted


from this and this, the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $2a(1-e^2)$ and $b^2=a^2(1-e^2)$ where $a$ is Semi major Axis, $b$ is the Semi-minor Axis and $e$ is the Eccentricity

and the length of the latus rectum of the parabola $y^2=4ax$ is $4a$

Can you utilize the relations to find the value of $e$

Check separately for cases when $a\ge b$ and when $a<b$

EDIT: after a drastic change in the question $y^2=4ax$ to $y^2=4cx$

Now WLOG, we can choose $a\ge b$

As the focus of the parabola is $(c,0)$ and those of the ellipse are $(\pm ae,0)$


Now, $2a(1-e^2)=4c=4ae\implies 1-e^2=2e\implies e^2+2e-1=0$

$\implies e=\frac{-2\pm{\sqrt{2^2-4(-1)1}}}{2\cdot1}=-1\pm\sqrt2$

As $0\le e<1, e=\sqrt2-1$

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Focus of ellipse must coincide with focus of parabola. And following your relations as well. But I am unable to produce a final answer – user78743 May 25 '13 at 12:36
@dkbose, if $a\ge b,2a(1-e^2)=4a\implies 1-e^2=2\implies e^2=-1$ which is impossible. If $a<b, 2b(1-e^2)=4a$ and $a^2=b^2(1-e^2)$ eliminate $b$ to get $1-e^2=2$ which is also impossible – lab bhattacharjee May 25 '13 at 12:42
I am extremely sorry , there was a typo in the problem which changed the whole meaning drastically . – user78743 May 25 '13 at 15:06
@dkbose, can you manage with the suggestions supplied already? – lab bhattacharjee May 25 '13 at 15:10
Basically , I am struggling with the answer . I can award you a bounty of 50 rep in 2 days to duly award your help . – user78743 May 25 '13 at 15:21

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