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 January 11th, 2012, 03:52 PM #1 Member   Joined: Dec 2011 Posts: 41 Thanks: 0 Ellipse question (eccentricity e.t.c) - pre exam Let $\epsilon=\ \frac{1}{4}\ and\ p\ =\frac{1}{2}$ be the eccentricity and the focal parameter of an ellipse in the plane. Find $a>b>o$ such that the equation of the ellipse becomes $\frac{x^2}{a^2}\ +\ \frac{y^2}{b^2}\=\ 1$ Find the two focus-directrix pairs. Direction on the directrix would be great cheers no but im pretty clueless on this all round so any help very much appreciated.
 January 11th, 2012, 06:11 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Ellipse question (eccentricity e.t.c) - pre exam The focal parameter of an ellipse p is given by: $p=\frac{b^2}{\sqrt{a^2-b^2}}$ The eccentricity e of an ellipse is given by: $e=\sqrt{1-\frac{b^2}{a^2}}$ So we have the non-linear system: (1) $\frac{1}{2}=\frac{b^2}{\sqrt{a^2-b^2}}$ (2) $\frac{1}{4}=\sqrt{1-\frac{b^2}{a^2}}$ We may arrange (2) as: $\frac{a}{4}=\sqrt{a^2-b^2}$ We may arrange (1) as $2b^2=\sqrt{a^2-b^2}$ Thus $a=8b^2$ Substituting for a into (2) gives: $\frac{1}{4}=\sqrt{1-\frac{b^2}{$$8b^2$$^2}}=\frac{\sqrt{64b^2-1}}{8b}$ $2b=\sqrt{64b^2-1}$ Since 0 < b, we may square: $4b^2=64b^2-1$ $b^2=\frac{1}{60}$ $b=\frac{1}{2\sqrt{15}}$ and so: $a=8$$\frac{1}{60}$$=\frac{2}{15}$ Since a > b, the major axis is on the x-axis, and the foci will be at the points $$$\pm\sqrt{a^2-b^2},0$$=$$\pm\frac{1}{30},0$$$ The directices will be the lines $x=\pm\frac{a}{e}=\pm\frac{\frac{2}{15}}{\frac{1}{4 }}=\pm\frac{8}{15}$ Note: the directrices may also be found with $x=\pm(|f|+p)=\pm$$\frac{1}{30}+\frac{1}{2}$$=\pm\f rac{16}{30}=\pm\frac{8}{15}$
 January 12th, 2012, 02:11 PM #3 Member   Joined: Dec 2011 Posts: 41 Thanks: 0 Re: Ellipse question (eccentricity e.t.c) - pre exam Thank you indeed. i will study what your post.
 January 13th, 2012, 08:24 AM #4 Member   Joined: Dec 2011 Posts: 41 Thanks: 0 Re: Ellipse question (eccentricity e.t.c) - pre exam shockingly bad post by me. forgot to take out the what.
 January 13th, 2012, 08:29 AM #5 Member   Joined: Dec 2011 Posts: 41 Thanks: 0 Re: Ellipse question (eccentricity e.t.c) - pre exam though im still a little unsure about the major axis on teh x axis. this is just far superior than any explanation that would have come from elsewhere. my exam is tomorrow, thank you so fucking much.

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