How do you find the eccentricity of 4x^2 + 16y^2 + 8x +64y +4 = 0

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- May 30th 2009, 08:34 AM #1

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- May 30th 2009, 11:05 AM #2
1. By completing the square you get:

$\displaystyle 4x^2 + 16y^2 + 8x +64y +4 = 0$

$\displaystyle 4(x^2+2x\bold{\color{blue}+1}) + 16(y^2 +4y \bold{\color{red}+4}) = -4\bold{\color{blue}+4}\bold{\color{red}+64}$

$\displaystyle \dfrac{(x+1)^2}{16}+\dfrac{(y+2)^2}{4}=1$

2. You are supposed to know that

$\displaystyle e^2+b^2=a^2$ Plug in the values you know and calculate e:

$\displaystyle e^2=16-4=12~\implies~\boxed{e=2\sqrt{3}}$

- May 30th 2009, 04:27 PM #3

- May 30th 2009, 04:38 PM #4
Sorry,

**earboth**the answer is wrong.The curve in question is an ellilse whose eccentricity**must**be less than $\displaystyle 1$.

The formula used by you for calculating eccentricity is wrong.It should be

$\displaystyle

a^2=b^2+a^2e^2

$

Thus $\displaystyle e=\frac{\sqrt{3}}{2}$

- May 30th 2009, 11:26 PM #5
Thanks for your reply - but in my opinion:

1. there exist a**linear eccentricity**$\displaystyle e^2=a^2-b^2$. That's the eccentricity I've calculated.

2. there exist a**numerical eccntricity**$\displaystyle \dfrac ea=\epsilon$. Obviously this is the eccentricity you've had in mind.

(Remark: I translated the technical terms literally from German so maybe the English expressions differ a bit)

- May 31st 2009, 04:59 AM #6

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To confirm the formula they gave you in your book for the eccentricity of an ellipse, and for worked examples of how to find the eccentricity, try

**some online lessons**.

Then make sure that you*memorize*the formula before your next test!!

- May 31st 2009, 06:42 AM #7

- May 31st 2009, 07:04 AM #8

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