Ecentricity

**jumpman23** · May 30th 2009, 08:34 AM

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

**earboth** · May 30th 2009, 11:05 AM

Originally Posted by jumpman23

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

1. By completing the square you get:

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

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

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

2. You are supposed to know that

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

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

**jumpman23** · May 30th 2009, 04:27 PM

thanks

**pankaj** · May 30th 2009, 04:38 PM

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

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

$<br /> a^2=b^2+a^2e^2<br />$

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

**earboth** · May 30th 2009, 11:26 PM

Originally Posted by pankaj

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

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

$<br /> a^2=b^2+a^2e^2<br />$

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

Thanks for your reply - but in my opinion:

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

2. there exist a numerical eccntricity $\dfrac ea=\epsilon$ . 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)

**stapel** · May 31st 2009, 04:59 AM

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!!

**pankaj** · May 31st 2009, 06:42 AM

Sorry.Actually I had never read about Linear and numerical eccentricity

**HallsofIvy** · May 31st 2009, 07:04 AM

Originally Posted by earboth

Thanks for your reply - but in my opinion:

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

2. there exist a numerical eccntricity $\dfrac ea=\epsilon$ . 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)

What you calculated would be called, in English, the "focal length"- the distance from the center of the ellipse to each focus.

Thread: Ecentricity

LinkBack

Thread Tools

Display

Ecentricity

Search Tags