# Ellipses homework help

1. Jun 5, 2007

### physicsgal

1. The problem statement, all variables and given/known data
an ellipse is represented by the equation:
x|^2 + 4y^2 - 4 x + 8y - 60 = 0

express the equation in standard form:
((x-2)^2 / 68) + ((y-4)^2/17) = 1

can anyone tell me if this is accurate? thanks

~Amy

2. Jun 5, 2007

### G01

That isn't what I'm getting. Can I see your work?

3. Jun 5, 2007

### physicsgal

thanks!

i wont show all of it (half page long), but here's the main parts:

x^2 - 4x + 4y^2 + 8y = 60

(x^2 - 4X + 4) + (4(y^2 - 2y + 1) = 68
(i got the 68 by added 60 + 4 + 4(1))

~Amy

4. Jun 5, 2007

### G01

OK, I get:

$$\frac{(x-2)^2}{68}+\frac{(y+1)^2}{17}=1$$

I think your mistake lies when you factored out that four and simplified, check your work.

Last edited: Jun 5, 2007
5. Jun 5, 2007

### physicsgal

thank you very much!

i have another quick question:
an elliptical pool table is 5m at its longest point, and 3m wide at its widest point. the pool table has two holes at the position of the foci.

so for the equation i have figured out:
x^2/6.25 + y^2/2.25 = 1.

and the foci are at (-1, 0), (1,0)
(im pretty sure all this is accurate, but anythings possible).

"if a pool ball is hit from one focus, it will bounce once and enter the hole at the other focus. will the ball always travel the same distance if it is hit from one focus? if so, what is the distance it will travel? prove your statement by selecting any point (except one of the vertices) on the edge of the pool table."
i know how to figure out how to get one of the points on the edge of the pool, but what next?

~Amy

6. Jun 5, 2007

### symbolipoint

Good work, ~Amy,
your solution must have been equivalent to:
$$% MathType!MTEF!2!1!+- % feqaeaartrvr0aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaaaamaaaOabaeqabaGaamiEamaaCaaaleqabaGaaGOmaa % aakiabgkHiTiaaisdacaWG4bGaey4kaSIaaGinaiaadMhadaahaaWc % beqaaiaaikdaaaGccqGHRaWkcaaI4aGaamyEaiabgkHiTiaaiAdaca % aIWaGaeyypa0JaaGimaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHsislcaaI0aGaamiEaiabgUcaRiaaisdadaqadaqaaiaadMhada % ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyEaaGaayjkaiaa % wMcaaiabgkHiTiaaiAdacaaIWaGaeyypa0JaaGimaaqaamaabmaaba % GaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaisdacaWG4bGa % ey4kaSYaaeWaaeaacaaI0aGaai4laiaaikdaaiaawIcacaGLPaaada % ahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacqGHsisldaqadaqa % aiaaisdacaGGVaGaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaaisdadaqadaqaaiaadMhadaahaaWcbeqaaiaa % ikdaaaGccqGHRaWkcaaIYaGaamyEaiabgUcaRmaabmaabaGaaGOmai % aac+cacaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGc % caGLOaGaayzkaaGaeyOeI0IaaGinamaabmaabaGaaGOmaiaac+caca % aIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0Ia % aGOnaiaaicdacqGH9aqpcaaIWaaabaWaaeWaaeaacaWG4bGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHi % TiaaisdacqGHRaWkcaaI0aWaaeWaaeaacaWG5bGaey4kaSIaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaisda % cqGHsislcaaI2aGaaGimaiabg2da9iaaicdaaaaa!8D64! $\begin{array}{l} x^2 - 4x + 4y^2 + 8y - 60 = 0 \\ x^2 - 4x + 4\left( {y^2 + 2y} \right) - 60 = 0 \\ \left( {x^2 - 4x + \left( {4/2} \right)^2 } \right) - \left( {4/2} \right)^2 + 4\left( {y^2 + 2y + \left( {2/2} \right)^2 } \right) - 4\left( {2/2} \right)^2 - 60 = 0 \\ \left( {x - 2} \right)^2 - 4 + 4\left( {y + 1} \right)^2 - 4 - 60 = 0 \\ \end{array}$$$
GO1 put his information up before I could.

7. Jun 5, 2007

### symbolipoint

I tried to post a solution process but the LaTex did not work (it used to work)

8. Jun 5, 2007

### symbolipoint

% MathType!MTEF!2!1!+- % feqaeaartrvr0aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaaaamaaaOabaeqabaGaamiEamaaCaaaleqabaGaaGOmaa % aakiabgkHiTiaaisdacaWG4bGaey4kaSIaaGinaiaadMhadaahaaWc % beqaaiaaikdaaaGccqGHRaWkcaaI4aGaamyEaiabgkHiTiaaiAdaca % aIWaGaeyypa0JaaGimaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHsislcaaI0aGaamiEaiabgUcaRiaaisdadaqadaqaaiaadMhada % ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyEaaGaayjkaiaa % wMcaaiabgkHiTiaaiAdacaaIWaGaeyypa0JaaGimaaqaamaabmaaba % GaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaisdacaWG4bGa % ey4kaSYaaeWaaeaacaaI0aGaai4laiaaikdaaiaawIcacaGLPaaada % ahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacqGHsisldaqadaqa % aiaaisdacaGGVaGaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaaisdadaqadaqaaiaadMhadaahaaWcbeqaaiaa % ikdaaaGccqGHRaWkcaaIYaGaamyEaiabgUcaRmaabmaabaGaaGOmai % aac+cacaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGc % caGLOaGaayzkaaGaeyOeI0IaaGinamaabmaabaGaaGOmaiaac+caca % aIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0Ia % aGOnaiaaicdacqGH9aqpcaaIWaaabaWaaeWaaeaacaWG4bGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHi % TiaaisdacqGHRaWkcaaI0aWaaeWaaeaacaWG5bGaey4kaSIaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaisda % cqGHsislcaaI2aGaaGimaiabg2da9iaaicdaaaaa!8D64!  \eqalign{ & x^2 - 4x + 4y^2 + 8y - 60 = 0 \cr & x^2 - 4x + 4\left( {y^2 + 2y} \right) - 60 = 0 \cr & \left( {x^2 - 4x + \left( {4/2} \right)^2 } \right) - \left( {4/2} \right)^2 + 4\left( {y^2 + 2y + \left( {2/2} \right)^2 } \right) - 4\left( {2/2} \right)^2 - 60 = 0 \cr & \left( {x - 2} \right)^2 - 4 + 4\left( {y + 1} \right)^2 - 4 - 60 = 0 \cr}

9. Jun 5, 2007

### G01

10. Jun 5, 2007

### physicsgal

thanks for your help ) talk to you tomorrow.

~Amy

11. Jun 5, 2007

### symbolipoint

~Amy
b^2 = a^2 - c^2

Yes, the ball will travel the same distance from one focus to the other nomatter from which direction it leaves the initial focus.

12. Jun 6, 2007

### chaoseverlasting

I guess 2a=5 and 2b=3. So the equation pans out to $$\frac{x^2}{2.5^2}+\frac{y^2}{1.5}^2=1$$, and what they're asking for is that the sum of the distances from the foci... this is a constant, you can prove this by finding the distance of any arbitrary point from the two directrixes, and using the relation ps=e.pm, where s is the focus, and M is a point such that pm is perpendicular to the directrix.... this works out to be 2ae, where e is the eccentricity of the ellipse.

13. Jun 6, 2007

### physicsgal

thanks chaoeverlasting, but i didnt quite understand everything you said.

here's what ive done:
if x = 1, y = 1.37 (1, 1.37)
foci (-2, 0), (2, 0)

d = *square root*(x2 - x1)^2 + (y2 - y1)2

( i do this for (1, 1.37) and (-2, 0). and then for (1, 1.37) and (2, 0))

= 3.30, and 1.70

but they should both equal the same thing.. does anyone see where i went wrong?

thanks

`Amy