## frx 2 years ago The equation for an ellipse is $\frac{ x ^{2} }{ a ^{2} }+\frac{ y ^{2} }{ b ^{2}}=1$ and the foci is given by (c,0) where $c ^{2}=a ^{2}-b ^{2}$ and my question is if b>a is the foci given by (0,c) where $c ^{2}=a ^{2}-b ^{2}$

1. asnaseer

yes, if a>b then foci are at $$(\pm c, 0)$$ otherwise they are at $$(0, \pm c)$$. see here for more details: http://hotmath.com/hotmath_help/topics/ellipse.html

2. henpen

I suppose you could try to derive the equation as a locus from the two foci (c,0) and (-c,0).|dw:1354462354704:dw| $\sqrt{y^2+(x-c)^2}+\sqrt{y^2+(x+c)^2}=k$

3. Chlorophyll

It seems you're confused with the notation: Foci always belong to major axis, and major axis always named a!

4. henpen

$y^2+(x-c)^2+y^2+(x+c)^2+\sqrt{y^2+(x-c)^2}\sqrt{y^2+(x+c)^2}=k^2$ $y^2+(x-c)^2+y^2+(x+c)^2-k^2=-\sqrt{y^2+(x-c)^2}\sqrt{y^2+(x+c)^2}$ $(y^2+(x-c)^2+y^2+(x+c)^2-k^2)^2=(y^2+(x-c)^2)(y^2+(x+c)^2)$ans so on

5. henpen

*and

6. frx

|dw:1354462531860:dw|

7. frx

So thats the deal when a>b, right?

8. frx

|dw:1354462661891:dw| And that b>a

9. frx

oh wrong it should be rotated 90 degrees

10. frx

|dw:1354462770084:dw|

11. asnaseer

yes, 'a' always represents the "major axis"

12. frx

Ok great thank you guys!

13. asnaseer

yw :)