## anonymous one year ago help me plz

1. anonymous

2. anonymous

@Directrix

3. anonymous

brb in 20

4. anonymous

im back

5. Nnesha

first of all you should know is it horizontal or vertical a always greater than b a>b if a is under the x variable then it would be horizontal equation $\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1$ if a is under the y variable then it would be vertical equation $\rm \frac{ (x-h)^2 }{ b^2 }+ \frac{ (y-k)^2 }{\color{reD}{ a^2} }=1$

6. Nnesha

so is it vertical or horizontal ?

7. anonymous

i think its vertical

8. Nnesha

foci for horizontal $\large\rm (h, k \pm c)$ and for vertical equation $\large\rm (h \pm c , k)$ use the equation to find c $\huge\rm c^2 =a^2-b^2$

9. anonymous

ok so which one do i use

10. Nnesha

well a is bigger number which is under the x variable so is it vertical or horizontal ?

11. Nnesha

12. anonymous

so it is horizontal right

13. Nnesha

yes right $\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1$ now use the equation posted above to find c

14. Nnesha

c^2=a^2-b^2 this one

15. anonymous

wait what

16. Nnesha

foci for horizontal $\large\rm (h, k \pm c)$ and for vertical equation $\large\rm (h \pm c , k)$ those two are foci pair we need so use this equation to find c $\huge\rm c^2 =a^2-b^2$

17. Nnesha

a^2 and b^2 are in the given equation

18. anonymous

wow ok so umm can you start me off plz

19. Nnesha

horizontal equation for ellipse $\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1$ the number at the denominator are a^2 and b^2 and like i said bigger number equal to a^2

20. Nnesha

what two numbers are replaced with a^2 and b^2 in this equation $\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1$ ??

21. anonymous

brb i gtg cook dinner

22. anonymous

36 and 11

23. Nnesha

right which one is a^2 and which one is b^2 ?

24. anonymous

a^2 is 36 b^2 is 11

25. anonymous

@IrishBoy123

26. anonymous

{ (x-h)^2 }{36} }+/{ (y-k)^2 }{11}=1 is this the correct formula

27. IrishBoy123

this ellipse is centred on the origin yep?

28. anonymous

yep

29. IrishBoy123

|dw:1444085734817:dw|

30. anonymous

yep

31. IrishBoy123

Para-phrasing Wiki: The distance from the origin O to either focus is f where: $f = \sqrt{a^2-b^2}.$

32. anonymous

ok she gave me a diffrent formula soo what do i do im so confused now

33. IrishBoy123

here, what is a and what is b? general equation for ellipse centred on Origin is: $$\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1$$

34. IrishBoy123

a = ?? b = ??

35. IrishBoy123

then |dw:1444086259207:dw|

36. anonymous

(x-h)^2 }{36} }+/{ (y-k)^2 }{11}= ok so i got this

37. IrishBoy123

so that means you have finished it?!

38. anonymous

no i have not thats the first equation

39. anonymous

(y-k)^2 {11} this is my answer

40. IrishBoy123

OK for an equation centred on O, we have general equation: $\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1$ The focii are at a distance f from O, where $f = \sqrt{a^2-b^2}$ |dw:1444086808815:dw|

41. IrishBoy123

soz, we crossed posts. i understood the question to be finding the focii of the ellipse

42. anonymous

yes idk what im doing im so cunfused here im sorrry

43. IrishBoy123

no worries you have here: $$\large\rm \frac{ x^2 }{ \color{ReD}{6^2} } +\frac{ y^2 }{(\sqrt{\color{red} 1\color{red}1 })^2}=1$$ and generally speaking: $$\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1$$ and: $$\large f = \sqrt{a^2-b^2}$$

44. IrishBoy123

so f =????

45. anonymous

omg i have no clue on how to do this like wth

46. anonymous

does f equal -4

47. IrishBoy123

$$a = 6, \; b = \sqrt{11}$$ $$f = \sqrt{ 6^2 - (\sqrt{11})^2}$$ $$f = 5$$

48. anonymous

atleast i was close haha

49. IrishBoy123

finished ?? :p

50. anonymous

i think so

51. IrishBoy123

cool!! well done

52. anonymous

53. IrishBoy123

mp

54. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Tazmaniadevil ok she gave me a diffrent formula soo what do i do im so confused now $$\color{blue}{\text{End of Quote}}$$ the formula is same difference is f and c variables f= foci $\huge\rm f = \sqrt{ a^2 -b^2}$ is same as $\huge\rm c^2= a^2-b^2 ~~~~~c=\sqrt{a^2-b^2}$

55. Nnesha

a^2 and b^2 are already given so would be easy you write it as c= sqrt{36-11} :=)

56. anonymous

OK

Find more explanations on OpenStudy