## anonymous one year ago Anyone here good with ellipses? If so, please help!! Find the center, vertices, and foci of the ellipse with equation x squared divided by four hundred plus y squared divided by two hundred and fifty six = 1 A. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-12,0), (12, 0) B. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-16,0), (16, 0) C. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-16), (0, 16) D. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-12), (0, 12) I know the answer is either B or C

1. anonymous

And also for Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct?

2. Nnesha

I have few minutes lets see if i can help u :=)

3. Nnesha

1st) is it vertical or horizontal ?

4. anonymous

Ok thank you! And for the first one it is horizontal. @Nnesha

5. Nnesha

yes right so formula for foci $\huge\rm (h \pm c , k)$

6. Nnesha

first of all you need to find c

7. Nnesha

use this equation $\huge\rm c^2=a^2-b^2$

8. anonymous

Would c = SqRt656? @Nnesha

9. Nnesha

remember (h,k) is the center point it's horizontal so only x values would change y would stay same no change in y values

10. Nnesha

no what is a^2 and b^2 ?

11. Nnesha

$\huge\rm \frac{ (x-h)^2 }{ a^2 }+\frac{ (y-k)^2 }{ b^2 }$ standard form of ellipse

12. anonymous

Wouldn't a^2 = 256 and b^2 = 400?

13. anonymous

Oh wait.

14. anonymous

12 = c?

15. Nnesha

okay for ellipse a= bigger number if a is under the x then it's horizontal and if a is under y then it would be vertical

16. Nnesha

yes !

17. Nnesha

so now what are the foci points ?

18. anonymous

+/12, 0?

19. anonymous

Forgot the ( ) sorry

20. Nnesha

$(\pm 12,0)$ looks great should be plus minus sign

21. anonymous

Yes that's what I meant, sorry.

22. anonymous

So my answer would actually be A. Ok, thanks! Could you help me with the 2nd one?

23. anonymous

And also with this one: The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides. I believe my answer would be: x^2/25 + y^2/4 = 1 But I just wanted to make sure.

24. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @jdoherty And also for Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct? $$\color{blue}{\text{End of Quote}}$$ *vertical*

25. anonymous

? @Nnesha

26. Nnesha

if ellipse is vertical then bigger number supposed to be under x or y ?

27. anonymous

Y?

28. Nnesha

yes right try to draw points on the paper

29. Nnesha

gtg sorry. :(

30. anonymous

Ok :( Are you free at all tomorrow?

31. anonymous

@saseal Could you help with the other two questions?

32. anonymous

ok

33. anonymous

Thank you! @saseal

34. anonymous

1. The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides. I believe my answer would be: x^2/25 + y^2/4 = 1 But I just wanted to make sure. 2. Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct?

35. anonymous

@saseal ^

36. anonymous

first one is correct...2nd one is debatable because the question is not very specified

37. anonymous

if they specifed the major axis is x and minor is y then you are correct

38. anonymous

Hmm...how could I determine for sure?

39. anonymous

hmm you are wrong...it says vertical major axis

40. anonymous

Oh ok. So that would make the y axis the major axis?

41. anonymous

Oh wait. would it be x^2/4 + y^2/9 = 1? @saseal

42. anonymous

yea

43. anonymous

Ok

44. anonymous

Could you help me a couple more questions? I kind of am just getting this stuff. Lol. @saseal

45. anonymous

ok

46. anonymous

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1 This is for Hyperbolas...

47. anonymous

$\frac{ x^2 }{ 3^2 } - \frac{ y^2 }{ 4^2 } = 1$

48. anonymous

now it becomes y^2/4^2 - x^2/3^2

49. anonymous

$\frac{ y^2 }{ 4^2 } - \frac{ x^2 }{ 3^2 } =1$

50. anonymous

Ok so then what can I do?

51. anonymous

you know where thid hyperbola opens open?

52. anonymous

Not really like when it's graphed?

53. anonymous

|dw:1439004618140:dw|

54. anonymous

Ok so they open up where the two points are? Like the end points of the opening?

55. anonymous

kinda...abit like parabola x2