## anonymous one year ago I WILL GIVE MEDAL AND FAN: Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1

1. anonymous

@nincompoop

2. anonymous

@ash2326

3. anonymous

@ganeshie8

4. UsukiDoll

standard ellipse equation For a wider-than-tall ellipse with center at (h, k), having vertices a units to either side of the center and foci c units to either side of the center, the ellipse equation is: $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ For a taller-than-wide ellipse with center at (h, k), having vertices a units above and below the center and foci c units above and below the center, the ellipse equation is: $\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1$ standard foci equation $a^2-c^2=b^2$ standard center $(h,k)$

5. UsukiDoll

based on your equation $\frac{x^2}{100}+\frac{y^2}{64}=1$ a > b there's a horizontal major axis

6. UsukiDoll

EXAMPLE: suppose we have the equation $\frac{x^2}{4}+\frac{y^2}{1}=1$

7. UsukiDoll

since a > b we have a horizontal major axis so we re-write this equation as where a = 2 and b = 1 $\frac{x^2}{2^2}+\frac{y^2}{1^2}=1$

8. UsukiDoll

before we start graphing, we want the center. Recall that the standard ellipse equation is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ For this equation we just have $x^2$ and $y^2$ $\frac{x^2-0}{2^2}+\frac{y^2-0}{1^2}=1$ which means that our center $(h,k) \rightarrow (0,0)$

9. UsukiDoll

|dw:1437650242909:dw|

10. UsukiDoll

for vertices we need (-a,0), (a,0), (-b,0),(0,b) since a = 2 and b = 1 our vertex points are (-2,0) (2,0) (-1,0)(0,1)

11. UsukiDoll

|dw:1437650445276:dw|

12. UsukiDoll

Foci : $a^2-c^2=b^2$ since a = 2 and b = 1 $2^2-c^2=1^2$ $4-c^2=1$ $-c^2=1-4$ $-c^2=-3$ $c^2=3$ $\sqrt{c^2}=\sqrt{3}$ $c=\sqrt{3}$

13. anonymous

lol nm i never saw that someone replied.

14. anonymous

@phi can you actually help a little more on explaining this a bit better

15. phi

I would hope that is enough info to answer the question

16. phi

try reading through the posts. 1) can you figure out the center? 2) the vertices?

17. UsukiDoll

don't forget the foci

18. anonymous

yeah hold on

19. UsukiDoll

alright. :)

20. anonymous

the center is (0,0) right?

21. UsukiDoll

yes

22. anonymous

for the foci: 10^2 - 8^2 = b^2 100 - 64 = b^2 36 = b^2 b = 6, so the foci is (-6,0) and (6,0)?

23. anonymous

how's that?

24. UsukiDoll

errr. $a^2-c^2=b^2$ we are looking for c a = 10 and b = 6

25. anonymous

oops, so: 100 - c^2 = 36 so c is 8?

26. UsukiDoll

but the vertices for b is (-6,0) (6,0)

27. UsukiDoll

we need vertices for a though (-a,0), (a,0)

28. UsukiDoll

yes c is 8

29. anonymous

vertices for a = (-10,0) and (10,0)

30. UsukiDoll

yes. now we can use that information to graph. though...if we draw it with os tools it's gonna look ugly. I'll attach a file

31. UsukiDoll

32. anonymous

ok so is this right? all together: center: (0,0) vertices: (-10,0); (10,0) foci: (-6,0); (6,0)

33. UsukiDoll

yeah

34. anonymous

awesome thanks, help on another one?

35. UsukiDoll

sure

36. anonymous

Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18

37. UsukiDoll

alright... we don't have the ellipse in standard form. so we need to divide 18 all over the equation

38. anonymous

ok so...: (3/18)x^2 + (6/18)y^2 that looks weird

39. UsukiDoll

yeah but we need a 1 and dividing 18 all over the equation makes that =1 happen

40. anonymous

oh so i did it right?

41. UsukiDoll

yeah

42. anonymous

so now what

43. UsukiDoll

I'm just trying to think what's going on... I don't have perfect squares in here.

44. anonymous

ok

45. UsukiDoll

find the center.. that's the easiest part >_<

46. anonymous

actually not sure how, i know it is (h,k) but where is that in our equation

47. anonymous

is it (0,0) again. i went on wolphram alpha and it simplifies the equation to $\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1$

48. UsukiDoll

If our equation was similar to $7x^2 +16(y+2)^2=112$ and then divide 112 all over then our center would be (0,-2)

49. anonymous

ok actually, looking at my answer choices, the center can only be (0,0) lol so moving on to the vertices.

50. UsukiDoll

but when we reduced that fraction from earlier $\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1$ there is no k and no h so center is (0,0) the vertices is next

51. anonymous

ok

52. UsukiDoll

and I just figured it out in my noggin we have square roots this time

53. UsukiDoll

$\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1$ $\frac{ x^2 }{ (\sqrt{6})^2 } + \frac{ y^2 }{ (\sqrt{3})^2 } = 1$

54. anonymous

so i got vertices: $(-\sqrt6,0); (\sqrt6,0)$ foci: $(-\sqrt3,0); (\sqrt3,0)$

55. UsukiDoll

yes!

56. anonymous

is that right?

57. anonymous

yay!

58. anonymous

one last one i have to work backwards on i thin. can you help?

59. UsukiDoll

I guess x.x ok y not

60. anonymous

Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. No clue how to work this one. Sorry for being such a pain...

61. UsukiDoll

vertical major axis -> 18 minor axis of length -> 16

62. UsukiDoll

minor would be the foci and the major would be the vertices since we have vertical major axis, our equation should be in the form of $\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1$

63. anonymous

ok so 18 ^2 for a^2? and 16 ^2 for b^2

64. UsukiDoll

yeah.....

65. anonymous

so how do I get the equation plug in these numbers so I get: $(\frac{ x^2 }{ 18 })+(\frac{ y^2 }{ 16}) = 1$is that right?

66. UsukiDoll

I haven't done this in a while, but these sites are confusing me... this vertical formula is switched $\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1$

67. UsukiDoll

our center is still (0,0) that's one good part

68. anonymous

ok so we are going to have x^2 and y^2 by themselves on the top but what about the bottom.

69. UsukiDoll

vertical major axis -> 18 minor axis of length -> 16

70. UsukiDoll

a = length of semi-major axis b = length of semi-minor axis vertices: (h, k + a), (h, k - a) co-vertices: (h + b, k), (h - b, k) [endpoints of the minor axis]

71. UsukiDoll

now I'm confusing myself.. but .. a =18 and b = 16

72. UsukiDoll

yeah if the number under the y part is larger then it's going to be vertical

73. UsukiDoll

so a has to be 18 and b has to be 16

74. UsukiDoll

but ... we're still not done we have to write those numbers to the second power

75. UsukiDoll

$\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1$

76. anonymous

(x^2/16^2) + (y^2/18^2) = 1?

77. UsukiDoll

yup I think you may have to expand like 16 x 16 and 18 x 18

78. anonymous

do you want me to give the answer choices? doesn't look like any of them are working.

79. UsukiDoll

yeah what are they

80. anonymous

$\frac{ x^2 }{64 }+\frac{ y^2 }{ 81 } = 1$ $\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 8 } = 1$ $\frac{ x^2 }{ 81 }+\frac{ y^2 }{ 64 } = 1$ $\frac{ x^2 }{ 8 }+\frac{ y^2 }{ 9 } = 1$

81. anonymous

that took a while lol. so how do we get one of those.

82. UsukiDoll

what the heck that doesn't make sense.

83. anonymous

did we do anything wrong?

84. UsukiDoll

on one choice that would mean a = 9, b = 8 and on the other a = 8 and b = 9 last choice won't work

85. UsukiDoll

second choice won't work either.

86. UsukiDoll

did you copy the right numbers?

87. anonymous

yep

88. anonymous

that's why i'm confused

89. UsukiDoll

then that would mean that you were given the wrong numbers

90. anonymous

here i will get phi back in here maybe: @phi

91. UsukiDoll

@Astrophysics help!

92. Astrophysics

What's up?

93. UsukiDoll

this doesn't match... I mean @twistnflip tell her about the last problem

94. anonymous

ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do?

95. phi

remember that a and b are the "semi-major" and "semi-minor" axes. (comparable to the radius) in other words, divide the given axes lengths by 2

96. UsukiDoll

*facepalm* missed that part.

97. phi

thus you want a=9 and b=8

98. UsukiDoll

k... I see it now 18/2 = 9 16/2 = 8

99. UsukiDoll

and since it's vertical that means the denominator in the y portion has to be larger

100. anonymous

so i get the equation: $\frac{ x^2 }{ 8} + \frac{ y^2 }{ 9 }$?

101. Astrophysics

Haha, well here is a good reference on the standard form of an ellipse, and will help you from not making these kinds of mistakes :P |dw:1437655357843:dw|

102. UsukiDoll

getting there. there's still one more step

103. anonymous

which is?

104. phi

you want a^2 and b^2 in your equation (not just a and b)

105. Astrophysics

|dw:1437655407458:dw|

106. UsukiDoll

since a = 9 and that formula has a^2 9^2 = ?

107. UsukiDoll

umm @Astrophysics it's vertical

108. UsukiDoll

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @twistnflip ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do? $$\color{#0cbb34}{\text{End of Quote}}$$ vertical major axis

109. Astrophysics

I know, I just put that to show it's squared, didn't really pay attention to the vertical/ horizontal :P

110. anonymous

81 and 8 ^2 = 64 so i have $\frac{ x^2 }{ 64} + \frac{ y^2 }{81 } = 1$

111. UsukiDoll

that's better.

112. anonymous

so that's it?

113. phi

you want it too be harder?

114. anonymous

haha no thanks so much guys! you da bomb