## anonymous one year ago Anyone good with hyperbolas? Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1

1. anonymous

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

2. anonymous

Yes this was the problem we were working on last night. (: @saseal

3. anonymous

|dw:1439046272960:dw|

4. anonymous

$c=\sqrt{a^2+b^2}$

5. anonymous

Ok, so would it be the SqRt of 9+16? so the SqRt of 27?

6. anonymous

yea

7. anonymous

um no

8. anonymous

how did you get 9+16=27?

9. anonymous

Isn't a^2 = 9 and b^2 = 16?

10. anonymous

9+16=25

11. anonymous

Oh my bad. Stupid arithmetic problem. Ok, so c = 5?

12. anonymous

yea

13. anonymous

its +/- 5 actually because hyperbola goes both ways, just like vertices

14. anonymous

Oh ok!

15. anonymous

So is that for foci or vertices? @saseal

16. anonymous

c is for foci and a is your vertices

17. anonymous

Ok...so would that make the foci: (+/-5, 0) And vertices: (+/-3, 0) @saseal

18. anonymous

@saseal Could you help me with a couple more?

19. anonymous

yea

20. anonymous

21. anonymous

So the answer is correct with the foci and vertices? How do I know where to put the 0? Meaning, the zero always goes in the y values?

22. anonymous

@saseal

23. anonymous

|dw:1439048329264:dw|this is for $\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1$

24. anonymous

|dw:1439048416352:dw|this is for $\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1$

25. anonymous

Ok... @saseal

26. anonymous

the center of your hyperbola is deternined by the x^2 like $\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1$ has center at origin while$\frac{ (x-1)^2 }{ a^2 } - \frac{( y+1)^2 }{ b^2 } =1$ has a center of (1,-1)

27. anonymous

Ok I get that

28. anonymous

its actually determined by x^2 and y^2 not just x^2, got distacted dem parents

29. anonymous

Oh ok, still makes sense. (:

30. anonymous

So my answer would be correct? @saseal

31. anonymous

yea

32. anonymous

Ok so my next question is: Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

33. anonymous

$\frac{ (x+1)^2 }{ 4^2 } + \frac{ (y+5)^2 }{ 3^2 } = 1$ where do you think is the center of the hyperbola?

34. anonymous

I graphed it and got the center to be ( -1, -5 ) @saseal

35. anonymous

you should be able to get the center without even graphing it

36. anonymous

you see the pattern there with the (x+1)^2 and the (y+5)^2 and the center?

37. anonymous

Oh ok. Yes. -1, -5

38. anonymous

I get the center is (-1, -5), so now what? @saseal

39. anonymous

you know wheres your a and b?

40. anonymous

My a is 16, b is 9. So would that make c = 5?

41. anonymous

a^2 is 16 not a. a is 4, dont be mistaken.

42. anonymous

Oh ok! So would the c still be 5?

43. anonymous

yes

44. anonymous

now where do you think the hyperbola will open out to? left/right or up/down

45. anonymous

left/right

46. anonymous

correct

47. anonymous

|dw:1439050835472:dw|

48. anonymous

where do you think the vertices are?

49. anonymous

(-5, -5) and (3, 5) ?

50. anonymous

there you have it

51. anonymous

now we solve for c to get your foci

52. anonymous

$c=\sqrt{a^2+b^2}$

53. anonymous

c = +/- 5

54. anonymous

correct, now put that into the coords just like you did for vertices

55. anonymous

(-6, -5) and (4, -5) ?

56. anonymous

yup

57. anonymous

Ok, can you help me with one more? And check one?

58. anonymous

sure

59. anonymous

For: Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7). Would the answer be: y^2/4 - x^2/45 = 1

60. anonymous

the x^2/45 part is wrong

61. anonymous

you need to meddle a bit with the foci formula, its not so straightforward

62. anonymous

Ok so how do I do that?

63. anonymous

ok if c^2 = a^2 + b^2 then c^2 - a^2 = b^2

64. anonymous

so for the x^2 would we subtract x^2/49?

65. anonymous

no

66. anonymous

foci is c which is 7 and your 4 is a^2

67. anonymous

so 33?

68. anonymous

$7^2-2^2=?$

69. anonymous

so yea its 45 lol

70. anonymous

i didnt square the stuff i got which is 6.708 thats why lol

71. anonymous

Oh ok!

72. anonymous

And the last problem on this lesson is: Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by six times x.

73. anonymous

im used to putting em with the ^2 on

74. anonymous

Oh ok

75. anonymous

this is definitely a hyperbola opening up/down

76. anonymous

|dw:1439051936537:dw|

77. anonymous

I agree

78. anonymous

$\frac{ y^2 }{ 10^2 } - \frac{ x^2 }{ b^2 } = 1$$y=\pm \frac{ 5 }{ 6 }x$

79. anonymous

So...how could I get it into standard form equation?

80. anonymous

and gradient is $y=\frac{ b }{ a } x$

81. anonymous

$\frac{ b }{ a } = \frac{ 5 }{ 6 } = \frac{ b }{ 10 }$

82. anonymous

Ok. So how can I solve for the b and a terms?

83. anonymous

multiply through$\frac{ 5 }{ 6 } = \frac{ 50 }{ 60 } = \frac{ 6b }{ 60 }$$6b=50$$b=8.333333333333$

84. anonymous

Ok that's what I thought.

85. anonymous

lemme recheck

86. anonymous

So would my answer be: y^2/64 - x^2/100 ? Because I just checked and b should equal 8...

87. anonymous

yup i made a mistake at the slope part b/a is for hyperbola opening left/right b/a is for opening up/down

88. anonymous

Ok so my final answer is y^2/64 - x^2/100 = 1?

89. anonymous

no lemme redo the last part

90. anonymous

cause isn't a^2 = 10 and b^2 = 8 ?

91. anonymous

i used the wrong formula for the asymtote lol

92. anonymous

lemme explain to ya

93. anonymous

|dw:1439052898040:dw|

94. anonymous

|dw:1439052978322:dw|

95. anonymous

see the difference?

96. anonymous

Yes

97. anonymous

One is going up/down, the other left/right

98. anonymous

$\frac{ a }{ b } = \frac{ 5 }{ 6 } = \frac{ 10 }{ b }$ so b is 12

99. anonymous

Oh ok. Well I was just looking on open study, to see what others did, and someone else listed the answer as: y^2/64 - x^2/100 = 1 But I understand also what you're saying.

100. anonymous

same question?

101. anonymous

yes

102. anonymous

he prolly flipped the gradient like i did

103. anonymous

Oh ok. So would the answer be: y^2/144 - x^2/100 = 1?

104. anonymous

$\frac{ y^2 }{ 10^2 }-\frac{ x^2 }{ 12^2 } = 1$

105. anonymous

Oh ok! So simplified: y^2/100 - x^2/144 = 1

106. anonymous

Correct? @saseal

107. anonymous

yea

108. anonymous

Ok thanks!

109. anonymous

@saseal could you help me again? so sorry!!!

110. anonymous

ok

111. anonymous

maybe just open a new one lol this one is getting really long

112. anonymous

Ok lol