## Babynini one year ago Identity and graph, Hyperbola?

1. Babynini

2. Babynini

@jim_thompson5910

3. jim_thompson5910

it's not a hyperbola you can use something like geogebra (which is what I just used) or desmos to check https://www.geogebra.org/ https://www.desmos.com/calculator

4. Babynini

oo thanks.

5. Babynini

ellipse?

6. jim_thompson5910

yep it's an ellipse

7. Babynini

fabulous, so...what next? I grouped stuff and factored out but i'm not sure how to finish the squares.

8. Babynini

4(x^2-8x ) 25(y^2+6y )=-189

9. jim_thompson5910

the x coordinate is -8 cut that in half to get -4 then square it to get 16 you will add and subtract 16 inside the parenthesis $\Large 4(x^2-8x \ \ \ \ \ \ \ \ ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189$ $\Large 4(x^2-8x {\color{red}{+16-16}} ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189$ the +16-16 is to make sure that the expression doesn't change (since we're effectively adding 0). Then notice how x^2-8x+16 factors to (x-4)^2

10. Babynini

hmm ok yeah

11. Babynini

so now we've got 4(x-4)^2 ?

12. Babynini

and is the right side 25(x-3)^2

13. Babynini

(sorry, by right I meant the y stuff)

14. jim_thompson5910

you're forgetting about the -16 in the parenthesis

15. Babynini

how do I write that?

16. jim_thompson5910

$\Large 4(x^2-8x + \underline{ \ \ \ \ \ \ \ \ } \ ) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189$ $\Large 4(x^2-8x+16-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189$ $\Large 4((x^2-8x+16)-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189$ $\Large 4((x-4)^2-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189$ $\Large 4(x-4)^2-64 + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189$

17. jim_thompson5910

what goes in the blank for the y terms?

18. Babynini

ahh ok. y^2-6y+9

19. Babynini

25((x-3)^2-9)

20. Babynini

25(x-3)^2-225 ?

21. jim_thompson5910

yeah +9-9 goes in the blank on line 1 for the y terms

22. jim_thompson5910

be careful it is NOT y^2 - 6y it's y^2 + 6y

23. Babynini

24. jim_thompson5910

so you should have this $\Large 4(x-4)^2-64 + 25(y+3)^2 - 225=-189$

25. Babynini

yeah. ok so since we're adding and doing stuff to the left side we need to do the same on the right, yeah?

26. jim_thompson5910

no need

27. jim_thompson5910

that's why I had +16 and -16 to balance things out

28. jim_thompson5910

same for +9-9

29. Babynini

oh! that's fabulous :) I was wondering because in class I remember the +9 but and doing stuff to the left side also but I like this better haha

30. jim_thompson5910

well you could add things to both sides, but you'd have to be careful

31. Babynini

Yep yep. So now what do we do?

32. Babynini

add 64 and 225 to both sides?

33. jim_thompson5910

yep

34. Babynini

so 100 on the right side :P

35. Babynini

then divide the whole thing by 100 so we get 0 on the right again?

36. jim_thompson5910

you mean 1 on the right side

37. jim_thompson5910

but yeah

38. Babynini

$\frac{ 4(x-4)^2 }{ 100 }+\frac{ 25(y+3)^2 }{ 100 }=1$

39. Babynini

haha yeah sorry.

40. jim_thompson5910

and you can rewrite that into $\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1$

41. Babynini

perfect :)

42. Babynini

so now we get all the other stuff from there?

43. jim_thompson5910

yeah you can find the center, foci, vertices, co-vertices, etc

44. Babynini

a = 5 b = 4 c= sq29

45. jim_thompson5910

how did you get c?

46. Babynini

oops b = 2

47. Babynini

a^2+b^2=c^2

48. Babynini

25+4=c^2

49. jim_thompson5910
50. Babynini

ou. then c = sqrt 21

51. jim_thompson5910

yep

52. Babynini

kks, and then the ellipse will be sideways yeah?

53. Babynini

so foci: $(0,\pm \sqrt{21})$

54. jim_thompson5910

not quite

55. jim_thompson5910

the foci will shift along with the center

56. jim_thompson5910

|dw:1433386754426:dw|

57. jim_thompson5910

|dw:1433386763450:dw|

58. jim_thompson5910

|dw:1433386776494:dw|

59. Babynini

so we've got to find the center first.

60. jim_thompson5910

yes

61. jim_thompson5910

in this case, the major axis is horizontal, so the foci + center all lay on the same horizontal level. Only the x values will change

62. jim_thompson5910

oh nearly forgot the vertices |dw:1433387027602:dw|

63. Babynini

wouldn't want to forget those! :P

64. Babynini

ok. that makes sense.

65. jim_thompson5910

for vertical ellipses (where b > a) $\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ we have this schematic |dw:1433387213931:dw|

66. Babynini

this one isn't horizontal? o.0

67. jim_thompson5910

no I'm saying IF it were vertical, then you'd have what you see above

68. Babynini

ooh right right

69. jim_thompson5910

the ellipse you have is horizontal, like this |dw:1433387651593:dw|

70. Babynini

so how do we find the center? :)

71. jim_thompson5910

Compare $\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1$ with $\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

72. jim_thompson5910

The center of $\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ is (h,k)

73. Babynini

so (4,-3)

74. jim_thompson5910

yes

75. Babynini

so foci : (4- sq21, -3) , (4sq21, -3)

76. Babynini

?

77. jim_thompson5910

I think you meant to say $\Large \left(4 {\color{red}{ \textbf{+}}}\sqrt{21},-3\right)$ for the second focus

78. Babynini

yeah

79. jim_thompson5910

you have the correct foci

80. Babynini

haha yay! Ok, vertex: (+/-5, -3) ?

81. jim_thompson5910

no

82. jim_thompson5910

did you mean $\Large \left(4 \pm 5, -3\right)$??

83. Babynini

ai ai ai yes sorry!

84. Babynini

(9,-3) (-1,-3)

85. jim_thompson5910

both correct

86. Babynini

phew.

87. Babynini

Length of major axis: 2a? minor axis 2b?

88. jim_thompson5910

yes correct

89. jim_thompson5910

only because a > b

90. Babynini

so major: 50 minor: 8

91. jim_thompson5910

50?

92. Babynini

lol I did (a^2)2 major = 10

93. jim_thompson5910

major is 10, yep

94. jim_thompson5910

minor isn't 8

95. Babynini

4

96. jim_thompson5910

yep minor = 4

97. Babynini

98. Babynini

all good? :)

99. jim_thompson5910

everything is perfect

100. Babynini

thanks so much! a thousand medals for your efforts :)

101. jim_thompson5910

you're welcome