Question

Someone help please !
(Picture Below)

Answer 1

it's hard to say where the vertices and co-vertices are exactly...but my best guess is that the vertices are at the points (-5,0) and (5,0)

Answer 2

oops i meant (-6, 0) and (6, 0)

and my best guess is that the co-vertices are at the points (-5, 0) and (5,0)

Answer 3

oh that helps a lot, thx

Answer 4

I thought that to until i saw my answer choice !

Answer 5

well we can eliminate choices B and C because those are hyperbolas, NOT ellipses

Answer 6

Okay thanks. do we have to sqaure ?

Answer 7

the length of the semi-major axis is 6 units long

the semi-major axis lies on the x-axis, so it corresponds to the x term

Answer 8

this means that the value of 'a' in (x^2)/(a^2) + (y^2)/(b^2) = 1 is a = 6

Answer 9

so a = 6

a^2 = 36

Answer 10

giving us the equation (x^2)/36 + (y^2)/25 = 1

Answer 11

i got it so would my answer be D ?

Answer 12

because i squared .

Answer 13

@jim_thompson5910

Answer 14

yes, the answer is D

Answer 15

Thank you :) Can you help me with more please

Answer 16

sure

Answer 17

Graph the conic section.

25x2 – 16y2 = 400

Answer 18

25x^2 – 16y^2 = 400

25x^2/400 – 16y^2/400 = 400/400

x^2/16 – y^2/25 = 1

does this look familiar?

Answer 19

A little bit. My answer choices are graphs so by looking at that and looking at my graph how am i supposed to know lol

Answer 20

well you can use a graphing calculator to get the following (see attached)

Answer 21

That's PERFECT ! Thank you :)

Answer 22

ok great

Answer 23

What are the vertices of the hyperbola with equation 4y2 – 25x2 = 100? Graph the hyperbola.

Answer 24

Could you graph this for me please ?

Answer 25

btw I used this graphing calculator

it's completely free

http://www.geogebra.org/cms/

Answer 26

I recommend downloading it if you can

Answer 27

sure thing, one sec

Answer 28

Thank you and okay will do.

Answer 29

look at the graph and you'll see that the vertices are (0,5) and (0, -5)

Answer 30

This is only one similar to that one. Could this one be my answer. Your says -5 , 5 mines is -4 , 4

Answer 31

that one you're showing me right now has vertices at (0,5) and (0, -5)

Answer 32

this is because one of the vertices are between 4 and 6 on the y axis

Answer 33

So does that mean that was is my answer ?

Answer 34

yes it looks like it is

Answer 35

Uh oh ! lol thats not a good answer.

Answer 36

@jim_thompson5910

Answer 37

hmm ok what are your other choices?

Answer 38

Ill show you.

Answer 39

alright thx

Answer 40

A B C D

Answer 41

and you went with D?

Answer 42

@jim_thompson5910

Answer 43

or B?

Answer 44

I went with b because it looked similar but i want to be 100% sure and right with my answer.

Answer 45

so how do you know B is wrong?

Answer 46

oh i gotcha, well this graph in the attachment I posted goes through (2,7)

Answer 47

@jim_thompson5910

Answer 48

the graph in choice D does not go through (2,7)

however, choice B does

so that's why B look like the best choice

Answer 49

I agree. I'll go with b

Answer 50

that's what I would pick

Answer 51

Haha thank you. I have 5 more can you help me please

Answer 52

sure

Answer 53

What are the asymptotes of the hyperbola with equation 9y2 – 4x2 = 36 ? Graph the hyperbola.

Answer 54

here's the graph

Answer 55

Thank you

Answer 56

the asymptotes are

y = (a/b)x or y = -(a/b)x (for hyperbolas that open up/down)

y = (4/9)x or y = -(4/9)x

Answer 57

Next question Suppose that the path of a newly discovered comet could be modeled by using one branch of the equation x^2/4 - y^2/9 = 1, where where distances are measured in astronomical units. Name the vertices of the hyperbola and then graph the hyperbola.

Answer 58

we can see from the graph (that's attached) that the vertices are at (-2,0) and (2,0)

Answer 59

you can definitely see how useful a tool geogebra is lol

Answer 60

Thank you :)

Answer 61

A hyperbola has vertices (±5, 0) and one focus (6, 0). What is the standard-form equation of the hyperbola?

Answer 62

np

Answer 63

sry was looking up formulas

The center is (0,0) since the vertices are plus/minus some number

so (h,k) = (0,0)

h = 0
k = 0

The value of 'a' is a = 5 because you go 5 units in either direction to get to a vertex

the value of b can be found through the following equation

c^2 = a^2 + b^2

6^2 = 5^2 + b^2

36 = 25 + b^2

36 - 25 = b^2

11 = b^2

b^2 = 11

b = sqrt(11)

So b = sqrt(11) and b^2 = 11

-------------------------------

(x^2)/(a^2) - (y^2)/(b^2) = 1

(x^2)/(25) - (y^2)/(11) = 1

Answer 64

so the answer is (x^2)/(25) - (y^2)/(11) = 1

Answer 65

THANK YOU !!!!!!!!!!!!! :)

Answer 66

@jim_thompson5910

Answer 67

you're welcome

Answer 68

I have two more :)

Answer 69

What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2 – 4y2 = 64?

Answer 70

16x^2 – 4y^2 = 64

16x^2/64 – 4y^2/64 = 64/64

x^2/4 - y^2/16 = 1

a^2 = 4
a = 2

b^2 = 16
b = 4


Vertices: (-2, 0) and (2, 0)

You start at the center (0,0) and you go 2 units in both directions along the x axis

Answer 71

foci

c^2 = a^2 + b^2

c^2 = 2^2 + 4^2

c^2 = 4 + 16

c^2 = 20

c = sqrt(20)

c = 2*sqrt(5)

so the foci are at ( -2*sqrt(5), 0 ) and (2*sqrt(5), 0), you just go out c units in each direction along the x axis

Answer 72

asymptotes:

y = (b/a)x or y = -(b/a)x

y = (4/2)x or y = -(4/2)x

y = 2x or y = -2x

Answer 73

Wow all of that ?

Answer 74

lol yep

Answer 75

it's a lot, conic sections have a lot to them

Answer 76

Last one. Thank you soooooooooooooooooooooo much . You are a very good helper.

Answer 77

@jim_thompson5910

Answer 78

ok go for it

Answer 79

Find an equation that models the path of a satellite if its path is a hyperbola, a = 55,000 km, and c = 81,000 km. Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

Answer 80

Thank you for this.

Answer 81

using the same equation as above

c^2 = a^2 + b^2

81^2 = 55^2 + b^2

6561 = 3025 + b^2

6561 - 3025 = b^2

3536 = b^2

b^2 = 3536

b = sqrt(3536 )

b = 59.464274989274

----------------------------------------------


(x^2)/(a^2) - (y^2)/(b^2) = 1


(x^2)/(55^2) - (y^2)/((59.464274989274)^2) = 1


(x^2)/(3025) - (y^2)/(3536) = 1


So the equation that models the path of flight is (x^2)/(3025) - (y^2)/(3536) = 1 where each unit represents 1000 km


You could optionally make each unit 1 km to get


(x^2)/(3025000) - (y^2)/(3536000) = 1


but that seems a bit too much in my book

Answer 82

so that's why I would go with (x^2)/(3025) - (y^2)/(3536) = 1

Answer 83

Wow lol thank you.

Answer 84

yeah again, it's a lot

Answer 85

Thank you so much for your help

Answer 86

sure thing

Answer 87

glad to be of help to you