Question

How does an ellipse work?

Answer 1

both our profile pics are cars

Answer 2

@saifoo.khan help

Answer 3

alright now.

Answer 4

do you have a certain question

Answer 5

in ellipse?

Answer 6

How can you tell what they will look like? For example, number 20, i have no idea

Answer 7

wow... that's the one I did not do

Answer 8

u got me there

Answer 9

let me get help

Answer 10

@mattvlaw HELP US!!

Answer 11

more info?

Answer 12

What can you tell about \[x ^{2}/144 + y ^{2}/9 = 1\]

Answer 13

related to ellipses
can you plz show on graph and explain

Answer 14

http://mathworld.wolfram.com/Ellipse.html
a is the square of the horizontal distance from the geometric center
b is the square of the vertical distance from the geometric center

Answer 15

Thank you

Answer 16

sorry I meant a^2 and b^2
standard equation is x^2/a^2 + y^2/b^2 = 1

Answer 17

Best source I know is this:
http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/MITRES_18_001_strang_3538.pdf

Answer 18

what is the horizontal distance and the geometric center?

Answer 19

I'll try to draw a picture, OK. Take me a minute

Answer 20

ok thank you, I am having a test on this tomorrow and I still dont quite understand the full concept. I understand how my teacher derived it, but now how to manipulate the formula to solve for certain parts.

Answer 21

same we are in the same clas

Answer 22

I dont under stand it at all

Answer 23

wait how do u find the foci?

Answer 24

what is this a and b thing

Answer 25

Standard equation is
(x/a)^2 + (y/b)^2 = c^2
OK?

Answer 26

So if y = 0 (and for your example c^2 = 1)
then x = +/- a.

Answer 27

how do you find foci?

Answer 28

In your case you had a^2 = 144 and b^2 =9

Answer 29

foci are at +/- sqrt(a^2 - b^2)
check out the reference, the relevant part is really short.

Answer 30

what is the foci of this question?

Answer 31

(+/- sqrt (144-9), 0)

Answer 32

Do you have a list of answers or something?

Answer 33

Ellipse is just like a circle except x is rescaled by the factor a and y is rescaled by the factor b.

Answer 34

If you need to do offsets from the origin, put them inside a parenthesis just like with a circle (or parabola).

Answer 35

yes that was the question: \[x ^{2}/144 + y ^{2}/9 =1\]

Answer 36

can you plz show how to graph it and find the foci?

Answer 37

MAybe it would help us if you go through step by step on how to solve this one practice problem that we got for homework: 20): What can you tell about (x^2/144)+(y^2/9)=1

Answer 38

please

Answer 39

Since it's in standard form with no offset, nothing like (x-h)^2, then it's at the origin. OK? And a = sqrt(144) = 12 and b = sqrt(9) = 3. OK so far?

Answer 40

what is offset?

Answer 41

If the center of the ellipse (halfway between the foci) was not at the origin but shifted.

Answer 42

forget about if the center is not the origin

Answer 43

My advice too

Answer 44

@Calcmathlete

Answer 45

Now, with your example, the ellipse will cross the x-axis at (12,0) and (-12,0)
and it will cross the y-axis (0,3) and (0,-3). OK?

Answer 46

Sorry I couldn't help you.

Answer 47

how!!?

Answer 48

its k :)

Answer 49

What...?

Answer 50

help us on the same question: What can you tell about (x^2/144)+(y^2/9)=1

Answer 51

plz show how to find the foci and how to graph it

Answer 52

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]or\[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\]

Answer 53

huh?

Answer 54

Let's start off with how to graph it.

Answer 55

what do you mean? here, try to consider we know nothing about ellipses

Answer 56

What I just posted above is the standard form of an ellipse.

Answer 57

ok

Answer 58

An ellipse has several parts. They have vertices, covertices, major axes, minor axes, center, and foci. There are other parts as well, but at this point, these are all you need to know about.

Answer 59

So, since our example equal 1, you would just look at the x axis and see that the square root of 144 is 12, so it would cross the x axis at (-12, 0) and (12, 0)?

Answer 60

can you plz explain how to find those starting with how to graph it and how to find the foci

Answer 61

THis is a horizontally oriented ellipse that has vertices at (12, 0) and (-12, 0).

Answer 62

Ok, thank you, keep on explaining calmathlete...

Answer 63

ok, but how can you tell

Answer 64

how?

Answer 65

The co-vertices are (0, 3) and (0, -3)
THis works by figuring out which ones a and b are.

Answer 66

a is always the bigger number and b is always the smaller number.

Answer 67

how did you find that?

Answer 68

For instance, we have 144 and 9. One of these is a^2 and one of these are b^2 right?

Answer 69

ok, i see...

Answer 70

ok

Answer 71

go on

Answer 72

So, find the square root of both and see which one is bigger.
You get 12 and 3. Which of these is bigger?

Answer 73

a^2

Answer 74

12

Answer 75

yup

Answer 76

Alright. So a = 12 and be = 3. That also means that a^2 = 144 and b^2 = 9

Answer 77

*b = 3

Answer 78

right

Answer 79

ok

Answer 80

Now, the major axis is the segment connecting the two vertices and the minor axis is the segment connecting the two co-vertices.

Answer 81

what vertices and how did you knwo that?

Answer 82

So minor is horizontal and major is vertical?

Answer 83

Oh yeah, before I move on, sinc ea is under the x, this is a horizontally oriented ellipse since it's wider going left and right.

Answer 84

Can you picture it?

Since the a is under x, meaning that it stretches wider than it does vertically, it's horizontally oriented.

Answer 85

right, I understand how it looks, just how to get there from the formula

Answer 86

Alright. The major axis is represented by 2a and the minor axis is represented by 2b.

Answer 87

how would tell which is a and which is b on a question?

Answer 88

Like I said, a is always bigger than b.

Answer 89

That's how you'd figure it out.

Answer 90

ok

Answer 91

LOok at the frst, well, second thing he said Anikate

Answer 92

Look at the first*

Answer 93

Its derived in the formula

Answer 94

That's why when you start graphing ellipses, you go to the left and right of the origin by the square root of the number under the x and y. For this problem, since a = 12 and it's under x, you go 12 units to the right of the center and plot a point. Then, go 12 points to the left of the center and plot a point. Now, since b = 3 and it's under the y, you go up 3 from the center and plot a point. Then go down three from the center and plot a point. These are your vertices and co-vertices.

Answer 95

Thank you, I just understood it. That is a perfect explanation