Question

Find the equation of each ellipse

Answer 1

Foci at (+-2,0) Length of the Major axis is 6

@jim_thompson5910

Answer 2

Foci at (+-2,0) means we have what you see as attached

Answer 3

If "Length of the Major axis is 6" and those two focii are shown there, where is the center and vertices?

Answer 4

Umm idkk

Answer 5

the center is the midpoint of the foci F1 and F2

Answer 6

what is the midpoint of F1 and F2?

Answer 7

+-(-2)

Answer 8

add up the coordinates and divide by 2

Answer 9

0

Answer 10

you should get (0,0)

Answer 11

the center is at (0,0)

where are the vertices?

Answer 12

(4,0)"?

Answer 13

but the length of the major axis is 6 units

half of this length is 3

so going from the center to either vertex is 3 units

Answer 14

3,0

Answer 15

yes so we have this so far

Answer 16

Yes

Answer 17

So now do you put it as the equation

Answer 18

the value of a = 3 since the distance from center C to vertex (V1 or V2) is 3 units

the distance is a horizontal distance

the 'a' corresponds to x, so 'a' is the horizontal component

Answer 19

c = 2 since the distance from center to focus is 2 units

Answer 20

use

b^2 = a^2 - c^2

to find b

Answer 21

9-4?

Answer 22

a^2 = 3^2 = 9
b^2 = 5
h = 0
k = 0

so this means

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

becomes

\[\Large \frac{x^2}{9}+\frac{y^2}{5}=1\]

Answer 23

when we get to b^2 = 5 there is no need to solve for b itself. This is because the formula has b^2 instead of b

Answer 24

oh Can i try one and you correct me?

Answer 25

sure

Answer 26

Foci(0,+-2) length of 8

\[\frac{ x ^{2} }{ 16 }+y ^{2}=1\]

Answer 27

major axis is of length 8?

Answer 28

yea

Answer 29

your answer isn't correct

Answer 30

What is wrong

Answer 31

did you manage to find out where the vertices go?

Answer 32

and where the center is?

Answer 33

Center (0,0) F(0,2) V(0,4)

Wait isnt it V(0,3)

Answer 34

the two vertices are

V1 = (0,-4)
V2 = (0,4)

Answer 35

yea

Answer 36

what's the distance from C to V1?

Answer 37

4

Answer 38

this is a vertical distance so it corresponds to the vertical component

that makes b = 4

agreed?

Answer 39

Yea

Answer 40

http://www.mathwords.com/f/foci_ellipse.htm

use the vertical form

b^2 = a^2+c^2

solve for a^2 to get

a^2 = b^2 - c^2

plug in b = 4 and c = 2. Then evaluate to get a^2 = ???

Answer 41

2

Answer 42

a^2 = b^2 - c^2

a^2 = 4^2 - 2^2

a^2 = 16 - 4 = ???

Answer 43

12

Answer 44

a^2 = 12
b^2 = 4^2 = 16
h = 0
k = 0

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

turns into

\[\Large \frac{x^2}{12}+\frac{y^2}{16}=1\]

Answer 45

Oh i get it :D Thanks

Answer 46

np