Question

Find the following values to the equation of the ellipse =(x+7)^2/36+(y-2)^2/121=1
a. Center
b. Vertices
c. Length of minor axis
d. Length of the major axis
e. Foci

Answer 1

do you need a ellipse formula?

Answer 2

well, i just need someone to walk me through this..would you be willing to help me?

Answer 3

sure, from the formula sheet, as you can see, your "a" and "b" values will come from the denominators, the higher one being "a"

Answer 4

ok im here with ya :)

Answer 5

so, as you can see, 36 can be written \(6^2\), and 121 as \(11^2\), leaving you with \[\frac{(x+7)^2}{6}+\frac{(y-2)^2}{11}=1\]

Answer 6

ok :) im here

Answer 7

from there, you can pretty much get the (h,k) for the center of the ellipse and the "a" and "b" values

Answer 8

still here :)

Answer 9

woops I meant \(6^2 \ and \ 11^2\) as denominators :)

Answer 10

yes yes :)

Answer 11

hmmm well lets see

Answer 12

keep in mind that the \((y-k)^2\) expression has the 'major' number and thus the 'major' axis, so the ellipse is upwardly

Answer 13

or vertical I should say

Answer 14

wait..how do you find c?

Answer 15

shouldnt the center be 0,0?

Answer 16

the center of it is at (h,k) off => \((x-h)^2 \ and \ (y-k)^2\)

Answer 17

\((x+7)^2 \implies (x-(-7))^2\) and \((y-2)^2 \implies (y-(+2))^2\)

Answer 18

hmm...sorry if im stupid..but are you saying the center is 2,7

Answer 19

(h, k), those stand for the ordered pair of (x,y) coordinates, and is -7, no 7

Answer 20

(2, -7) will be in the IV quadrant

Answer 21

IV?

Answer 22

that'd be the fourth(IV) quadrant

Answer 23

im so so sorry..but i may sound dumb..but i learn betetr when the work is worked out. may you please walk through it and let me review?

Answer 24

my graph is a bit out of whack, is supposed to be -7 over the X :S

Answer 25

so, that'd be the II quadrant

Answer 26

hmm...may you start with the center?

Answer 27

yes

Answer 28

your center is at (-7,2), draw it on a graphic
from that point go UPWARDS 11 units, so 2+11 = 13 = y-coord.
from that point go DOWNWARDS 11 units, so 2-11 = -9 = y-coord.
13 and -9 are your vertices, since they're the outter bounds of the major axis

Answer 29

your center is at (-7,2)
from that point go LEFTWARDS 6 units, -7-6 = -13 = x-coord.
from that point go RIGHTWARDS 6 units, -7+6 = -1 = x-coord.
that'd be your minor axis

Answer 30

ahh alright!! that makes it much more clear. may you work down the list? this seems to be a great way for me to undrstand :)

Answer 31

a) (h,k)

c) the minor of "a" and "b"

d) the major of "a" and "b"

b) vertices are at: k + "a"; because in this case is a vertical ellipse

e) the foci come from: k + \(\sqrt{a^2-b^2} \implies k+\sqrt{11^2-6^2}\) in your case k=2
because the ellipse is vertical and thus the major is going up/down and thus Y-coord from (h,k) is used which is k

Answer 32

alright im understanding..

Answer 33

k + \(\sqrt{a^2-b^2} \implies k-\sqrt{11^2-6^2}\) gives you the UPPER focus
k - \(\sqrt{a^2-b^2} \implies k-\sqrt{11^2-6^2}\) gives you the LOWER focus

Answer 34

woops, I meant
k + \(\sqrt{a^2-b^2} \implies k+\sqrt{11^2-6^2}\) gives you the UPPER focus
k - \(\sqrt{a^2-b^2} \implies k-\sqrt{11^2-6^2}\) gives you the LOWER focus

Answer 35

so you add that amount to K and it gives one focus, substract it from it and you'd get the other one

Answer 36

I'll need to dash for now :), but I'll be here tomorrow

Answer 37

thanks man..you did amazing!