Question

Can someone please help me find the center, vertices, foci, eccentricity, lengths of minor and major axis of \[\frac{ (x-5)^2 }{ 4 }+\frac{ (y+3)^2 }{ 16 }=1\]

Answer 1

oh no not again ...

Answer 2

thats what i said :( last problem though so thats my motivation! :p

Answer 3

you know what this is right?

Answer 4

is it a parabola?

Answer 5

oh no an ellips!

Answer 6

yes it is
do you know what it looks like (we need that to start)

Answer 7

looks like that :)

Answer 8

wow that is ugly but yes
you got the center?

Answer 9

sorry lol, i think the center is (5,-3) but im not sure

Answer 10

yes you are right (not sure why you were not sure)

Answer 11

im just never sure of myself in math :(

Answer 12

i guess that is good
ok next for the vertices which is very much like the last one

Answer 13

\[\frac{ (x-5)^2 }{ 4 }+\frac{ (y+3)^2 }{ 16 }=1\]in this case since \(16>4\) we have \[\frac{ (x-h)^2 }{ b^2 }+\frac{ (y-k)^2 }{ a^2 }=1\] in other words the \(a\) goes with the larger denominator
since \(a^2=16\) we have \(a=4\) and since you know how it is oriented you have the vertices are \(4\) units above and below \((5,-3)\)

Answer 14

you got those?

Answer 15

would that be (5,1) (5,-7)?

Answer 16

you sure?

Answer 17

just kidding, it is right

Answer 18

haha i was about to go and change my mind again :p

Answer 19

ok now for the foci
this time it is a little different we need \(c\) but here we use \(c^2=a^2-b^2\)

Answer 20

that gives \(c^2=16-4=12\) so \(c=\sqrt{12}=2\sqrt3\)

Answer 21

c^2=12?

Answer 22

yes

Answer 23

but the square root of 12, so the answer would actually be 2 square root 3?

Answer 24

again above and below \((5,-3)\) and again you just write it

Answer 25

nevermind i forgot we have to find the square of 12!

Answer 26

yeah it is what you said \(2\sqrt3\)

Answer 27

so foci are \[(5,-3-2\sqrt3),(5,-3+2\sqrt3)\]

Answer 28

how about the eccentricity?

Answer 29

e=c/a, so would that be 2√3/4?

Answer 30

yeah, but if you don't want your teacher to think you are challenged in the arithmetic department, you might want to cancel a 2 and write \(\frac{\sqrt3}{2}\)

Answer 31

LOL good idea :D

Answer 32

last part is nothing
length of major axis is \(2a=8\) and length of minor axis is \(2b=4\) since \(a=4\) and \(b=2\)

Answer 33

how do i find the length of the major and minor axis?

Answer 34

see above

Answer 35

oh ocay that is easy! so minor is 4 an

Answer 36

and major is 8

Answer 37

yes

Answer 38

thank you SO much, i really appreciate all of your help!!!!

Answer 39

yw
wanna check?

http://www.wolframalpha.com/input/?i=ellipse+%28x-5%29^2%2F4+%2B+%28y%2B3%29^2%2F16%3D1

Answer 40

awesome, thank you :)

Answer 41

don't be confused by the "semimajor" axis etc
that is just half of the major axis
good luck on your exam

Answer 42

thanks i truly appreciate it:)