Question

Find the vertices and foci of (x+2)over 4^2+(y-1)0ver 25^2=1

Answer 1

\[(x+2)^{2}/4+(y-1)^{2}/25=1\]

Answer 2

@ShailKumar

Answer 3

This is an ellipse and the vertex is (-2,1)

Answer 4

for the foci i got (-2, 1+or- square root 21)

Answer 5

and for the vertices i got (-2,6) (-2,-4) but when i put it in a graphing calculator i get different vertices and foci

Answer 6

https://www.desmos.com/calculator/46lmaqxorx

Answer 7

(2,-1) is the center of the ellipse. How did you find the vertices ?

Answer 8

I used the formula (h, K+a) (h,k-a) because the major axis is vertical

Answer 9

(-2,1) is the center of the ellipse

Answer 10

Actually you need to calculate eccentricity first

Answer 11

its parallel to the y-axis because a is always biggest number and its under y^2

Answer 12

oh, is that where e=c/a

Answer 13

yes

Answer 14

is c square root 21 though?

Answer 15

because then e=square root 21/5

Answer 16

Right

Answer 17

yes right

Answer 18

ok why is eccentricity important in finding the vertices and foci?

Answer 19

If center is (h,k), then foci are (h, k+c) and (h, k-c)

Answer 20

ok thats what i did and i got (-2,1+square root 21) (-2,1-square root 21)

Answer 21

eccentricity decide which conic section (parabola, ellipse or hyperbola) you are talking about. So their properties should be dependent on eccentricity.

Answer 22

Nice Question ShailKumar

Answer 23

You did it right for foci.

Answer 24

ok maybe i was interpreting the graph incorrectly

Answer 25

BTW, for this ellipse e = c/b. That plot is showing a parabola.

Answer 26

on the graph the vertices were like (-4,0) and (0,0) though

Answer 27

isnt e=c/a not b?

Answer 28

Forget that. you've done correctly.

Answer 29

oooh the graph on the link was showing a parabola?! ok that explains a lot

Answer 30

Could you find vertices?

Answer 31

No

Answer 32

I thought they were (-2,6) and (-2,-4)

Answer 33

but then those points arent on the x-axis...

Answer 34

Wait...

Answer 35

Vertices are (h,k+b) and (h,k-b).

Answer 36

arent they (h,k+a) (h,k-a)?

Answer 37

In your equation of ellipse: you have taken a^2 = 25 and b^2 = 4. right ?

Answer 38

yep

Answer 39

Then you are right. It should be (h,k+a) (h,k-a).

Answer 40

BTW, you've got correct vertices (-2,6) and (-2,-4)

Answer 41

it looks so weird lmost like a circle when i graph it though

Answer 42

and it doesnt make sense because then all of the points are on x=-2 and how am i supposed to draw that? i dont know any x-coordinates

Answer 43

https://www.desmos.com/calculator/fwn45vpf2z
this graph is showing different vertices

Answer 44

ok yea that one was a parabola graph i just posted a new one though and it still shows different vertices

Answer 45

You have interchanged the roles of a and b.

Answer 46

what do you mean?

Answer 47

In the plot you've got, you have put a>b. But in the problem a<b

Answer 48

a=25 and b=4 so a>b always in an ellipse

Answer 49

Not always. You post standard ellipse equation.

Answer 50

oh ok so im just confused about how to graph this problem because how do i know how wide the ellipse is when i dont have any x-coordinates?

Answer 51

For width you don't need x coordinates. Actually width of this ellipse is 2a = 4. Length is 2b = 10 !

Answer 52

Did you see the attached image?

Answer 53

which attached image?

Answer 54

Attaching again

Answer 55

Ummm I have (x+2)^2/4+(y-a)^2/25=1 is this the same as you have?

Answer 56

Sorry thats (y-1)^2/25

Answer 57

Actually I have (x+4)^2/4+(y-1)^2/25=1.

Answer 58

it should be (x+2)^2/4

Answer 59

Yep.

Answer 60

im still confused about how to graph it...I can plot the vertex, the vertices but how can i draw an ellipse without knowing how wide to draw it?

Answer 61

Do you wanna draw it by hand ?

Answer 62

yes i have to draw it by graphing paper

Answer 63

how do i plot 2a=4 and 2b=10?

Answer 64

Ellipse has two perpendicular axes called major axis and minor axis. First, check a>b or b>a. If a>b, major axis is parallel x axis otherwise parallel to y axis.