Question

Find the coordinates of the vertices of the ellipse
PLEASE!!

Answer 1

\[(x-3)^2/9+y^/25=1\]

Answer 2

(x-3)^2/9+y^2/25=1

Answer 3

do you know where the center of this ellipse is?

Answer 4

isn"t it (h,k)

Answer 5

from the center, you'll either go up/down OR left/right "a" units from the center.

Answer 6

yes, what's (h, k) in your equation?

Answer 7

(3,0)

Answer 8

good....

Answer 9

now, to find the vertices, what's the value of "a" in your equation?

Answer 10

3

Answer 11

is the ellipse elongated vertically or horizontally....???

Answer 12

horizontally?

Answer 13

no...
since the bigger denominator is under the y^2, this is elongated vertically and therefore a = 5.

so from the center, go UP/DOWN 5 units to get your vertices.

Answer 14

oh ok

Answer 15

think of it this way... whichever fraction has the bigger denominator, it will be elongated under that variable....
since your denominator is 25 and is under the y^2, the ellipse wll be elongated vertically.

Answer 16

so can you give me the coordinates of the vertices?

Answer 17

would they be
(3,0), (5,0), (0,4), (0,-1)

Answer 18

wait...wold they be (6,0), (3,0), (0,5), (0,-5) instead....?

Answer 19

when you asked for vertices, these points are actually the endpoints of the MAJOR axis.
from the center (3, 0), go up 5 units and down 5 units.

Answer 20

why are you giving 4 points? did you also want the co-vertices?

Answer 21

well there are my options....
a. (6,0), (3,0), (0,5), (0,-5)
b. (3,0), (5,0), (0,4), (0,-1)
c. (2,0), (8,0), (3,3), (3,-3)
d, (6,0), (0,0), 3,5), (3,-5)
.

Answer 22

from the center, (3, 0)
vertices: go up/down 5 units
co-vertices: go right/left 3 units.

Answer 23

here is the drawing using the center (3, 0):