find center,vertices,and foci?

Question

x^2/64+y^2/144=1

Answer 1

this is the equation of an ellipse

it is a vertical ellipse
centre=(0,0)
vertices are (8,0),(-8,0) and (12,0),(-12,0)
a^2=144 b^2=64
b^2=a^2(1-e^2)
64=144(1-e^2)
1-e^2=64/144=4/9
e^2=5/9
e=rt5/3
foci=(0,ae),(0,-ae)
=(0,4rt5),(0,-4rt5)

Answer 2

Recall that the equation of an ellipse goes as follows:

([x - h]^2)/(a^2) + ([y - k]^2)/(b^2) = 1

The above is the general form of an ellipse. The foci is located at (h,k).

The horizontal vertices (rightmost and leftmost portions of the ellipse) are located at (h + a, k), (h - a, k) and
The vertical vertices (highest and lowest points) are located at (h, k + b), (h, k - b).

To avoid that verbiage, all you have to do is find the foci (similar to the center of a circle, but in this case it's not a circle), and move "a" units to the right of the foci, draw a dot, "a" units to the left of the foci, draw a dot, "b" units up from the foci, draw a dot, "b" units down from the foci, draw a dot, and then form an ellipse when connecting the dots.

In our above case, our (h,k) is (0,0), because x^2 is the same as (x - 0)^2. Similarly, y^2 is the same as (y - 0)^2. This is the foci.

64 represents our a^2, therefore
a^2 = 64, and we take the square root to obtain a = 8 (we reject the negative root a = -8 because we're letting "a" represent how much we shift from the foci). Now, from (0,0), draw a dot 8 units above and 8 units below the foci. These are the vertices, and should be at the points (8,0) and (-8,0).

144 represents our b^2, so
b^2 = 144, and b = 12.
Therefore, our vertices are at (0,12) and (0,-12).

Unfortunately, I mistook "center" and "foci", so I did not solve for the foci.

Answer 3

x^2/64+y^2/144=1
THIS IS THE EQUATION OF AN ELLIPSE.
When y=0, x= +/- 8 So vertices are at (-8,) and (8,0)
The foci are at (-c,0) and (c,0) wher c=sqrt(144 -64)=
sqrt(80) = 4sqrt(5)
The center is at the origin (0,0).

Answer 4

The equation of an ellipse with

Center (h,k)
Semi major axis a
Semi minor axis b

is:

(x - h)^2 / b^2 + (y - k)^2 / a^2 = 1

So the center (h,k) = (0,0) is at the origin.
a = 12
b = 8

The vertical vertices are:
(0,0+a) and (0,0-a) = (0,12) and (0,-12)

The horizontal vertices are:
(0+b,0) and (0-b,0) = (8,0) and (-8,0)


c^2 = a^2 - b^2 = 144 - 64 = 80
c = √80 = 4√5

The foci are:
(0,0+c) and (0,0-c) = (0,4√5) and (0,-4√5)