Find the center, vertices and foci for the ellipse 4x^2 + 16y^2 = 64

Given the equation of an ellipse is 4x^2 + 16y^2 =
64.

We need to find the center and
foci.

First we need to re-write the equation into the
standard form of the ellipse.

==> x^2/a^2 + y^2 /
b^2 = 1..............(1)

==> 4x^2 + 16y^2 =
64.

Let us divide by 64
.

==> 4x^2/64 + 16y^2/64 =
1

==> x^2/ (64/4)  + y^2/ (64/16) =
1

==> x^2 / (16)  + y^2 / 4 =
1

==> x^2/ (4^2)  + y^2 / (2^2) =
1............(2)

Comparing equation (1) and (2) we conclude
that:

a = +-4 and b=+- 2.

Then
the vertices's are:

(4,0) , (-4,0), ( 0,2)
and ( 0,-2).

==> Then
the center is (0, 0).

==> C^2 = a^2 -
b^2

==> C^2 = 16 - 4 =
12

==> c= +-sqrt12 =
+-2sqrt3

==> The foci is ( 2sqrt3,0 )
and (
-2sqrt3,0)