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

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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)

 

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