Question 132267
Choose the correct coordinates of the foci

x2 + 4y2 - 16 = 0

We are dealing with an ellipse.

The first thing to do is to write this in standard form.

x^2 + 4y^2 - 16 = 0

Add 16 on both sides.

x^2 + 4y^2 = 16

We now divide all three terms by 16 to get 1 on the left side because we want the equation to be in standard form.

x^2/16 + 4y^2/16 = 16/16

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

Since 16 is bigger than 4, the x^2 terms has the BIGGER denominator.

This means our general form for this equation is

x^2/a^2 + y^2/b^2 = 1

The foci is the following sets of points:

(-c, 0) and (c, 0)

To find c, use this:

c^2 = a^2 - b^2

The square root of 16 = 4 = a^2.

The square root of 4 = 2 = b^2

Then:

c^2 = 4 - 2

c^2 = 2

Take square root of both sides:

c = +2, -2

I got the following two points for the foci:

(2, 0) and (-2, 0)

Recheck your work to make sure.  Also, reject my work for more practice on your part.