Solution
We have the following equation:
[tex]\frac{x^2}{16}+\frac{y^2}{4}=1[/tex]For this case the center is: C(0,0)
[tex]a=\sqrt[]{16}=4,b=\sqrt[]{4}=2[/tex]We can find the value of c like this:
[tex]c=\sqrt[]{16-4}=\sqrt[]{12}=2\sqrt[]{3}[/tex]Since x is the bigger axis we have the following coordinates for the focci:
[tex](0+2\sqrt[]{3},0),(0-2\sqrt[]{3},0)=(2\sqrt[]{3},0),(-2\sqrt[]{3},0)[/tex]Then the answer is:
b. (±2√3, 0)
