Answer:
The foci are located at [tex](0,\pm \sqrt{5})[/tex]
Step-by-step explanation:
The standard equation of an ellipse with a vertical major axis is [tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex] where [tex]a^2\:>\:b^2[/tex].
The given equation is [tex]9x^2+4y^2=36[/tex].
To obtain the standard form, we must divide through by 36.
[tex]\frac{9x^2}{36}+\frac{4y^2}{36}=\frac{36}{36}[/tex]
We simplify by canceling out the common factors to obtain;
[tex]\frac{x^2}{4} +\frac{y^2}{9}=1[/tex]
By comparing this equation to
[tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex]
We have [tex]a^2=9,b^2=4[/tex].
To find the foci, we use the relation: [tex]a^2-b^2=c^2[/tex]
This implies that:
[tex]9-4=c^2[/tex]
[tex]c^2=5[/tex]
[tex]c=\pm\sqrt{5}[/tex]
The foci are located at [tex](0,\pm c)[/tex]
Therefore the foci are [tex](0,\pm \sqrt{5})[/tex]
Or
[tex](0,-\sqrt{5})[/tex] and [tex](0,\sqrt{5})[/tex]
