Answer:
Option A.
Step-by-step explanation:
The given equation of ellipse is
[tex]4x^2+9y^2=36[/tex]
Divide both sides by 36.
[tex]\dfrac{4x^2}{36}+\dfrac{9y^2}{36}=1[/tex]
[tex]\dfrac{x^2}{9}+\dfrac{y^2}{4}=1[/tex]
[tex]\dfrac{x^2}{3^2}+\dfrac{y^2}{2^2}=1[/tex] ...(1)
The standard form of an ellipse is
[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex] ...(2)
where, (h,k) is center, (h±a,k) are vertices and (h±c,k) are foci.
On comparing (1) and (2), we get
[tex]h=0,k=0,a=3,b=2[/tex]
Now,
Center [tex]=(h,k)=(0,0)[/tex]
Vertices [tex]=(h\pm a,k)=(0\pm 3,0)=(3,0),(-3,0)[/tex]
We know that
[tex]c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}=\sqrt{5}[/tex]
Foci [tex]=(h\pm c,k)=(0\pm \sqrt{5},0)=(\sqrt{5},0),(-\sqrt{5},0)[/tex]
Therefore, the correct option is A.