Answer:
The center, vertices and foci of the ellipse with equation [tex]4 x^{2}+9 y^{2}=36 is (0,0),(\pm 3,0),(\pm \sqrt{5}, 0)[/tex] respectively
Solution:
The equation of ellipse with centre (0, 0) in the form of [tex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1[/tex] --- eqn 1
Where,
x is the major axis
Centre (0, 0)
Vertices is [tex](\pm \mathrm{a}, 0)[/tex]
Foci is [tex](\pm \mathrm{c}, 0)[/tex] where [tex]c=\sqrt{a^{2}-b^{2}}[/tex]
Now given that the equation of ellipse is
[tex]4 x^{2}+9 y^{2}=36[/tex] --- eqn 2
On dividing equation (2) by 36,
[tex]\frac{x^{2}}{9}+\frac{y^{2}}{4}=1[/tex]
On comparing equations (1) and (2),
We get a = 3, b= 2
[tex]c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}[/tex]
So centre of [tex]\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (0, 0)[/tex]
Vertices of [tex]\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm 3,0)[/tex]
Foci of [tex]\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm \sqrt{5}, 0)[/tex]
