Question

Find the center, vertices, and foci of the ellipse with equation 4x^2 + 9y^2 =36

Answer 1

Answer:

The center, vertices and foci of the ellipse with equation $4 x^{2}+9 y^{2}=36 is (0,0),(\pm 3,0),(\pm \sqrt{5}, 0)$ respectively

Solution:

The equation of ellipse with centre (0, 0) in the form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$   --- eqn 1

Where,

x is the major axis  

Centre (0, 0)

Vertices is $(\pm \mathrm{a}, 0)$      

Foci is $(\pm \mathrm{c}, 0)$ where $c=\sqrt{a^{2}-b^{2}}$

Now given that the equation of ellipse is

$4 x^{2}+9 y^{2}=36$ --- eqn 2

On dividing equation (2) by 36,

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

On comparing equations (1) and (2),  

We get a = 3, b= 2  

$c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$

So centre of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (0, 0)$

Vertices of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm 3,0)$

Foci of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm \sqrt{5}, 0)$