# How do you find the center, vertices, foci and eccentricity of x^2/4+y^2/9=1?

Dec 13, 2015

Foci: " (0, -sqrt(5)) ; (0, sqrt(5))

Vertices: " (0, -3) ; (0, 3)

Co-vertices: (-2, 0) ; (2, 0)

Ecentricity: $e = \frac{\sqrt{5}}{3}$

#### Explanation:

General formula for vertical ellipse

Remember: $\text{ " " a" >b "> "0 " " }$ ${c}^{2} = {a}^{2} - {b}^{2}$

${\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

Center: $\left(h , k\right) \text{ " " " " }$ Foci: $\left(h , k \pm c\right)$

Vertices $\left(h , k \pm a\right) \text{ " " " }$Co-vertices: $\left(h \pm b , k\right)$

Eccentricity: $e = \frac{c}{a}$

Given: ${x}^{2} / 4 + {y}^{2} / 9 = 1$

Let's identify the center : (0, 0) ;" " h= 0; " " " k= 0

b^2 = 4; " "color(red)( b= 2) " " " "   a^2 = 9; " "color(red)( a= 3)

Solve for $c$ , ${c}^{2} = {a}^{2} - {b}^{2}$

c^2 = 9-4 +> c^2 = 5 => color(red)(c= +-sqrt(5)

Use the equation provided above, and "plug it" in

Foci: " (0, -sqrt(5)) ; (0, sqrt(5))

Vertices: " (0, -3) ; (0, 3)

Co-vertices: (-2, 0) ; (2, 0)

Ecentricity: $e = \frac{\sqrt{5}}{3}$