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

1 Answer
Dec 13, 2015

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

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

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

Ecentricity: e= sqrt(5)/3

Explanation:

General formula for vertical ellipse

Remember: " " " a" >b "> "0 " " " c^2= a^2 -b^2

(x-h)^2/b^2 + (y-k)^2/a^2 = 1

Center: (h, k)" " " " " " Foci: (h, k+-c)

Vertices (h, k+-a)" " " " "Co-vertices: (h+-b, k)

Eccentricity: e= 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= sqrt(5)/3