Question

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

Answer 1

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`