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

1 Answer
Vikki
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#

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