How do you find the center, vertices, foci and eccentricity of #9x^2 + 25y^2 = 225#?

1 Answer
Narad T.
Nov 3, 2016

The center is #=(0,0)#
The vertices are #(5,0)(-5,0)(0,3)(0,-3)#
The foci are F#=(4,0)# and F'#=(-4,0)#
The eccentricity #e=4/5#

Explanation:

Let's start by rewriting the equation
#9x^2+25y^2=225#
By dividing by 225
#x^2/(225/9)+y^2/(225/25)=1#
#x^2/25+y^2/9=1#
this is the equation of an ellipse
#(x-x_0)^2/a^2+(y-y_0)^2/b^2=1#
Here #x_0=0# and #y_0=0#
the vertices are #(5,0)(-5,0)(0,3)(0,-3)#
#c^2=a^2-b^2#
#c=sqrt(25-9)=sqrt16=4#
So the foci are #(x_0+c,y_0)# and #(x_0-c,y_0)#
so F#=(4,0)# and F'#=(-4,0)#
The eccentricity #e=c/a=4/5#
graph{x^2/25+y^2/9=1 [-10, 10, -5, 5]}

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