Question

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

Answer 1

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]}