Question

What are the vertices and foci of the ellipse `9x^2-18x+4y^2=27`?

Answer 1

The vertices are `(3,0), (-1,0), (1,3), (1,-3)`
The foci are `(1,sqrt5)` and `(1,-sqrt5)`

Explanation:

Let's rearrange the equation by completing the squares

`9x^2-18x+4y^2=27`

`9(x^2-2x+1)+4y^2=27+9`

`9(x-1)^2+4y^2=36`

Dividing by `36`

`(x-1)^2/4+y^2/9=1`

`(x-1)^2/2^2+y^2/3^2=1`

This is the equation of an ellipse with a vertical major axis

Comparing this equation to

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

The center is `=(h,k)=(1,0)`

The vertices are A`=(h+a,k)=(3,0)` ; A'`=(h-a,k)=(-1,0)` ;

B`=(h.k+b)=(1,3)` ; B'`=(h,k-b)=(1,-3)`

To calculate the foci, we need

`c=sqrt(b^2-a^2)=sqrt(9-4)=sqrt5`

The foci are F`=(h.k+c)=(1, sqrt5)` and F'`=(h,k-c)=(1,-sqrt5)`

graph{(9x^2-18x+4y^2-27)=0 [-7.025, 7.02, -3.51, 3.51]}