Question

How do you find the center, vertices, foci and eccentricity of `4x² + 9y² - 8x - 36y + 4 = 0`?

Answer 1

The centre is at `(1,2)`
The vertices are at `(1/2,2 ), (3/2,2), (1,2*1/3)` and `(1,1*2/3)`
The foci are at `(1-sqrt(5)/6, 2)` and `(1+sqrt(5)/6,2)`

The eccentricity is `(sqrt(5)/6)`

Explanation:

Rearrange the elements to get the standard form of an ellipse
`(x-h)^2/a^2 + (y-k)^2/b^2 = 1` where `(h,k)` is the centre and `a` and `b` are the semi-minor and semi-major axes respectively.

So in this example
`4x^2 -8x +9y^2 - 36y +4 = 0`

Using the method of completing squares

`4(x^2 - 2x) + 9 (y^2-4y) +4 = 0`
`4(x-1)^2 -1 +9(y-2)^2 - 4 + 4 = 0`
`4(x-1)^2 +9(y-2)^2 = 1`
`(x-1)^2/(1/4) +(y-2)^2/(1/9)=1`

Therefore the centre is at `(1,2)`
The semi-major axis has length `1/2` and the semi-minor axis has length `1/3` so the vertices are at `(1/2,2 ), (3/2,2), (1,2*1/3)` and `(1,1*2/3)`

If `c` is the distance between the foci,
`1/4-c^2=1/9`
`:.c^2=5/36` and `c = sqrt(5)/6`

The foci are at `(1-sqrt(5)/6, 2)` and `(1+sqrt(5)/6,2)`

The eccentricity is `c/a = (sqrt(5)/6)`