# Ellipse componentsFind the center, vertices, and foci of the ellipse 9x^2 + 4y^2 + 36x -8y + 4 =0 .

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### 2 Answers

You need to create groups that form a perfect square, such that:

`(9x^2 + 36x) + (4y^2 - 8y) + 4 = 0`

You need to notice that in each group misses one term that makes the perfect square, hence, you may add the missing terms both sides, such that:

`(9x^2 + 36x + 36) + (4y^2 - 8y + 4) + 4 = 0 + 36 + 4`

Each group makes a perfect square now, such that:

`(3x + 6)^2 + (2y - 2)^2 + 4 = 0 + 36 + 4`

Reducing duplicate members yields:

`(3x + 6)^2 + (2y - 2)^2 + 4 = 36`

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

You need to convert the original form of equation in standard form of equation of ellipse, centered at `(h,k)` , such that:

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

Reasoning by analogy yields:

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

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

Evaluating the foci of ellipse yields:

`(-2,1 - c),(-2,1 + c)`

`c = sqrt(a^2 - b^2) => c = sqrt(9 - 4) => c = sqrt 5`

**Hence, the ellipse is centered at `(-2,1)` , it has the foci **`(-2,1 - sqrt 5),(-2,1 + sqrt 5).`

The equation of the ellipse is :

x^2/a^2 + y^2/b^2 = 1

We'll re-write the equation:

x^2*b^2 + y^2*a^2 - a^2*b^2 = 0

9x^2 + 4y^2 + 36x -8y + 4 =0

We'll complete the squares:

9x^2 + 36x + 36 - 36 + 4y^2 - 8y + 4 - 4 + 4 = 0

(x+6)^2 + (2y-2)^2 - 36 = 0

(x+6)^2 + (2y-2)^2 = 36

We'll divide by 36:

(x+6)^2/6^2 + (y-1)^2/3^2 = 1

semi-major axis is:a = 6

semi-minor axis is:b=3