Question

How do you find the center, foci and vertices of `9x^2+4y^2+36x-24y+36=0`?

Answer 1

`C: (-2, 3)`
`V: (-2, 3 +- 3)`
`f: (-2, 3 +- sqrt5)`

Explanation:

First off, group terms with the same variables together

`9x^2 + 4y^2 + 36x - 24y + 36 = 0`

`=> (9x^2 + 36x) + (4y^2 - 24y) + 36 = 0`

`=> 9(x^2 + 4x) + 4(y^2 - 6y) + 36 = 0`

Next, we "complete" the square.
Add a third term to each of the grouped terms such that it will be a perfect square trinomial.

`(x + a)^2 = x^2 + 2a + a^2`

However, note that the equality must be maintained. To do this, we need to either add the same value on the other side of the equation or subtract the same value on the same side of the equation.

`=> 9(x^2 + 4x + 4) + 4(y^2 - 6y + 9) + 36 - 9(4) - 4(9) = 0`

`=> 9(x + 2)^2 + 4(y - 3)^2 + 36 - 36 -36 = 0`

`=> 9(x + 2)^2 + 4(y - 3)^2 - 36 = 0`

Move the constant to the other side

`=> 9(x + 2)^2 + 4(y - 3)^2 = 36`

We want the other side of the equation to be 1, so divide both sides of the equation by the moved constant

`=> (9(x + 2)^2 + 4(y - 3)^2 = 36)/36`

`=> (x + 2)^2/4 + (y - 3)^2/9 = 1`

Now that we have transform the equation into the standard equation for ellipse, we can easily get the desired properties

`C: (-2, 3)`

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Major axis is vertical

`=> V: (-2, 3 +- 3)`

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`a^2 - b^2 = c^2`

`=> 9 - 4 = c^2`

`=> sqrt5 = c`

`=> f: (-2, 3 +- sqrt5)`

Answer 2

Rearrarange the equation thus:
`9(x+2)^2-36+4(y-3)^2-36+36=0`

Explanation:

Then re-arrange to standard form:
`((x+2)/2)^2+((y-3)/3)^2=1`
which is an ellipse centered at `(-2,3)`, vertices `(0,3)`, `(-4,3)`,`(-2,0)`, `(-2,6)`. Its foci are possibly `(-2,3+-sqrt(5))`.