Question

How do you find the vertex, focus and directrix of `y^2+6y+12x+33=0`?

Answer 1

The vertex is `(-2,-3)`
The focus is `(-5,-3)`
The diretrix is `x=1`

Explanation:

Let's put `y` on one side of the equation and `x` on the other.
`y^2+6y=-12x-33` We complete the square for the left side of the equation.
`y^2+6y+9-9=-12x-33`
`(y+3)^2-9=-12x-33`
`(y+3)^2=-12x-24`
`(y+3)^2=-12(x+2)` Since this is in the form of `(y-k)^2=4p(x-h)`, we know this is a horizontal parabola.

Now, we can figure out that: `h=-2, k=-3, p=-3`
Also, we know that the vertex is `(x,h)`, the focus is `(h+p,k)`, and that the diretrix is `x=h-p`

Therefore, the vertex is `(-2,-3)`
The focus is `(-5,-3)`
The diretrix is `x=1`