Question

find the vertex and focus of the parabola: y² + 12y - 4x + 48 = 0 vertex = (?, ) focus = (, )

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for the $y$ - terms.
\[

$$\begin{align*} y^{2}+12y - 4x+48&=0\\ y^{2}+12y&=4x - 48\\ y^{2}+12y + 36&=4x-48 + 36\\ (y + 6)^{2}&=4(x - 3) \end{align*}$$

\]

Step2: Identify the vertex

For a parabola of the form $(y - k)^{2}=4p(x - h)$, the vertex is $(h,k)$.
Comparing $(y + 6)^{2}=4(x - 3)$ with $(y - k)^{2}=4p(x - h)$, we have $h = 3$ and $k=-6$. So the vertex is $(3,-6)$.

Step3: Identify the value of $p$ and the focus

For the equation $(y - k)^{2}=4p(x - h)$, we have $4p = 4$, so $p = 1$.
The focus of a parabola of the form $(y - k)^{2}=4p(x - h)$ is $(h + p,k)$.
Substituting $h = 3$, $k=-6$ and $p = 1$ into the formula for the focus, we get $(3 + 1,-6)=(4,-6)$.

Answer:

Vertex = $(3,-6)$
Focus = $(4,-6)$