Question

find the vertex and focus of the parabola: y² + 16y - 4x + 32 = 0 vertex = ( ?, ) focus = (, )

Explanation:

Step1: Rewrite the equation in standard form

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

$$\begin{align*} y^{2}+16y - 4x+32&=0\\ y^{2}+16y&=4x - 32\\ y^{2}+16y + 64&=4x - 32+64\\ (y + 8)^{2}&=4x + 32\\ (y + 8)^{2}&=4(x + 8) \end{align*}$$

\]

Step2: Identify the vertex

For a parabola of the form $(y - k)^{2}=4p(x - h)$, the vertex is $(h,k)$.
Comparing $(y + 8)^{2}=4(x + 8)$ with $(y - k)^{2}=4p(x - h)$, we have $h=-8$ and $k = - 8$. So the vertex is $(-8,-8)$.

Step3: Identify the value of $p$

For the equation $(y - k)^{2}=4p(x - h)$, from $(y + 8)^{2}=4(x + 8)$, we have $4p = 4$, so $p = 1$.

Step4: Find the focus

The focus of the parabola $(y - k)^{2}=4p(x - h)$ is $(h + p,k)$.
Substituting $h=-8$, $k=-8$ and $p = 1$ into the formula for the focus, we get $(-8+1,-8)=(-7,-8)$.

Answer:

Vertex = $(-8,-8)$
Focus = $(-7,-8)$