Question

question 2 of 20 what are the coordinates of the focus of the parabola shown below? write your answer without using spaces. $y^{2}+16y + 4x+4 = 0$

Explanation:

Step1: Rewrite the equation in standard form

First, complete the square for the $y$ - terms.
\[

$$\begin{align*} y^{2}+16y + 4x+4&=0\\ y^{2}+16y&=-4x - 4\\ y^{2}+16y + 64&=-4x - 4+64\\ (y + 8)^{2}&=-4x+60\\ (y + 8)^{2}&=-4(x - 15) \end{align*}$$

\]

Step2: Identify the parameters of the parabola

The standard - form of a parabola opening to the left is $(y - k)^{2}=-4p(x - h)$, where $(h,k)$ is the vertex and $p$ is the distance from the vertex to the focus.
Comparing $(y + 8)^{2}=-4(x - 15)$ with $(y - k)^{2}=-4p(x - h)$, we have $h = 15$, $k=-8$, and $4p = 4$, so $p = 1$.

Step3: Find the focus of the parabola

For a parabola of the form $(y - k)^{2}=-4p(x - h)$ opening to the left, the focus is given by the point $(h - p,k)$.
Substituting $h = 15$, $k=-8$, and $p = 1$ into the formula for the focus, we get $(15-1,-8)=(14,-8)$.

Answer:

(14,-8)