Answer:
(14, -8)
Explanation:
Given the function:
[tex]y^2+16y+4x+4=0[/tex]
We want to find the coordinates of the focus of the parabola.
A parabola is the locus of points such that the distance to the focus equals the distance to the directrix. For a right-left facing parabola with vertex at (h, k), and a focal length |p|, the standard equation is:
[tex]4p\left(x-h\right)=\left(y-k\right)^2[/tex]
First, we rewrite the given equation in the standard equation form:
[tex]\begin{gathered} -4x-4=y^2+16y \\ -4x-4+64=y^2+16y+64 \\ -4x+60=(y+8)^2 \\ -4(x-15)=(y+8)^2 \\ \implies4\left(-1\right)\left(x-15\right)=\left(y-\left(-8\right)\right)^2 \\ \implies\left(h,k\right)=\left(15,-8\right),p=-1 \end{gathered}[/tex]
Next, our parabola is symmetric around the x-axis and so the focus lies at a distance p from the center (15,-8) along the x-axis. Thus, the coordinates of the focus is:
[tex](15+p,-8)=(15+(-1),-8)=(14,-8)[/tex]
The focus is at (14, -8).
