Question

What are the coordinates of the focus of the conic section shown below? Y^2+16y-4x+4=0

Answer 1

come help me on the other i accidentaly closed it

Answer 2

it doesn't matter, I'l do it here.

Answer 3

here is the parabola... \(\normalsize\color{blue}{ y^2+16y-4x+4=0 }\)

\(\normalsize\color{blue}{ y^2+16y-4x+4=0 }\)
\(\normalsize\color{blue}{ y^2+16y+4\color{red}{+60-60}-4x=0 }\)
\(\normalsize\color{blue}{ \underline{y^2+16y+4+60}-60-4x=0 }\)
\(\normalsize\color{blue}{ \underline{y^2+16y+64}-60-4x=0 }\)
\(\normalsize\color{blue}{ (y+8)^2-60-4x=0 }\)
\(\normalsize\color{blue}{ (y+8)^2-4x-60=0 }\)
\(\normalsize\color{blue}{ (y+8)^2-60=4x }\)
\(\normalsize\color{blue}{ \frac{1}{4}(y+8)^2-15=x }\)

Answer 4

\(\normalsize\color{blue}{ x=\frac{1}{4}(y+8)^2-15 }\)

\(\normalsize\color{green}{ x=a(y-h)^2+k }\) horizontal parabola.

Answer 5

so we just have to put it in parabola form and then solve for (h,k)?

Answer 6

yes, it is a parabola shaped as U that opens to the right. h and k are there (pretty much) ....

good luck