Question

Identify each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

$4x^2-16x+16y+32=0$

Short answer

The vertex is $\left(2,-1\right)$, focus is $\left(2,-2\right)$ and directrix is $y=0$.

Step-by-step solution

Step 1. Given Information 

We are given the equation $4x^2-16x+16y+32=0$.

We need to find the type of conic.

Step 2. Determining the conic 

Rewriting the equation we get:

$$\begin{gathered} 4x^2-16x+16y+32=0 \\ \Rightarrow 4(x^2-4x)+16(y+2)=0 \\ \Rightarrow (x^2-4x)+4(y+2)=0 \\ \Rightarrow (x^2-4x)=-4(y+2) \\ \Rightarrow (x^2-4x+4)=-4(y+2-1) \\ \Rightarrow {(x-2)}^2=-4(y+1) \end{gathered}$$

This is the equation of a parabola. Comparing it with the standard equation ${\left(x-h\right)}^2=-4a(y-k)$we get:

$$\begin{gathered} h=2, \\ -4a=-4 \\ \Rightarrow a=1 \\ k=-1 \end{gathered}$$

Step 3. Finding vertex, focus, and directrix  

Vertex is $\left(h,k\right)=\left(2,-1\right)$.

Focus is $\left(h,k-a\right)=\left(2,-1-1\right)=\left(2,-2\right)$.

Directrix is

$$\begin{gathered} y=k+a \\ \Rightarrow y=-1+1 \\ \Rightarrow y=0 \end{gathered}$$