Question

find the vertex and focus of the parabola: x² + 4x - 8y + 12 = 0 vertex = (?, ) focus = (, )

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for the $x$ - terms.
Given $x^{2}+4x - 8y+12 = 0$.
We have $x^{2}+4x=(x + 2)^{2}-4$.
So the equation becomes $(x + 2)^{2}-4-8y+12 = 0$, which simplifies to $(x + 2)^{2}=8y - 8$, and further to $(x + 2)^{2}=8(y - 1)$.

Step2: Identify the vertex

For a parabola of the form $(x - h)^{2}=4p(y - k)$, the vertex is $(h,k)$.
Comparing $(x + 2)^{2}=8(y - 1)$ with $(x - h)^{2}=4p(y - k)$, we have $h=-2,k = 1$. So the vertex is $(-2,1)$.

Step3: Find the value of $p$

Since $(x + 2)^{2}=8(y - 1)$ and $(x - h)^{2}=4p(y - k)$, then $4p=8$, so $p = 2$.

Step4: Find the focus

For a parabola of the form $(x - h)^{2}=4p(y - k)$, the focus is $(h,k + p)$.
Here, $h=-2,k = 1,p = 2$. So the focus is $(-2,1 + 2)=(-2,3)$.

Answer:

Vertex = $(-2,1)$
Focus = $(-2,3)$