Question

find the vertex and focus of the parabola: x² + 6x - 8y + 1 = 0 vertex = (?, ) focus = (, )

Explanation:

Step1: Rewrite the given equation in standard form

First, complete the square for the $x$ - terms.
Starting with $x^{2}+6x - 8y+1 = 0$.
We know that $(x + a)^{2}=x^{2}+2ax + a^{2}$, for $x^{2}+6x$, we have $2a = 6$ (so $a = 3$) and $x^{2}+6x=(x + 3)^{2}-9$.
The equation becomes $(x + 3)^{2}-9-8y + 1=0$.
Then $(x + 3)^{2}=8y + 8$, or $(x + 3)^{2}=8(y + 1)$.

Step2: Identify the vertex

The standard - form of a parabola opening upwards is $(x - h)^{2}=4p(y - k)$, where $(h,k)$ is the vertex.
Comparing $(x + 3)^{2}=8(y + 1)$ with $(x - h)^{2}=4p(y - k)$, we have $h=-3$ and $k=-1$.
So the vertex is $(-3,-1)$.

Step3: Find the value of $p$

From $(x - h)^{2}=4p(y - k)$ and $(x + 3)^{2}=8(y + 1)$, we know that $4p = 8$, so $p = 2$.

Step4: Identify the focus

The focus of a parabola of the form $(x - h)^{2}=4p(y - k)$ is given by the point $(h,k + p)$.
Since $h=-3$, $k=-1$ and $p = 2$, the focus is $(-3,-1 + 2)=(-3,1)$.

Answer:

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