Question

find the vertex and focus of the parabola:
x² - 10x - 12y - 23 = 0
vertex = (?, )
focus = (, )

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for the $x$ - terms.
\[

$$\begin{align*} x^{2}-10x - 12y-23&=0\\ x^{2}-10x&=12y + 23\\ x^{2}-10x+25&=12y + 23+25\\ (x - 5)^{2}&=12y+48\\ (x - 5)^{2}&=12(y + 4) \end{align*}$$

\]

Step2: Identify the vertex

For a parabola of the form $(x - h)^{2}=4p(y - k)$, the vertex is $(h,k)$.
Comparing $(x - 5)^{2}=12(y + 4)$ with $(x - h)^{2}=4p(y - k)$, we have $h = 5$, $k=-4$. So the vertex is $(5,-4)$.

Step3: Identify the value of $p$

Since $(x - 5)^{2}=12(y + 4)$ and $(x - h)^{2}=4p(y - k)$, then $4p = 12$, so $p = 3$.

Step4: Find the focus

For a parabola of the form $(x - h)^{2}=4p(y - k)$, the focus is $(h,k + p)$.
Here, $h = 5$, $k=-4$, $p = 3$. So the focus is $(5,-4 + 3)=(5,-1)$.

Answer:

Vertex = $(5,-4)$
Focus = $(5,-1)$