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

Explanation:

Step1: Rewrite the equation in standard form

First, 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

The standard - form of a parabola opening upwards or downwards is $(x - h)^{2}=4p(y - k)$, where $(h,k)$ is the vertex. Comparing $(x - 5)^{2}=12(y + 4)$ with $(x - h)^{2}=4p(y - k)$, we have $h = 5$ and $k=-4$. 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 $(x - h)^{2}=4p(y - k)$ opening upwards, the focus is given by the point $(h,k + p)$. Here, $h = 5$, $k=-4$, and $p = 3$. So the focus is $(5,-4 + 3)=(5,-1)$.

Answer:

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