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.

$$
LATEXBLOCK0
$$

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: Find the value of \(p\)

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

Step4: Find the focus

The focus of a parabola \((x - h)^{2}=4p(y - k)\) is \((h,k + p)\).
Substituting \(h = 5\), \(k=-4\), \(p = 3\), we get \((5,-4+3)=(5,-1)\)

Answer:

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