Question

find the vertex and focus of the parabola: x² - 8x + 8y - 32 = 0 vertex = (?, ) focus = (, )

Explanation:

Step1: Complete the square for x-terms

Given equation: \(x^2 - 8x + 8y - 32 = 0\)
Rearrange: \(x^2 - 8x = -8y + 32\)
Complete square for \(x^2 - 8x\): \(x^2 - 8x + 16 = (x - 4)^2\), so add 16 to both sides:
\((x - 4)^2 = -8y + 32 + 16\)
Simplify RHS: \((x - 4)^2 = -8y + 48\)
Factor RHS: \((x - 4)^2 = -8(y - 6)\)

Step2: Identify vertex form of parabola

The standard form of a vertical parabola is \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex.
Compare \((x - 4)^2 = -8(y - 6)\) with \((x - h)^2 = 4p(y - k)\):
\(h = 4\), \(k = 6\), \(4p = -8 \implies p = -2\)

Step3: Find vertex

Vertex is \((h, k) = (4, 6)\)

Step4: Find focus

For a vertical parabola, focus is \((h, k + p)\).
Substitute \(h = 4\), \(k = 6\), \(p = -2\):
Focus = \((4, 6 + (-2)) = (4, 4)\)

Answer:

Vertex = \((4, 6)\)
Focus = \((4, 4)\)