Question

find the vertex and focus of the parabola: x² + 4x - 8y + 12 = 0 vertex = (?, ) focus = (, )

Explanation:

Step1: Rewrite the equation in standard form

First, complete the square for the \(x\) terms.

$$
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$$

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

Step2: Find the vertex and focus

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

Answer:

Vertex \(=(-2,1)\), Focus \(=(-2,3)\)