Question

for the parabola: $y^{2}+4y - 12x + 16 = 0$ find the vertex and focus. make sure to enter your answers as coordinates. vertex: focus: write the equation of the directrix, and then sketch the parabola, the directrix and place a dot on the vertex and the focus. directrix: graphing tool tutorial +

Explanation:

Step1: Rewrite the equation in standard form

Given \(y^{2}+4y - 12x + 16 = 0\).
Complete the square for the \(y\) - terms.

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The standard form of a parabola of the form \((y - k)^{2}=4p(x - h)\)

Step2: Identify \(h,k\) and \(p\)

For the equation \((y + 2)^{2}=12(x - 1)\), we have \(h = 1,k=-2\)
Since \(4p=12\), then \(p = 3\)

Step3: Find the vertex

The vertex of the parabola \((y - k)^{2}=4p(x - h)\) is \((h,k)\).
So, the vertex is \((1,-2)\)

Step4: Find the focus

The focus of the parabola \((y - k)^{2}=4p(x - h)\) is \((h + p,k)\)
Substitute \(h = 1,k=-2,p = 3\)
The focus is \((1+3,-2)=(4,-2)\)

Step5: Find the directrix

The directrix of the parabola \((y - k)^{2}=4p(x - h)\) is \(x=h - p\)
Substitute \(h = 1,p = 3\)
The directrix is \(x=1-3=-2\)

Answer:

Vertex: \((1,-2)\)
Focus: \((4,-2)\)
Directrix: \(x=-2\)