Question

the equation of a parabola is given.
$x^{2}-6x - 4y+13 = 0$
(a) write the equation of a parabola in standard form.
(b) identify the vertex, focus, and focal diameter.
express numbers in exact, simplest form.

Explanation:

Step1: Rewrite the equation in standard form

For a parabola of the form \((x - h)^2=4p(y - k)\), start with the given equation \(x^{2}-6x - 4y+13 = 0\).
Complete the square for the \(x\) - terms:

$$
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Step2: Identify the vertex

For the parabola \((x - h)^2=4p(y - k)\), the vertex is \((h,k)\).
Comparing \((x - 3)^{2}=4(y - 1)\) with \((x - h)^2=4p(y - k)\), we have \(h = 3\) and \(k = 1\). So the vertex is \((3,1)\)

Step3: Identify \(p\) and then the focus

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

Step4: Find the focal diameter

The focal diameter of the parabola \((x - h)^2=4p(y - k)\) is \(|4p|\)
Since \(p = 1\), the focal diameter is \(|4\times1|=4\)

Answer:

(a) The standard form of the parabola is \((x - 3)^{2}=4(y - 1)\)
(b) The vertex is \((3,1)\), the focus is \((3,2)\) and the focal diameter is \(4\)