Question

b. $x = \frac{1}{2}(y + 1)^{2}+2$ vertex: p - value: axis of symmetry: focus: directrix:

Explanation:

Step1: Identify the vertex form of the parabola

The equation of a parabola in vertex form is \(x = a(y - k)^{2}+h\), where \((h,k)\) is the vertex.
For the equation \(x=\frac{1}{2}(y + 1)^{2}+2\), comparing with \(x = a(y - k)^{2}+h\), we have \(h = 2\) and \(k=-1\).
So, the vertex is \((2,-1)\).

Step2: Find the \(p -\)value

The formula for \(a\) in terms of \(p\) for the parabola \(x=a(y - k)^{2}+h\) is \(a=\frac{1}{4p}\).
Given \(a=\frac{1}{2}\), then \(\frac{1}{2}=\frac{1}{4p}\).
Cross - multiply: \(4p\times1=2\times1\), so \(p=\frac{1}{2}\).

Step3: Determine the axis of symmetry

For a parabola of the form \(x=a(y - k)^{2}+h\), the axis of symmetry is given by the equation \(y = k\).
Since \(k=-1\), the axis of symmetry is \(y=-1\).

Step4: Calculate the focus

The focus of a parabola \(x=a(y - k)^{2}+h\) is \((h + p,k)\).
We know \(h = 2\), \(p=\frac{1}{2}\), and \(k=-1\).
So, \(h + p=2+\frac{1}{2}=\frac{4 + 1}{2}=\frac{5}{2}\).
The focus is \((\frac{5}{2},-1)\).

Step5: Find the directrix

The directrix of a parabola \(x=a(y - k)^{2}+h\) is \(x=h - p\).
We know \(h = 2\) and \(p=\frac{1}{2}\).
So, \(h - p=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}\).
The directrix is \(x=\frac{3}{2}\).

Answer:

Vertex: \((2,-1)\)
\(P -\)value: \(\frac{1}{2}\)
Axis of symmetry: \(y = - 1\)
Focus: \((\frac{5}{2},-1)\)
Directrix: \(x=\frac{3}{2}\)