The matricial equation of the degenerate parabola is [tex]\left[\begin{array}{c}x'\\y'\end{array}\right] = \left[\begin{array}{c}\frac{ \sqrt{2} }{2}\cdot x + \frac{1}{8}\cdot x^{2}+1 \\\-\frac{\sqrt{2}}{2}\cdot x + \frac{1}{8}\cdot x^{2} + 1 \end{array}\right][/tex].
A parabola is degenerated when its axis of symmetry is not parallel to any orthogonal axis (i.e. x, y). First, we have to find an "equivalent" parabola whose axis of symmetry is parallel to the y-axis and whose vertex is along that axis.
Direction of the degenerate parabola
[tex]\tan \theta = \frac{2 - 1}{2 - 1}[/tex]
tan θ = 1
θ = 45°
This means that the degenerate parabola must be rotated 45° clockwise (- 45°) around the origin with respect to the "equivalent" parabola.
Distance from the vertex to the focus
[tex]p = \sqrt{(2 - 1)^{2}+(2 - 1)^{2}}[/tex]
[tex]p = \sqrt{2}[/tex]
Vertex of the "equivalent" parabola
[tex](h', k') = (0, \sqrt{2})[/tex]
Equation of the "equivalent" parabola
[tex]y = \frac{1}{4\cdot p}\cdot x^{2} + k'[/tex]
[tex]y = \frac{1}{4\sqrt{2}}\cdot x^{2}+\sqrt{2}[/tex]
Now we apply following rotation formula:
[tex]\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{cc}\cos (-45^{\circ})&-\sin (-45^{\circ})\\\sin (-45^{\circ})&\cos (-45^{\circ})\end{array}\right] \cdot \left[\begin{array}{c}x\\\frac{1}{4\sqrt{2}}\cdot x^{2}+\sqrt{2}} \end{array}\right][/tex]
[tex]\left[\begin{array}{c}x'\\y'\end{array}\right] = \left[\begin{array}{c}\frac{ \sqrt{2} }{2}\cdot x + \frac{1}{8}\cdot x^{2}+1 \\\-\frac{\sqrt{2}}{2}\cdot x + \frac{1}{8}\cdot x^{2} + 1 \end{array}\right][/tex]
A representation of the degenerate parabola is shown in the figure attached below.
To learn more on degenerate conics: https://brainly.com/question/10071979
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