Identify the conic section $$$- x^{2} \csc{\left(1 \right)} \sec{\left(1 \right)} = 2 \pi$$$
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Identify and find the properties of the conic section $$$- x^{2} \csc{\left(1 \right)} \sec{\left(1 \right)} = 2 \pi$$$.
Solution
The general equation of a conic section is $$$A x^{2} + B x y + C y^{2} + D x + E y + F = 0$$$.
In our case, $$$A = \frac{2}{\sin{\left(2 \right)}}$$$, $$$B = 0$$$, $$$C = 0$$$, $$$D = 0$$$, $$$E = 0$$$, $$$F = 2 \pi$$$.
The discriminant of the conic section is $$$\Delta = 4 A C F - A E^{2} - B^{2} F + B D E - C D^{2} = 0$$$.
Next, $$$B^{2} - 4 A C = 0$$$.
Since $$$\Delta = 0$$$, this is the degenerated conic section.
Since $$$B^{2} - 4 A C = 0$$$, the equation represents two nonreal lines.
Answer
$$$- x^{2} \csc{\left(1 \right)} \sec{\left(1 \right)} = 2 \pi$$$A represents two nonreal lines.
General form: $$$\frac{2 x^{2}}{\sin{\left(2 \right)}} + 2 \pi = 0$$$A.