Identify the conic section $$$4 y^{2} = 1$$$

Identify and find the properties of the conic section $$$4 y^{2} = 1$$$.

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, $$$A = 0$$$, $$$B = 0$$$, $$$C = 4$$$, $$$D = 0$$$, $$$E = 0$$$, $$$F = -1$$$.

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 parallel lines.

$$$4 y^{2} = 1$$$A represents a pair of the lines $$$y = - \frac{1}{2}$$$, $$$y = \frac{1}{2}$$$A.

General form: $$$4 y^{2} - 1 = 0$$$A.

Factored form: $$$\left(2 y - 1\right) \left(2 y + 1\right) = 0$$$A.

Graph: see the graphing calculator.