Identify the conic section $$$\sqrt{5} x^{2} = 62$$$

Identify and find the properties of the conic section $$$\sqrt{5} x^{2} = 62$$$.

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, $$$A = \sqrt{5}$$$, $$$B = 0$$$, $$$C = 0$$$, $$$D = 0$$$, $$$E = 0$$$, $$$F = -62$$$.

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.

$$$\sqrt{5} x^{2} = 62$$$A represents a pair of the lines $$$x = - \frac{5^{\frac{3}{4}} \sqrt{62}}{5}$$$, $$$x = \frac{5^{\frac{3}{4}} \sqrt{62}}{5}$$$A.

General form: $$$\sqrt{5} x^{2} - 62 = 0$$$A.

Factored form: $$$\left(5 x - 5^{\frac{3}{4}} \sqrt{62}\right) \left(5 x + 5^{\frac{3}{4}} \sqrt{62}\right) = 0$$$A.

Graph: see the graphing calculator.