Identify the conic section $$$- 25 x^{2} - 200 x + y^{2} - 12 y - 389 = 0$$$
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Identify and find the properties of the conic section $$$- 25 x^{2} - 200 x + y^{2} - 12 y - 389 = 0$$$.
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, $$$A = 25$$$, $$$B = 0$$$, $$$C = -1$$$, $$$D = 200$$$, $$$E = 12$$$, $$$F = 389$$$.
The discriminant of the conic section is $$$\Delta = 4 A C F - A E^{2} - B^{2} F + B D E - C D^{2} = -2500$$$.
Next, $$$B^{2} - 4 A C = 100$$$.
Since $$$B^{2} - 4 A C \gt 0$$$, the equation represents a hyperbola.
To find its properties, use the hyperbola calculator.
Answer
$$$- 25 x^{2} - 200 x + y^{2} - 12 y - 389 = 0$$$A represents a hyperbola.
General form: $$$25 x^{2} + 200 x - y^{2} + 12 y + 389 = 0$$$A.
Graph: see the graphing calculator.