Identify the graph of 9X^2+4xy+5y^2-40=0 and find theta to the nearest degree.

Answer:The answer is ellipse; 23° to the nearest degree ⇒ answer (d)Step-by-step explanation:* At first lets talk about the general form of the conic equation- Ax² + Bxy + Cy²  + Dx + Ey + F = 0∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse. ∵ B² - 4AC = 0 , if a conic exists, it will be a parabola. ∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.* Now we will study our equation:- 9x² + 4xy + 5y² - 40 = 0∵ A = 9 , B = 4 , 5 = 5∴ B² - 4 AC = (4)² - 4(9)(5) = -164 < 0∴ B² - 4AC < 0∴ If a conic exists, it will be either a circle or an ellipse.* To find the type of the graph lets check;- If A and C are nonzero, have the same sign, and are not  equal to each other, then the graph is an ellipse. - If A and C are equal and nonzero and have the same  sign, then the graph is a circle.∵ A and C have same signs and are not equal∴ The graph is an ellipse* If we have term xy ⇒ B ≠ 0∴ The graph is rotate by angle Ф* To find the angle of rotation use the rule:- cot(2Ф) = (A - C)/B∵ A = 9 , B = 4 , C = 5∴ cot(2Ф) = (9 - 5)/4 = 4/4 = 1∴ tan(2Ф) = 1∴ 2Ф = 45°∴ Ф = 22.5° ≅ 23° to the nearest degree* The answer is ellipse; with angle of rotation = 23°