Identify the graph of the equation. What is the angle of rotation for the equation?13x^2+6√3xy+7y^2-16=0

Answer:The answer is ellipse; 30° ⇒ 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:* 13x² + 6√3xy + 7y² - 16 = 0∵ A = 13 , B = 6√3 , C = 7∴ B² - 4 AC = (6√3)² - 4(13)(7) = -256∴ B² - 4AC < 0∴ The graph is ellipse or circle* 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 with different values∴ It is an ellipse* To find the angle of rotation use the rule:- cot(2Ф) = (A - C)/B∵ A = 13 , B = 6√3 , C = 7∴ cot(2Ф) = (13 - 7)/6√3 = 6/6√3 = 1/√3∵ tan(2Ф) = 1/cot(2Ф)∴ tan(2Ф) = √3 ⇒ 2Ф = [tex]tan^{-1}\sqrt{3}=60[/tex]∴ 2Ф = 60°∴ Ф = 30°* The answer is ellipse; with angle of rotation = 30°