Answer:
The answer is ellipse; 23° to the nearest degree ⇒ answer (d)
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°