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

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°

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