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

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 =

∴ 2Ф = 60°

∴ Ф = 30°

* The answer is ellipse; with angle of rotation = 30°

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step-by-step explanation: